Proof. By expanding in a power series and applying Fubini’s theorem (justified here since \(G_{T,N}\) is bounded, which implies that the absolute value of the term inside the expectation below is growing at most exponentially in \(n\)), \[\langle F\rangle_{T,N,\epsilon} =\sum_{n=0}^\infty \frac{(-\mu)^n}{n!}\int_{M_T^n} \mathbb{E}\biggl[:\mathrel{F(\mathrm{i}Z_0+ \mathrm{i}\varphi_N)}: e^{2\mathrm{i}n b Z_0}\prod_{l=1}^n :\mathrel{e^{2\mathrm{i}b\varphi_N(u_l)}}:\biggr]\, \mathrm{d}u_1\cdots \mathrm{d}u_n.\] Recalling the definitions 11 and 12 , we get \[\begin{align} &:\mathrel{F(\mathrm{i}Z_0+ \mathrm{i}\varphi_N)}: e^{2\mathrm{i}n b Z_0}\prod_{l=1}^n :\mathrel{e^{2b\mathrm{i}\varphi_N(u_l)}}: \\ &\qquad =\exp\biggl(2\mathrm{i}\biggl(\sum_{j=1}^k \alpha_j + nb\biggr) Z_0 + 2\mathrm{i}\sum_{j=1}^k \alpha_j \varphi_N(q_j) +2\sum_{j=1}^k\alpha_j^2 G_{T,N}(q_j, q_j) \\ &\qquad \qquad+2\mathrm{i}b\sum_{l=1}^n \varphi_N(u_l) +2b^2\sum_{l=1}^n G_{T,N}(u_l,u_l)\biggr). \end{align}\] Since \(\varphi_N\) is a centered Gaussian field with covariance \(G_{T,N}\) and \(Z_0\) is an independent Gaussian random variable with mean zero and variance \(\epsilon^{-1}\), the expected value of the above quantity is \[\begin{align} &\exp\biggl(-\frac{2b^2 (n-w)^2}{\epsilon} - 4\sum_{1\le j<j'\le k}\alpha_j \alpha_{j'}G_{T,N}(q_j, q_{j'})\\ &\qquad \qquad - 4b \sum_{j=1}^k \sum_{l=1}^n\alpha_j G_{T,N}(q_j,u_l) - 4b^2 \sum_{1\le l<l'\le n} G_{T,N}(u_l, u_{l'}) \biggr). \end{align}\] Plugging this into the first display of the proof yields the desired formula. ◻

Proof of Theorem 1 in the case \(w<0\). Suppose that \(w\) is a negative integer. In the expansion from Lemma 16, the zero-mode factor in the \(n\)th term is \[\exp\bigl(-2b^2(n-w)^2/\epsilon\bigr), \qquad n\in\mathbb{Z}_{\ge0}.\] Since \(n-w\ne0\) for every \(n\ge0\), each term tends to zero as \(\epsilon\downarrow0\). The estimate used in the proof of Lemma 17 dominates the whole series uniformly for \(0<\epsilon\le1\): after taking real parts, the insertion terms are \(O(n)\) for fixed \(T,N\), while the positive-semidefinite estimate for \(G_{T,N}\) controls the screening-screening pair sum by \(O(n)\). Thus the \(n\)th integral is bounded by \(C^n\) on the finite torus, and the factor \(1/n!\) makes the resulting majorant summable. Dominated convergence therefore gives \[\lim_{\epsilon\downarrow0}\langle F\rangle_{T,N,\epsilon}=0\] for every fixed \(T\) and \(N\). The subsequent limits \(N\to\infty\) and \(T\to\infty\) are therefore also zero. This proves the claim. ◻

Proof. If \(w=0\), the claim follows immediately from Lemma 17, Lemma 35, and Lemma 32. So assume \(w\ge 1\). Fix distinct points \(u_1,\dots,u_w\in M_T\). Throughout this proof, \(C,C_1,\ldots\) will denote positive constants that do not depend on \(u_1,\ldots,u_w\). By Lemma 35 and Lemma 32, \(G_{T,N}(u_l,u_{l'})\to G_T(u_l,u_{l'})\) for each \(l\neq l'\). To justify dominated convergence, first note that by Lemma 37, \[\begin{align} &\operatorname{Re}\biggl\{- 4b \sum_{j=1}^k \sum_{l=1}^w\alpha_j G_{T,N}(q_j,u_l) - 4b^2 \sum_{1\le l<l'\le w} G_{T,N}(u_l, u_{l'})\biggr\} \\ &\le C_1 -4b \sum_{j=1}^k\sum_{l=1}^w\operatorname{Re}(\alpha_j)G_{T,N}(q_j,u_l). \end{align}\] Now, if \(\operatorname{Re}(\alpha_j) \ge 0\) for some \(j\), then Lemma 37 gives \[-4b\operatorname{Re}(\alpha_j) G_{T,N}(q_j, u_l)\le C_2.\] On the other hand, if \(\operatorname{Re}(\alpha_j) < 0\), then Lemma 38 gives \[-4b\operatorname{Re}(\alpha_j) G_{T,N}(q_j, u_l)\le C_3 -4b\operatorname{Re}(\alpha_j) |\log d(q_j, u_l)|,\] where \(d(q_j, u_l)\) denotes the geodesic distance between \(q_j\) and \(u_l\) on \(M_T\). Since \(\operatorname{Re}(\alpha_j)> -\frac{1}{2b}\) for each \(j\), we conclude that there is some \(\delta>0\) such that for any \(u_1,\ldots,u_w\), \[\begin{align} &\biggl|\exp\biggl(- 4b \sum_{j=1}^k \sum_{l=1}^w\alpha_j G_{T,N}(q_j,u_l) - 4b^2 \sum_{1\le l<l'\le w} G_{T,N}(u_l, u_{l'}) \biggr)\biggr| \\ &= \exp\biggl(\operatorname{Re}\biggl\{- 4b \sum_{j=1}^k \sum_{l=1}^w\alpha_j G_{T,N}(q_j,u_l) - 4b^2 \sum_{1\le l<l'\le w} G_{T,N}(u_l, u_{l'})\biggr\}\biggr) \\ &\le \exp\biggl(C_1 +(2-\delta) \sum_{j=1}^k\sum_{l=1}^w|\log d(q_j,u_l)|\biggr). \end{align}\] Since \(M_T\) is a two-dimensional manifold and \(u_1,\ldots,u_w\) are distinct, it is straightforward to see that the above upper bound has a finite integral over \(M_T^w\). This allows us to apply the dominated convergence theorem and complete the proof. ◻

Proof. Recall the formula for \(\langle F \rangle_T\) from Theorem 18. Since \(G_T=\widetilde{G}_T+\frac{\pi}{6}T\), we may factor out the constant contribution: \[\begin{align} \langle F \rangle_T &= \frac{(-\mu)^w}{w!} \exp\biggl( -4\sum_{1\le j<j'\le k}\alpha_j\alpha_{j'} \widetilde{G}_T(q_j,q_{j'}) \biggr) \\ &\quad \cdot \exp\biggl\{ -\frac{2\pi T}{3} \biggl( \sum_{1\le j<j'\le k}\alpha_j\alpha_{j'} +bw\sum_{j=1}^k\alpha_j +b^2\binom{w}{2} \biggr) \biggr\} \\ &\qquad \qquad \cdot \int_{M^w}\Phi_T(u_1,\dots,u_w)\,\mathrm{d}u_1\cdots \mathrm{d}u_w . \end{align}\] Using \(\sum_{j=1}^k \alpha_j=-bw\), we obtain \[\begin{align} -\frac{2\pi T}{3} \biggl( \sum_{1\le j<j'\le k}\alpha_j\alpha_{j'} +bw\sum_{j=1}^k\alpha_j +b^2\binom{w}{2} \biggr) &= \frac{\pi T}{3}\biggl(\sum_{j=1}^k\alpha_j^2+b^2w\biggr)\\ &= -\frac{\pi T}{3}\sum_{j=1}^k \alpha_j(b - \alpha_j). \end{align}\] Therefore, \[\begin{align} &\exp\biggl\{ \frac{\pi T}{3}\sum_{j=1}^k\alpha_j(b -\alpha_j) \biggr\}\,\langle F \rangle_T \notag\\ &\qquad= \frac{(-\mu)^w}{w!} \exp\biggl( -4\sum_{1\le j<j'\le k}\alpha_j\alpha_{j'} \widetilde{G}_T(q_j,q_{j'}) \biggr) \int_{M^w}\Phi_T(u_1,\dots,u_w)\,\mathrm{d}u_1\cdots \mathrm{d}u_w. \end{align}\] We already know that \(\widetilde{G}_T(q_j,q_{j'}) \to \widetilde{G}(q_j,q_{j'})\) as \(T\to \infty\). So the proof will be complete if we can show that \[\begin{align} \label{eq:phittoshow} \lim_{T\to \infty}\int_{M^w}\Phi_T(u_1,\dots,u_w)\,\mathrm{d}u_1\cdots \mathrm{d}u_w = \int_{M^w}\Phi(u_1,\dots,u_w)\,\mathrm{d}u_1\cdots \mathrm{d}u_w. \end{align}\tag{17}\] Again, we know that the integrand on the left converges pointwise to the integrand on the right as \(T\to \infty\). Thus, we only need to show that the limit on the left can be moved inside the integral. To this end, fix \(R>0\), and let \[K_R:=([-R,R]\times[-1,1])^w.\] On \(K_R\), the integrand is dominated by the integrable bound from Lemma 47 of §5.7.2. So, \[\lim_{T\to\infty}\int_{K_R}\Phi_T(u_1,\ldots,u_w)\,\mathrm{d}u_1\cdots \mathrm{d}u_w = \int_{K_R}\Phi(u_1,\ldots,u_w)\,\mathrm{d}u_1\cdots \mathrm{d}u_w.\] Next, again by Lemma 47, there are positive constants \(T_0,C,\eta, \delta\), independent of \(R\), such that \[\begin{align} &\sup_{T\ge T_0}\int_{M^w\setminus K_R} |\Phi_T(u_1,\ldots,u_w)|\,\mathrm{d}u_1\cdots \mathrm{d}u_w \\ &\le \int_{M^w\setminus K_R} C \exp\biggl(-\eta\sum_{l=1}^w \|u_l\|\biggr) \prod_{j=1}^k\prod_{l=1}^w d(q_j,u_l)^{-(2-\delta)} \,\mathrm{d}u_1\cdots \mathrm{d}u_w =: \epsilon(R), \end{align}\] where \(d\) denotes geodesic distance on \(M\). From this, Fatou’s lemma shows that \[\int_{M^w\setminus K_R} |\Phi(u_1,\ldots,u_w)|\,\mathrm{d}u_1\cdots \mathrm{d}u_w\le \epsilon(R).\] Combining the last three displays, we get \[\begin{align} \limsup_{T\to\infty} \biggl|\int_{M^w}\Phi_T(u_1,\ldots,u_w)\, \mathrm{d}u_1 \cdots \mathrm{d}u_w - \int_{M^w} \Phi(u_1,\ldots,u_w)\, \mathrm{d}u_1\cdots \mathrm{d}u_w\biggr|\le \epsilon(R). \end{align}\] But the left side does not depend on \(R\), and the dominated convergence theorem shows that \(\epsilon(R)\to 0\) as \(R\to \infty\). This proves equation 17 , thereby completing the proof of the main claim of the theorem. For the finiteness of the integral appearing in the theorem statement, we simply invoke the bound from Lemma 47 and apply Fatou’s lemma. ◻

Proof. Throughout this proof, \(C_0, C_1,\ldots\) will denote positive universal constants whose values may change from line to line. Write \(\tau=a+\mathrm{i}b\) with \(a>0\), and define \[z_\pm:=1-e^{-\pi\tau\pm \pi \mathrm{i}x}.\] From equation 18 , \[g(\tau,x)+\frac{\pi\tau}{2}= -\frac{1}{2}\log z_+ -\frac{1}{2}\log z_-.\] For either sign, \[\operatorname{Re}(z_\pm)=1-e^{-\pi a}\cos(\pi(x\mp b))\ge 1-e^{-\pi a}>0.\] So \(z_\pm\) lies in the open right half-plane. Hence, for our branch of logarithm, \(|\arg(z_\pm)|<\frac{\pi}{2}\). Therefore \[\biggl|\operatorname{Im}\biggl(g(\tau,x)+\frac{\pi\tau}{2}\biggr)\biggr| =\frac{1}{2}|\arg z_++\arg z_-| \le \frac{1}{2}(|\arg z_+|+|\arg z_-|) \le \frac{\pi}{2}.\] This proves the bound on the imaginary part claimed in the lemma. For the lower bound on the real part, note that since \(|z_\pm|\le 1+|e^{-\pi\tau\pm \pi \mathrm{i}x}|\le 2\), we get \[\operatorname{Re}\biggl(g(\tau,x)+\frac{\pi\tau}{2}\biggr) =-\frac{1}{2}\log|z_+|-\frac{1}{2}\log|z_-| \ge -\log 2.\] Now we prove the upper bound on the real part. Let \[w_\pm:=\pi(\tau\mp \mathrm{i}x)=\pi a+\mathrm{i}\pi(b\mp x).\] Choose \(k_\pm\in\mathbb{Z}\) so that \[|\operatorname{Im}(w_\pm)-2\pi k_\pm|=\operatorname{dist}(\operatorname{Im}(w_\pm),2\pi\mathbb{Z}),\] and define \[\widetilde{b}_\pm:=\operatorname{Im}(w_\pm)-2\pi k_\pm\in[-\pi,\pi].\] Then \[|z_\pm| =|1-e^{-w_\pm}| =|1-e^{-\pi a+\mathrm{i}\widetilde{b}_\pm}|.\] Applying Lemma 45 with \(u=\pi a\) and \(v=\widetilde{b}_\pm\), we get \[|z_\pm|\ge C_0\min\biggl\{1,\sqrt{(\pi a)^2+\widetilde{b}_\pm^2}\biggr\}.\] But \[\sqrt{(\pi a)^2+\widetilde{b}_\pm^2}=|w_\pm-2\pi\mathrm{i}k_\pm|=|\pi(\tau\mp\mathrm{i}x-2\mathrm{i}k_\pm)|\ge \pi d(\tau,x).\] Thus, we conclude that \[|z_\pm|\ge C_1\min\{1,d(\tau,x)\}.\] Hence, \[\begin{align} \operatorname{Re}\biggl(g(\tau,x)+\frac{\pi\tau}{2}\biggr) &=-\frac{1}{2}\log|z_+|-\frac{1}{2}\log|z_-|\\ &\le -\log(\min\{|z_+|,|z_-|\})\\ &\le -\log(C_1\min\{1,d(\tau,x)\})\\ &\le C_2+|\log d(\tau,x)|. \end{align}\] This completes the proof. ◻

Proof. Fix \(t=(t_1,\ldots,t_k)\in\Omega_k\) and \((c_1,\ldots,c_w)\in\Delta_{n_1,\ldots,n_k}\). We prove, case by case, that every time argument in 20 has positive real part. Since \(t\in\Omega_k\), we have \[\operatorname{Re}(t_{r+1}-t_r)>0,\qquad 1\le r<k.\] Hence, for any \(p\ge q\), \[\operatorname{Re}(t_p-t_q)\ge 0,\] with strict inequality when \(p>q\). Also, by definition of \(\Delta_{n_1,\ldots,n_k}\), \[c_l<0\;\text{for }l<n_1,\qquad 0<c_l<1\;\text{for }n_1\le l<n_k,\qquad c_l>0\;\text{for }l\ge n_k.\] For the first sum in 20 , note that if \(l<n_1\), then \(a_l=t_1+c_l\); so for any \(1\le j\le k\), \[t_j-a_l=(t_j-t_1)-c_l.\] Here \(\operatorname{Re}(t_j-t_1)\ge 0\) and \(-c_l>0\), and hence \[\operatorname{Re}(t_j-a_l)>0.\] For the middle sum in equation 20 , fix \(1\le r<k\) and \(n_r\le l<n_{r+1}\), so that \(a_l=c_l t_{r+1}+(1-c_l)t_r\). If \(1\le j\le r\), then \[a_l-t_j=c_l(t_{r+1}-t_r)+(t_r-t_j).\] Since \(c_l>0\), we have \(\operatorname{Re}(c_l(t_{r+1}-t_r))>0\), and \(\operatorname{Re}(t_r-t_j)\ge 0\). Therefore, \[\operatorname{Re}(a_l-t_j)>0.\] If \(r+1\le j\le k\), then \[t_j-a_l=(t_j-t_{r+1})+(1-c_l)(t_{r+1}-t_r).\] Since \(1-c_l>0\), we have \(\operatorname{Re}((1-c_l)(t_{r+1}-t_r))>0\), and \(\operatorname{Re}(t_j-t_{r+1})\ge 0\). Hence, \[\operatorname{Re}(t_j-a_l)>0.\] For the third sum in equation 20 , note that if \(l\ge n_k\), then \(a_l=t_k+c_l\); so for any \(1\le j\le k\), \[a_l-t_j=(t_k-t_j)+c_l.\] Here \(\operatorname{Re}(t_k-t_j)\ge 0\) and \(c_l>0\), so \[\operatorname{Re}(a_l-t_j)>0.\] Finally, let us inspect the pair terms in equation 20 . For \(1\le l<l'\le w\), we need to show that \(\operatorname{Re}(a_{l'}-a_l)>0\). It is convenient to write each \(a_l\) in the form \[a_l=t_1+\xi_l+\sum_{r=1}^{k-1}\theta_{l,r}(t_{r+1}-t_r),\] where \[\xi_l:= \begin{cases} c_l, & l<n_1\;\text{or}\;l\ge n_k,\\ 0, & n_1\le l<n_k, \end{cases}\] and, for each \(1\le r\le k-1\), \[\theta_{l,r}:= \begin{cases} 0, & l<n_1,\\ 1, & n_j\le l<n_{j+1}\;\text{for some }j\in\{1,\ldots,k-1\}\text{ with }r<j,\\ c_l, & n_j\le l<n_{j+1}\;\text{for some }j\in\{1,\ldots,k-1\}\text{ with }r=j,\\ 0, & n_j\le l<n_{j+1}\;\text{for some }j\in\{1,\ldots,k-1\}\text{ with }r>j,\\ 1, & l\ge n_k. \end{cases}\] (For a fixed \(l\) in the middle region \(n_1\le l<n_k\), the index \(j\) above is the unique one with \(n_j\le l<n_{j+1}\).) Therefore, \[a_{l'}-a_l = (\xi_{l'}-\xi_l) +\sum_{r=1}^{k-1}(\theta_{l',r}-\theta_{l,r})(t_{r+1}-t_r).\] From the explicit formulas and the ordering \(c_1<\cdots<c_w\) on \(\Delta_{n_1,\ldots,n_k}\), we have \[\xi_{l'}-\xi_l\ge 0, \qquad \theta_{l',r}-\theta_{l,r}\ge 0\;\text{for all }r.\] Moreover, because \(l<l'\), at least one of these inequalities is strict. Since each \(\operatorname{Re}(t_{r+1}-t_r)>0\), it follows that \[\operatorname{Re}(a_{l'}-a_l)>0.\] This completes the proof. ◻

Proof. Let \(\boldsymbol{n}:=(n_1,\ldots,n_k)\) and write \[B_{\boldsymbol{n}}(t)=P_{\boldsymbol{n}}(t)\,I_{\boldsymbol{n}}(t),\] where \(t=(t_1,\ldots,t_k)\in\Omega_k\), \[P_{\boldsymbol{n}}(t):=w!\prod_{j=1}^{k-1}(t_{j+1}-t_j)^{n_{j+1}-n_j},\] and \(I_{\boldsymbol{n}}(t)\) is the \(\Delta_{\boldsymbol{n}}\times[-1,1]^w\) integral in equation 20 . Since each exponent \(n_{j+1}-n_j\) is a nonnegative integer, \(P_{\boldsymbol{n}}\) is entire. So it is enough to show that \(I_{\boldsymbol{n}}\) is analytic on \(\Omega_k\).

Fix \((c_1,\ldots,c_w)\in\Delta_{\boldsymbol{n}}\) and \((b_1,\ldots,b_w)\in[-1,1]^w\). By Lemma 21, the integrand is holomorphic in \(t\) pointwise. We need a suitable bound on the integrand to ensure that the holomorphy is retained after integration.

For \(\tau\) in the open right half-plane and \(x\in\mathbb{R}\), let \[R(\tau,x):=g(\tau,x)+\frac{\pi\tau}{2}.\] By Lemma 20, there are universal constants \(C_0,C_1,C_2>0\) such that \[\label{eq:bnk-rem-bounds} -C_0\le \operatorname{Re}R(\tau,x)\le C_1+|\log d(\tau,x)|, \qquad |\operatorname{Im}R(\tau,x)|\le C_2,\tag{21}\] where \(d\) is the functional defined in Lemma 20. Let \(E(t,c,b)\) be the exponent in the integrand, and decompose \[E=E_{\mathrm{lin}}+E_{\mathrm{rem}},\] where \(E_{\mathrm{lin}}\) is obtained by replacing each \(g(\tau,x)\) by \(-\pi\tau/2\), and \(E_{\mathrm{rem}}\) is the sum of the remainder terms.

Fix \(m\in\{1,\ldots,k\}\), fix \(t_j\) for \(j\neq m\), and let \(\gamma\) be the boundary of a triangle in the corresponding one-variable slice of \(\Omega_k\). We now bound the integrand uniformly for \(t_m\in\gamma\). Here and below, \(O_\gamma(1)\) denotes any quantity whose absolute value is bounded by a constant depending on \(\gamma\) (and the fixed data \(b,w,\alpha_1,\ldots,\alpha_k,x_1,\ldots,x_k,\boldsymbol{n}\)), but independent of \(t_m\in\gamma\) and of the integration variables \((c_1,\ldots,c_w,b_1,\ldots,b_w)\). Write \[L:=\{1,\ldots,n_1-1\},\qquad M:=\{n_1,\ldots,n_k-1\},\qquad R:=\{n_k,\ldots,w\}.\] On \(\Delta_{\boldsymbol{n}}\), we have \(0<c_l<1\) for \(l\in M\), so only the \(c_l\)’s with \(l\) in \(L\) or \(R\) are unbounded. Take \(l\in L\). In \(E_{\mathrm{lin}}\), the insertion part involving this \(l\) contributes \[2\pi b\sum_{j=1}^k\alpha_j(t_j-t_1-c_l) =(-2\pi b^2w)(-c_l)+O_\gamma(1),\] using \(\sum_{j=1}^k\alpha_j=-bw\). For pair terms, fix this \(l\) and write the pair indices as \(p<q\). Terms containing \(c_l\) are of two kinds: \(p=l<q\) and \(p<q=l\). If \(p=l<q\), then \(2\pi b^2(a_q-a_l)\) contributes \(+2\pi b^2(-c_l)\) (because \(l<n_1\) implies \(a_l=t_1+c_l\)), and there are exactly \(w-l\) such values of \(q\). If \(p<q=l\), then \(2\pi b^2(a_l-a_p)\) contributes \(-2\pi b^2(-c_l)\), and there are exactly \(l-1\) such values of \(p\). Therefore the net coefficient of \((-c_l)\) coming from pair linear terms is \[2\pi b^2((w-l)-(l-1))=2\pi b^2(w-2l+1).\] Thus, the net coefficient of \((-c_l)\) coming from all linear terms is \[2\pi b^2(w-2l+1)-2\pi b^2w = -2\pi b^2(2l-1).\] Similarly, for \(l\in R\), the linear parts of the insertion terms give \[2\pi b\sum_{j=1}^k\alpha_j(c_l+t_k-t_j) =-2\pi b^2w c_l+O_\gamma(1),\] and the net coefficient of \(c_l\) from the linear parts of the pair terms is \[2\pi b^2((l-1)-(w-l)) =2\pi b^2(2l-w-1).\] Thus, the net coefficient of \(c_l\) from all terms is \[-2\pi b^2w + 2\pi b^2(2l-w-1)= -2\pi b^2(2(w-l)+1).\] From the above discussion, we deduce that there exist constants \(\sigma>0\) and \(C_\gamma\), independent of the \(b\)’s and \(c\)’s, such that \[\label{eq:bnk-linear-tail} \operatorname{Re}(E_{\mathrm{lin}}) \le C_\gamma -\sigma\sum_{l\in L}(-c_l) -\sigma\sum_{l\in R}c_l.\tag{22}\] Let us now inspect \(E_{\mathrm{rem}}\) using equation 21 . Consider first one insertion remainder term. Write \[\alpha_j=\alpha_j^{\mathrm{R}}+\mathrm{i}\alpha_j^{\mathrm{I}},\qquad R(\tau,x)=U(\tau,x)+\mathrm{i}V(\tau,x).\] Then \[\operatorname{Re}(-4b\alpha_jR(\tau,x)) =-4b(\alpha_j^{\mathrm{R}}U-\alpha_j^{\mathrm{I}}V) \le -4b\alpha_j^{\mathrm{R}}U+4b|\alpha_j^{\mathrm{I}}|\,|V|.\] By equation 21 , \(|V|\le C_2\), so the second term is bounded by a constant. For the first term we split into two cases. If \(\alpha_j^{\mathrm{R}}\ge 0\), then \(U\ge -C_0\) gives \[-4b\alpha_j^{\mathrm{R}}U\le 4b\alpha_j^{\mathrm{R}}C_0.\] If \(\alpha_j^{\mathrm{R}}<0\), then \(U\le C_1+|\log d(\tau,x)|\) gives \[-4b\alpha_j^{\mathrm{R}}U \le -4b\alpha_j^{\mathrm{R}}C_1 +(-4b\alpha_j^{\mathrm{R}})|\log d(\tau,x)|.\] Combining the two cases and absorbing all constant terms into \(C_{j,\gamma}\) gives \[\operatorname{Re}(-4b\alpha_jR(\tau,x)) \le C_{j,\gamma}+p_j|\log d(\tau,x)|, \qquad p_j:=\max\{0,-4b\operatorname{Re}(\alpha_j)\}.\] Because \(\operatorname{Re}(\alpha_j)>-\frac{1}{2b}\) by assumption, we get \[0\le p_j<2.\] Now consider one pair remainder term. Note that \[\operatorname{Re}(-4b^2R(\tau,x))=-4b^2\operatorname{Re}R(\tau,x)\le 4b^2C_0,\] again by equation 21 . So pair remainder terms contribute only bounded constants and no logarithmic singular factors.

Summing all insertion and pair remainder terms, and absorbing all bounded contributions into a single constant, we obtain \[\label{eq:bnk-rem-est} \operatorname{Re}(E_{\mathrm{rem}}) \le C_\gamma+\sum_{j=1}^k\sum_{l=1}^w p_j\,|\log d_{j,l}|,\tag{23}\] where \(0\le p_j<2\) for each \(j\) and \(d_{j,l}\) denotes the corresponding \(d\)-quantity from Lemma 20. More explicitly, for each insertion factor we write it as \(g(\tau_{j,l},\chi_{j,l})\) and set \[d_{j,l}:=d(\tau_{j,l},\chi_{j,l}) =\min\{d_+(\tau_{j,l},\chi_{j,l}),d_-(\tau_{j,l},\chi_{j,l})\},\] with \[d_\pm(\tau,\chi):=\min_{q\in\mathbb{Z}}|\tau\pm \mathrm{i}\chi-2\mathrm{i}q|.\] Here the insertion arguments are exactly those from equation 20 : \[\tau_{j,l}= \begin{cases} t_j-a_l=t_j-t_1-c_l, & l<n_1,\\ a_l-t_j=c_l(t_{r+1}-t_r)+(t_r-t_j), & n_r\le l<n_{r+1},\;1\le j\le r,\\ t_j-a_l=(t_j-t_{r+1})+(1-c_l)(t_{r+1}-t_r), & n_r\le l<n_{r+1},\;r+1\le j\le k,\\ a_l-t_j=t_k-t_j+c_l, & l\ge n_k, \end{cases}\] and \[\chi_{j,l}= \begin{cases} x_j-b_l, & l<n_1,\\ b_l-x_j, & n_r\le l<n_{r+1},\;1\le j\le r,\\ x_j-b_l, & n_r\le l<n_{r+1},\;r+1\le j\le k,\\ b_l-x_j, & l\ge n_k. \end{cases}\] Exponentiating the inequality 23 , we get \[e^{\operatorname{Re}(E_{\mathrm{rem}})} \le e^{C_\gamma}\prod_{j=1}^k\prod_{l=1}^w e^{p_j|\log d_{j,l}|}.\] For any \(r>0\) and \(p\ge 0\), \[\begin{align} e^{p|\log r|} &= \begin{cases} r^{-p}, & 0<r\le 1,\\ r^{p}, & r\ge 1, \end{cases}\\ &= (r\wedge 1)^{-p}(r\vee 1)^p \le (r\wedge 1)^{-p}(1+r)^p. \end{align}\] Applying this with \(r=d_{j,l}\) and absorbing \(e^{C_\gamma}\) into \(C_\gamma\), we get \[e^{\operatorname{Re}(E_{\mathrm{rem}})} \le C_\gamma\prod_{j=1}^k\prod_{l=1}^w (d_{j,l}\wedge 1)^{-p_j}(1+d_{j,l})^{p_j}.\] The singular factor \((d_{j,l}\wedge 1)^{-p_j}\) depends only on the two real variables \((c_l,b_l)\) corresponding to that insertion. To show that the full product over \(j\) is locally integrable for fixed \(l\), we must rule out the possibility that several such factors become singular at the same point. Since \(\gamma\) is a compact triangle contained in the one-variable slice of \(\Omega_k\), there exists \(\eta_\gamma>0\) such that \[\operatorname{Re}(t_{r+1}-t_r)\ge \eta_\gamma \qquad\text{for all }1\le r<k\text{ and all }t_m\in\gamma.\] Consequently, \[\operatorname{Re}(t_p-t_q)\ge \eta_\gamma\qquad\text{whenever }p>q.\] Fix \(l\). We inspect the explicit formulas for \(\tau_{j,l}\) case by case. If \(l<n_1\), then \(\tau_{j,l}=t_j-t_1-c_l\) and \(c_l<0\). Hence for every \(j>1\), \[\operatorname{Re}\tau_{j,l}=\operatorname{Re}(t_j-t_1)-c_l\ge \eta_\gamma,\] so \(d_{j,l}\ge \eta_\gamma\) and the corresponding factor is uniformly bounded. Thus only the \(j=1\) factor can become singular. Its singular points satisfy \(c_l=0\) and \(b_l\equiv x_1\pmod 2\); because \(b_l\in[-1,1]\), only finitely many such points occur.

If \(n_r\le l<n_{r+1}\) for some \(1\le r<k\), then \(0<c_l<1\). For \(1\le j\le r\) we have \[\tau_{j,l}=c_l(t_{r+1}-t_r)+(t_r-t_j),\] while for \(r+1\le j\le k\), \[\tau_{j,l}=(t_j-t_{r+1})+(1-c_l)(t_{r+1}-t_r).\] If \(j<r\), then \(\operatorname{Re}(t_r-t_j)\ge \eta_\gamma\), so again \(d_{j,l}\ge \eta_\gamma\). If \(j>r+1\), then \(\operatorname{Re}(t_j-t_{r+1})\ge \eta_\gamma\), so \(d_{j,l}\ge \eta_\gamma\) as well. Therefore only the adjacent insertions \(j=r\) and \(j=r+1\) can be singular. Moreover, \[\tau_{r,l}=c_l(t_{r+1}-t_r)\] can approach the singular set only when \(c_l\to 0\), whereas \[\tau_{r+1,l}=(1-c_l)(t_{r+1}-t_r)\] can approach it only when \(c_l\to 1\). Hence these two possible singularities lie on different boundary faces of the strip \(0<c_l<1\), so they cannot pile up at the same \((c_l,b_l)\). Again, for each face there are only finitely many relevant values of \(b_l\) because \(b_l\in[-1,1]\).

If \(l\ge n_k\), then \(\tau_{j,l}=t_k-t_j+c_l\) and \(c_l>0\). For every \(j<k\), \[\operatorname{Re}\tau_{j,l}=\operatorname{Re}(t_k-t_j)+c_l\ge \eta_\gamma,\] so only the \(j=k\) factor can become singular, necessarily near points with \(c_l=0\) and \(b_l\equiv x_k\pmod 2\).

We have therefore shown that for each fixed \(l\), near any candidate singular point there is at most one unbounded factor among \(\{(d_{j,l}\wedge 1)^{-p_j}:1\le j\le k\}\); all the other factors are uniformly bounded there. For that single potentially singular factor, the map \[(c_l,b_l)\longmapsto \tau_{j,l}(c_l)\pm \mathrm{i}\chi_{j,l}(b_l)-2\mathrm{i}q\] is affine with invertible linear part (uniformly for \(t_m\in\gamma\)), so \(d_{j,l}\) is comparable to Euclidean distance from the relevant singular point. Since \(p_j<2\), this shows that the whole product \[\prod_{j=1}^k (d_{j,l}\wedge 1)^{-p_j}\] is locally integrable in \((c_l,b_l)\). Because different values of \(l\) involve disjoint variable pairs \((c_l,b_l)\), the full singular product \(\prod_{j=1}^k\prod_{l=1}^w(d_{j,l}\wedge1)^{-p_j}\) is locally integrable on \(\Delta_{\boldsymbol{n}}\times[-1,1]^w\).

For the growth factor, for \(t_m\in\gamma\) every relevant \(\tau\) is affine in \((c_l,t_m)\) with bounded coefficients, while \(b_l\in[-1,1]\) and the \(x_j\) are fixed. Choosing \(k=0\) in the definition of \(d_\pm\) gives \[d_{j,l}\le |\tau\pm \mathrm{i}x|\le C_\gamma(1+|c_l|),\] and so, \[(1+d_{j,l})^{p_j}\le C_\gamma'(1+|c_l|)^{p_j}.\] Thus the product of all growth factors has at most polynomial growth in the unbounded \(c_l\) variables. Combining this with the linear tail bound 22 yields \[|\text{integrand}(t,c,b)|\le \Psi_\gamma(t,c,b),\] where \(\text{integrand}(t,c,b)\) denotes the integrand in equation 20 , and \(\Psi_\gamma\) is an explicit product of:

• locally integrable singular factors \((d_{j,l}\wedge 1)^{-p_j}\) (with exponents \(<2\)),

• polynomial factors in \(|c_l|\), and

• exponential tails in the unbounded directions \(c_l\to -\infty\) for \(l\in L\) and \(c_l\to +\infty\) for \(l\in R\).

Now view \(\Psi_\gamma\) as a function of \((t_m,c,b)\), with \((t_j)_{j\ne m}\) fixed. The hidden constants in the above bounds depend continuously on \(t\) on \(\Omega_k\); restricting to the compact slice obtained by varying \(t_m\in\gamma\), they are uniformly bounded. Therefore the same singular-exponent and tail estimates imply \[\Psi_\gamma\in L^1(\gamma\times\Delta_{\boldsymbol{n}}\times[-1,1]^w),\] (with arc-length measure on \(\gamma\)), and \[|\text{integrand}(t,c,b)|\le \Psi_\gamma(t,c,b).\] Hence Fubini’s theorem applies, giving: \[\int_{\gamma} I_{\boldsymbol{n}}(t)\,\mathrm{d}t_m =\int_{\Delta_{\boldsymbol{n}}}\int_{[-1,1]^w} \biggl(\int_{\gamma}\text{integrand}(t,c,b)\,\mathrm{d}t_m\biggr) \mathrm{d}b\,\mathrm{d}c.\] The innermost integral on the right vanishes, since \(\text{integrand}(t,c,b)\) is analytic in \(t_m\). By Morera’s theorem, this shows that \(I_{\boldsymbol{n}}\) is holomorphic in \(t_m\). Since \(m\) was arbitrary, \(I_{\boldsymbol{n}}\) is separately holomorphic on \(\Omega_k\). Thus, Hartogs’ theorem [44] implies that \(I_{\boldsymbol{n}}\) is jointly holomorphic on \(\Omega_k\). This completes the proof. ◻

Proof. Fix \(\alpha_1,\ldots,\alpha_k\) and \(x_1,\ldots,x_k\), and write \[C(t):=C_{\alpha_1,\ldots,\alpha_k}(t_1,x_1;\ldots;t_k,x_k),\] where \(C_{\alpha_1,\ldots,\alpha_k}\) is the function defined in equation 24 . Let \(A(t)\) denote the free factor from equation 24 . Since the external points \(v_1,\ldots,v_k\) occur along \(\gamma\) in index order, this is explicitly \[A(t)= \exp\biggl( -4\sum_{1\le j<j'\le k}\alpha_j\alpha_{j'} g(t_{j'}-t_j,x_{j'}-x_j) \biggr).\] Let \[\begin{align} B^\gamma(t) &:= \int_{M_\gamma^w} \exp\biggl( -4b\sum_{j=1}^k\sum_{l=1}^w \alpha_j g(\mathcal{T}_\gamma(v_j-u_l)) \\ &\qquad \qquad -4b^2\sum_{1\le l<l'\le w}g(\mathcal{T}_\gamma(u_l-u_{l'})) \biggr) \,\mathrm{d}u_1\cdots \mathrm{d}u_w, \end{align}\] so that \[C(t)=\frac{(-\mu)^w}{w!}A(t)B^\gamma(t).\] Take \(t\) in the domain \(\Omega_k\) from the theorem (i.e., each \(t_{j+1}-t_j\) lies in the open right half-plane), and consider the contour \(\gamma\) from equation 24 . Write \[\gamma=\Gamma_0\cup\Gamma_1\cup\cdots\cup\Gamma_{k-1}\cup\Gamma_k,\] where \[\begin{align} &\Gamma_0:=\{t_1+c:\;c<0\},\\ &\Gamma_r:=\{(1-c)t_r+ct_{r+1}:\;0<c<1\}\quad (1\le r<k),\\ &\Gamma_k:=\{t_k+c:\;c>0\}. \end{align}\] Because \(\mathcal{T}_\gamma(u-u')=\mathcal{T}_\gamma(u'-u)\), the screening-screening part of the integrand is symmetric under permutations of the screening variables, and the mixed part is obviously symmetric as well. Hence we may restrict to screening points \(u_1,\ldots,u_w\) ordered along the orientation of \(\gamma\) and multiply by \(w!\). We then partition according to how many ordered points lie in each segment: for \(1\le n_1\le\cdots\le n_k\le w+1\), the indices \[l<n_1,\quad n_r\le l<n_{r+1}\;(1\le r<k),\quad l\ge n_k\] correspond exactly to \[u_l\in\Gamma_0\times[-1,1],\qquad u_l\in\Gamma_r\times[-1,1],\qquad u_l\in\Gamma_k\times[-1,1],\] respectively. Writing \(u_l=(a_l,b_l)\) and parameterizing the time coordinates by \[a_l= \begin{cases} t_1+c_l,& l<n_1,\\ (1-c_l)t_r+c_l t_{r+1},& n_r\le l<n_{r+1},\;1\le r<k,\\ t_k+c_l,& l\ge n_k, \end{cases}\] with \((c_1,\ldots,c_w)\in\Delta_{n_1,\ldots,n_k}\) gives exactly the change of variables 19 and Jacobian factor \[\mathrm{d}a_1\cdots\mathrm{d}a_w= \biggl(\prod_{r=1}^{k-1}(t_{r+1}-t_r)^{n_{r+1}-n_r}\biggr) \mathrm{d}c_1\cdots\mathrm{d}c_w.\] On such an ordered sector, the time-ordering is explicit: \[\mathcal{T}_\gamma(u_l-u_{l'})=u_{l'}-u_l, \qquad 1\le l<l'\le w.\] Moreover, \[\mathcal{T}_\gamma(v_j-u_l)= \begin{cases} v_j-u_l, & l<n_1,\\ u_l-v_j, & n_r\le l<n_{r+1},\;1\le j\le r,\\ v_j-u_l, & n_r\le l<n_{r+1},\;r+1\le j\le k,\\ u_l-v_j, & l\ge n_k, \end{cases}\] with \(1\le r<k\) in the middle cases. Since \[v_j-u_l=(t_j-a_l,x_j-b_l),\qquad u_l-v_j=(a_l-t_j,b_l-x_j),\] and \[u_{l'}-u_l=(a_{l'}-a_l,b_{l'}-b_l),\] substituting into the time-ordered contour integral yields, term by term in \((n_1,\ldots,n_k)\), precisely the formula 20 used in Lemma 22 (with \(g\) interpreted via its analytic continuation on \(\operatorname{Re}\tau>0\)). Therefore \[B^\gamma(t) = \sum_{1\le n_1\le\cdots\le n_k\le w+1}B_{n_1,\ldots,n_k}(t),\] where the right side is exactly the analytically continued Coulomb decomposition introduced in §4.4.

Now, the free part \(A\) is analytic on \(\Omega_k\) by the discussion in §4.3. Also, by Lemma 22, each \(B_{n_1,\ldots,n_k}\) is analytic on that same region. Since there are only finitely many index tuples \((n_1,\ldots,n_k)\), \(B^\gamma\) is analytic there. Therefore \(C\) is analytic on \(\Omega_k\).

Next, let \(s=(s_1,\ldots,s_k)\) with \(s_1<\cdots<s_k\) real, and let \(t^{(m)}\in\Omega_k\) satisfy \(t^{(m)}\to s\). For each index tuple \(\boldsymbol{n}=(n_1,\ldots,n_k)\), write \[B_{\boldsymbol{n}}(t)=P_{\boldsymbol{n}}(t)I_{\boldsymbol{n}}(t)\] as in the proof of Lemma 22. The factor \(P_{\boldsymbol{n}}\) is polynomial, hence continuous at \(s\). For \(I_{\boldsymbol{n}}\), the estimates proved in Lemma 22 give an \(L^1\) majorant uniform for \(t\) in a small compact neighborhood of \(s\) inside \(\Omega_k\): the singular exponents are still \(<2\), and the same exponential tails in the unbounded \(c_l\)-directions hold. Since each time argument in equation 20 stays in the open right half-plane for \(t\in\Omega_k\) and converges to a positive real value at \(t=s\), and \(g(\tau,x)\) is continuous up to that boundary, the integrand converges pointwise. Therefore, by dominated convergence, \[I_{\boldsymbol{n}}(t^{(m)})\to I_{\boldsymbol{n}}(s),\qquad B_{\boldsymbol{n}}(t^{(m)})\to B_{\boldsymbol{n}}(s).\] Summing over finitely many \(\boldsymbol{n}\) gives \[B^\gamma(t^{(m)})\to \sum_{\boldsymbol{n}} B_{\boldsymbol{n}}(s).\] At real \(s\), this finite sum is exactly the Euclidean Coulomb factor (the same partition and change of variables that produced equation 20 from the Euclidean integral), i.e., the \(B\)-term in equation 2 . Also \(A(t^{(m)})\to A(s)\). Hence \[C(t^{(m)})\to C_{\alpha_1,\ldots,\alpha_k}(s_1,x_1;\ldots;s_k,x_k)\] with the right-hand side given by equation 2 . Since the sequence \(t^{(m)}\to s\) was arbitrary in \(\Omega_k\), this proves the Euclidean boundary limit.

For uniqueness on \(\Omega_k\), let \(\widetilde{C}\) be any analytic function on \(\Omega_k\) with the same Euclidean boundary values, and set \[H:=\widetilde{C}-C.\] Then \(H\) is analytic on \(\Omega_k\) and has boundary limit \(0\) at every point of \[\mathcal{E}_k:=\{(r_1,\ldots,r_k)\in\mathbb{R}^k:r_1<\cdots<r_k\}.\] Introduce coordinates \[u_1:=t_1,\qquad u_j:=t_j-t_{j-1}\quad (2\le j\le k),\] which biholomorphically map \(\Omega_k\) onto \[D_k:=\mathbb{C}\times {\mathbb{H}}_+^{k-1},\qquad \mathbb{H}_+:=\{z\in\mathbb{C}:\operatorname{Re}z>0\},\] and map \(\mathcal{E}_k\) onto \[B_k:=\mathbb{R}\times(0,\infty)^{k-1}.\] Define \[\widehat H(u_1,\ldots,u_k):=H(u_1,u_1+u_2, \ldots, u_1+\cdots+u_k).\] Then \(\widehat H\) is analytic on \(D_k\) and is zero on \(B_k\). Therefore, Lemma 48 from §5.7.3 gives \(\widehat H\equiv 0\) on \(D_k\), and hence \(H\equiv 0\) on \(\Omega_k\). So the analytic continuation to \(\Omega_k\) is unique.

Finally, let \(s = (s_1,\ldots,s_k)\) be a real \(k\)-tuple satisfying the condition that \((s_j-s_{j'})\pm (x_j - x_{j'})\notin 2\mathbb{Z}\) for all \(1\le j<j'\le k\), and let \(t^{(m)}\in\Omega_k\) satisfy \(t^{(m)}\to \mathrm{i}s\) as \(m\to\infty\). Set \[v_j:=(\mathrm{i}s_j,x_j),\qquad 1\le j\le k,\] and let \(\widetilde{\gamma}\) be the contour from the definition 7 associated with \(v_1,\ldots,v_k\). For each index tuple \(\boldsymbol{n}=(n_1,\ldots,n_k)\), we again write \[B_{\boldsymbol{n}}(t)=P_{\boldsymbol{n}}(t)I_{\boldsymbol{n}}(t).\] As above, \(P_{\boldsymbol{n}}(t^{(m)})\to P_{\boldsymbol{n}}(\mathrm{i}s)\). For \(I_{\boldsymbol{n}}\), one must refine the estimate from Lemma 22 slightly and use the non-light-cone hypothesis on \(s\). The exponential tail bounds in the unbounded \(c_l\)-directions are unchanged. On the middle part of the limiting contour, after writing \(a=\mathrm{i}u\), the possible insertion singularities are governed by the affine lines \[u\mp b=s_j\mp x_j+2m,\qquad m\in\mathbb{Z}.\] Because \((s_j-s_{j'})\pm(x_j-x_{j'})\notin 2\mathbb{Z}\), no two parallel lines coming from different insertions coincide; any remaining intersections are transverse. Therefore, after an affine change of coordinates near any candidate singular point, the local singular factor is of the form \(|u|^{-q_r}\) or \(|u|^{-q_r}|v|^{-q_{r'}}\), where \[q_j:=\max\{0,-2b\operatorname{Re}(\alpha_j)\}<1.\] These model singularities are locally integrable, and by compactness they give an \(L^1\) majorant uniform for \(t\) in a small compact neighborhood of \(\mathrm{i}s\) inside \(\overline{\Omega}_k\). Moreover, by the boundary-value formula 5 , each \(g\)-factor in 20 converges pointwise as \(t^{(m)}\to \mathrm{i}s\).

The limiting argument is precisely the time-ordered difference determined by \(\widetilde{\gamma}\): if an ordered screening point lies before \(v_1\) on \(\widetilde{\gamma}\), then every mixed factor is of the form \(g(v_j-u_l)\); if it lies on the segment from \(v_r\) to \(v_{r+1}\), then the mixed factors are \(g(u_l-v_j)\) for \(j\le r\) and \(g(v_j-u_l)\) for \(j\ge r+1\); and if it lies after \(v_k\), then every mixed factor is of the form \(g(u_l-v_j)\). (When \(v_r=v_{r+1}\), that segment is empty and there is nothing to check.)

Likewise, for screening points ordered along the contour one has \(\mathcal{T}(u_{l'}-u_l)=u_{l'}-u_l\) whenever \(l<l'\). Therefore, by dominated convergence, each sector integral converges to the corresponding ordered-sector contribution in the contour integral from 7 . Performing on that contour integral the same ordered-sector decomposition as above shows that the sum of these limits is exactly its screening part. Also, since \(v_{j'}\) occurs later than \(v_j\) on \(\widetilde{\gamma}\) whenever \(j<j'\), we have \(\mathcal{T}(v_j-v_{j'})=v_{j'}-v_j\), and so the continuity of 5 gives convergence of the free factor to the free factor in 7 . Summing over finitely many \(\boldsymbol{n}\), and combining with the convergence of the free factor \(A(t^{(m)})\), we obtain \[C(t^{(m)})\longrightarrow C^{\mathrm L}_{\alpha_1,\ldots,\alpha_k}(s_1,x_1;\ldots;s_k,x_k),\] which is exactly equation 7 . This proves the Lorentzian limit.

It remains to justify absolute convergence of the integral in 7 , and continuity of the resulting boundary function on its domain. For any target point satisfying the same non-light-cone condition, the very same sectorwise argument applies in a sufficiently small neighborhood: the tail bounds are unchanged, parallel singular lines stay separated by the non-light-cone hypothesis, and the only remaining intersections are transverse, hence locally integrable. Evaluating the resulting majorant at the target point gives absolute convergence of the contour integral, and dominated convergence for nearby boundary points shows that the value of the contour integral in 7 depends continuously on \((t_1,x_1),\ldots,(t_k,x_k)\). Hence the analytic continuation extends continuously to the full domain of \(C^{\mathrm L}_{\alpha_1,\ldots,\alpha_k}\). ◻

Proof. On a generator, \((\mathcal{O}_{f,n}^*)^*=\mathcal{O}_{\overline{\overline{f}},n}=\mathcal{O}_{f,n}\). Since the involution was extended conjugate-linearly and anti-multiplicatively to the tensor algebra, the identities \((AB)^*=B^*A^*\) and \((A^*)^*=A\) follow for all \(A,B\in\mathcal{A}\). ◻

Proof. By linearity, it is enough to consider a monomial \[A=\mathcal{O}_{f_1,n_1}\cdots\mathcal{O}_{f_k,n_k}.\] Set \(\boldsymbol{n}=(n_1,\ldots,n_k)\), write \(\boldsymbol{n}^\#=(n_k,\ldots,n_1)\), set \(\alpha_j:=bn_j\), and \(w:=-\sum_{j=1}^k n_j\). If \(w<0\), then both \(\omega(A)\) and \(\omega(A^*)\) vanish, because reversing the labels leaves \(w\) unchanged. Thus assume \(w\ge0\). We first claim that whenever \((u_1,\ldots,u_k)\) satisfies the non-light-cone condition 6 , \[\begin{align} \label{eq:conjclaim} C_{b\boldsymbol{n}}^{\mathrm L}(u_1,\ldots,u_k) = \overline{C_{b\boldsymbol{n}^\#}^{\mathrm L}(u_k,\ldots,u_1)}. \end{align}\tag{28}\] To prove this, fix \(x_1,\ldots,x_k\), and for \((t_1,\ldots,t_k)\in \Omega_k\), define \[F(t_1,\ldots,t_k) := C_{\alpha_1,\ldots,\alpha_k}(t_1,x_1;\ldots;t_k,x_k),\] and \[G(t_1,\ldots,t_k) := \overline{C_{\overline{\alpha}_k,\ldots,\overline{\alpha}_1}(-\overline{t_k},x_k;\ldots; -\overline{t_1},x_1)}.\] If \((t_1,\ldots,t_k)\in\Omega_k\), then \((-\overline{t}_k,\ldots,-\overline{t}_1)\in\Omega_k\) as well, and so \(G\) is well-defined. Moreover, if \[\sum_{n_1,\ldots,n_k\in \mathbb{Z}_{\ge0}} a_{n_1,\ldots,n_k} (\tau_1-t_1^0)^{n_1}\cdots(\tau_k - t_k^0)^{n_k}\] is the power series expansion of the analytic function \[(\tau_1,\ldots,\tau_k)\mapsto C_{\overline{\alpha}_k,\ldots,\overline{\alpha}_1}(-\tau_k,x_k;\ldots; -\tau_1,x_1)\] in a neighborhood of a point \((t_1^0,\ldots,t_k^0)\in \Omega_k\), then for any \((t_1,\ldots,t_k)\) in the corresponding neighborhood of \((\overline{t_1^0},\ldots,\overline{t_k^0})\), \[\begin{align} G(t_1,\ldots,t_k) &= \overline{C_{\overline{\alpha}_k,\ldots,\overline{\alpha}_1}(-\overline{t_k},x_k;\ldots; -\overline{t_1},x_1)}\\ &=\overline{\sum_{n_1,\ldots,n_k\in \mathbb{Z}_{\ge0}} a_{n_1,\ldots,n_k} (\overline{t_1}-t_1^0)^{n_1}\cdots(\overline{t_k} - t_k^0)^{n_k}}\\ &= \sum_{n_1,\ldots,n_k\in \mathbb{Z}_{\ge0}} \overline{a_{n_1,\ldots,n_k}} (t_1-\overline{t_1^0})^{n_1}\cdots(t_k - \overline{t_k^0})^{n_k}. \end{align}\] This shows that \(G\) is analytic on \(\Omega_k\). Next, equation 2 and Theorem 1 show that complex conjugation of the Euclidean correlator simply replaces each charge by its conjugate, because all coefficients and the kernel \(g\) are real on real arguments. Therefore, for real numbers \(r_1<\cdots<r_k\), \[G(r_1,\ldots,r_k) = C_{\alpha_k,\ldots,\alpha_1}(-r_k,x_k;\ldots; -r_1,x_1).\] Because the Euclidean correlator is invariant under simultaneous permutation of the labeled triples \((\alpha_j,t_j,x_j)\), we get \[G(r_1,\ldots,r_k) = C_{\alpha_1,\ldots,\alpha_k}(-r_1,x_1;\ldots; -r_k,x_k).\] Changing variables \(a_l\mapsto -a_l\) in equation 2 and using the real-axis symmetry \(g(t,x)=g(-t,x)\), we get \[C_{\alpha_1,\ldots,\alpha_k}(-r_1,x_1;\ldots; -r_k,x_k) = C_{\alpha_1,\ldots,\alpha_k}(r_1,x_1;\ldots; r_k,x_k) = F(r_1,\ldots,r_k).\] Thus \(F\) and \(G\) are analytic on \(\Omega_k\) and coincide there on every strictly increasing real \(k\)-tuple. Therefore, by the uniqueness statement in Theorem 23, they agree on all of \(\Omega_k\). In other words, \[F(t_1,\ldots,t_k) = \overline{C_{\overline{\alpha}_k,\ldots,\overline{\alpha}_1}(-\overline{t_k},x_k;\ldots; -\overline{t_1},x_1)}.\] Now let \((s_1,\ldots,s_k)\in\mathbb{R}^k\) satisfy condition 6 for our chosen \(x_1,\ldots,x_k\), and choose \(t^{(m)}\in\Omega_k\) with \(t^{(m)}\to(\mathrm{i}s_1,\ldots,\mathrm{i}s_k)\). Then \((-\overline{t_k^{(m)}},\ldots,-\overline{t_1^{(m)}})\in\Omega_k\) and converges to \((\mathrm{i}s_k,\ldots,\mathrm{i}s_1)\). Applying the Lorentzian boundary value statement from Theorem 23 and letting \(m\to\infty\) proves the claim 28 . Since the light cone boundary set has measure zero, this identity holds almost everywhere on \(M^k\). Since \[A^*=\mathcal{O}_{\overline{f}_k,n_k}\cdots\mathcal{O}_{\overline{f}_1,n_1},\] we have \[\begin{align} \omega(A^*) &=\int_{M^k}\prod_{j=1}^k\overline{f_{k+1-j}(u_j)}\, C_{b\boldsymbol{n}^\#}^{\mathrm L}(u_1,\ldots,u_k)\,\mathrm{d}u_1\cdots\mathrm{d}u_k\\ &=\overline{\int_{M^k}\prod_{j=1}^k f_{k+1-j}(u_j)\, \overline{C_{b\boldsymbol{n}^\#}^{\mathrm L}(u_1,\ldots,u_k)}\,\mathrm{d}u_1\cdots\mathrm{d}u_k}\\ &=\overline{\int_{M^k}\prod_{j=1}^k f_{k+1-j}(u_j)\, C_{b\boldsymbol{n}}^{\mathrm L}(u_k,\ldots,u_1)\,\mathrm{d}u_1\cdots\mathrm{d}u_k}\\ &=\overline{\int_{M^k}\prod_{j=1}^k f_j(u_j)\, C_{b\boldsymbol{n}}^{\mathrm L}(u_1,\ldots,u_k)\,\mathrm{d}u_1\cdots\mathrm{d}u_k} =\overline{\omega(A)}. \end{align}\] This completes the proof of the lemma. ◻

Proof of Theorem 5. By Lemma 24 and Lemma 25, \[\langle B,A \rangle_\mathcal{A}= \omega(B^*A) = \omega((A^*B)^*) = \overline{\omega(A^*B)} = \overline{\langle A,B \rangle_\mathcal{A}}.\] This proves the claim. ◻

Proof of Theorem 7. By definition, \[\mathcal{H}=\overline{\mathcal{H}_0},\] so \(\mathcal{H}_0\) is dense in \(\mathcal{H}\). For each \(A\in\mathcal{A}\), let \(p_A:\mathcal{H}\to\mathbb{C}\) denote the coordinate projection \[p_A((z_B)_{B\in\mathcal{A}}) := z_A.\] By the definition of the product topology, each \(p_A\) is continuous. We first prove that the pairing is well-defined. Suppose \(\Phi(A_1)=\Phi(A_2)\). Then, by definition of \(\Phi\), \[\langle C^*A_1 \rangle=\langle C^*A_2 \rangle \qquad\text{for all } C\in\mathcal{A}.\] Hence, for every \(C\in\mathcal{A}\), \[\begin{align} p_{A_1}(\Phi(C)) &=\langle A_1^*C \rangle = \overline{\langle C^*A_1 \rangle} = \overline{\langle C^*A_2 \rangle} \\ &=\langle A_2^*C \rangle = p_{A_2}(\Phi(C)), \end{align}\] where we used Theorem 5. Therefore \(p_{A_1}-p_{A_2}\) vanishes on the dense subset \(\Phi(\mathcal{A})\subseteq\mathcal{H}\). Since \(p_{A_1}-p_{A_2}\) is continuous, it vanishes on all of \(\mathcal{H}=\overline{\Phi(\mathcal{A})}\). Therefore \(z_{A_1}=z_{A_2}\) for all \(z\in\mathcal{H}\), proving that \[(\Phi(A),z):=z_A\] is well-defined.

Next, fix \(z\in\mathcal{H}\). Choose a net \((C_\nu)\) in \(\mathcal{A}\) such that \(\Phi(C_\nu)\to z\) in the product topology; see §5.9 for background on nets and their relation to closure. For any \(A,B\in\mathcal{A}\) and \(\lambda,\mu\in\mathbb{C}\), \[\begin{align} z_{\lambda A+\mu B} &= \lim_\nu \Phi(C_\nu)_{\lambda A+\mu B} = \lim_\nu \langle (\lambda A+\mu B)^* C_\nu \rangle \\ &= \overline{\lambda}\lim_\nu \langle A^*C_\nu \rangle+\overline{\mu}\lim_\nu \langle B^*C_\nu \rangle = \overline{\lambda} z_A+\overline{\mu} z_B. \end{align}\] So \(A\mapsto z_A\) is conjugate linear on \(\mathcal{A}\). Also, \(\Phi\) is linear, since \[\Phi(\lambda A+\mu B)=\lambda\Phi(A)+\mu\Phi(B).\] Consequently, for \(u_j=\Phi(A_j)\in\mathcal{H}_0\), \[(\lambda u_1+\mu u_2,z) =(\Phi(\lambda A_1+\mu A_2),z) =z_{\lambda A_1+\mu A_2} =\overline{\lambda} z_{A_1}+\overline{\mu} z_{A_2} =\overline{\lambda}(u_1,z)+\overline{\mu}(u_2,z).\] Hence the pairing is conjugate linear in the first argument.

For linearity in the second argument, fix \(u=\Phi(A)\in\mathcal{H}_0\) and take any \(z_1,z_2\in\mathcal{H}\) and \(\lambda,\mu\in\mathbb{C}\). Then \[\begin{align} (u,\lambda z_1+\mu z_2) &=(\lambda z_1+\mu z_2)_A\\ &=\lambda (z_1)_A+\mu (z_2)_A=\lambda(u,z_1)+\mu(u,z_2). \end{align}\] So \(z\mapsto (u,z)\) is linear, and the pairing is sesquilinear.

Now let \(u=\Phi(A)\) and \(v=\Phi(B)\) be in \(\mathcal{H}_0\). Then \[(u,v) =\Phi(B)_A =\omega(A^*B) =\langle A,B\rangle_{\mathcal{A}}.\] By Theorem 5, this shows that the restriction of the pairing to \(\mathcal{H}_0\times\mathcal{H}_0\) is Hermitian.

Next, suppose that \((u,z)=0\) for all \(u\in\mathcal{H}_0\). Taking \(u=\Phi(A)\) gives \[0=(\Phi(A),z)=z_A \qquad\text{for all } A\in\mathcal{A}.\] Hence every coordinate of \(z=(z_A)_{A\in\mathcal{A}}\) is zero, so \(z=0\) in \(\mathbb{C}^{\mathcal{A}}\) (and therefore in \(\mathcal{H}\)). The converse implication is immediate. In particular, if \(v\in\mathcal{H}_0\) and \((u,v)=0\) for all \(u\in\mathcal{H}_0\), then \(v=0\). Thus the Hermitian form on \(\mathcal{H}_0\) is nondegenerate. ◻

Proof of Proposition 8. Let \[u_+:=\Omega=\Phi(1).\] Then, by Theorem 7, \[(u_+,u_+)=(\Phi(1),\Phi(1))=\omega(1)=1>0.\] To construct a negative vector, fix a nonnegative function \(\rho\in C_c^\infty((-1,1)^2,\mathbb{R})\) with \[\int_{\mathbb{R}\times[-1,1]} \rho(t,x)\,\mathrm{d}t\,\mathrm{d}x=1.\] For \(0<\epsilon<\frac{1}{2}\), define \[f_\epsilon(t,x):=\epsilon^{-2}\rho\biggl(\frac{t}{\epsilon},\frac{x}{\epsilon}\biggr),\] viewed as a smooth compactly supported function on the cylinder \(M\). Then \(f_\epsilon\ge0\), \(f_\epsilon\in C_c^\infty(M,\mathbb{R})\), \[\int_M f_\epsilon(u)\,\mathrm{d}u=1,\] and the support of \(f_\epsilon\) shrinks to \((0,0)\) as \(\epsilon\downarrow0\). Set \[A_\epsilon:=\mathcal{O}_{f_\epsilon,-b}\in\mathcal{A}, \qquad u_\epsilon:=\Phi(A_\epsilon)-\omega(A_\epsilon)\Omega\in\mathcal{H}_0.\] We first compute the one-point function. Taking \(k=1\) and \(\alpha_1=-b\) in Theorem 1 (which is allowed since \(-b > -\frac{1}{2b}\), because \(b <\frac{1}{\sqrt{2}}\)), we obtain a Euclidean one-point correlator that is independent of the insertion point and equal to \[c_b := -\mu\int_M e^{4b^2 g(u)}\,\mathrm{d}u.\] The integral is finite by Theorem 1, and the integrand is strictly positive. Thus, \(c_b<0\). Since a constant analytic continuation remains the same constant, Theorem 2 gives the same value for the Lorentzian one-point function. Therefore \[\omega(A_\epsilon)=c_b\int_M f_\epsilon(u)\,\mathrm{d}u=c_b<0\] for every \(\epsilon\).

Next, we prove the integrated two-point estimate \[\begin{align} \label{eq:5x} \lim_{\epsilon\to 0} \omega(A_\epsilon^*A_\epsilon)= 0. \end{align}\tag{29}\] Note that \(A_\epsilon^* = A_\epsilon\), and thus, \[\begin{align} \omega(A_\epsilon^* A_\epsilon) &= \int_{M^2} f_\epsilon(u)f_\epsilon(v) C_{-b,-b}^{\mathrm{L}}(u,v) \, \mathrm{d}u\, \mathrm{d}v. \end{align}\] Since the right side is a weighted average of \(C_{-b,-b}^{\mathrm{L}}(u,v)\) in a neighborhood of radius \(O(\epsilon)\) around zero in \(M^2\), it suffices to show that \[\begin{align} \label{eq:5x0} \lim_{u,v\to 0} C_{-b,-b}^{\mathrm{L}}(u,v) = 0. \end{align}\tag{30}\] Now, with \(u=(t,x)\) and \(v= (s,y)\), we have \[\begin{align} C_{-b,-b}^{\mathrm{L}}(u,v) &= \frac{\mu^2}{2} e^{-4b^2 g(\mathrm{i}(t-s), x-y)}I(u,v), \end{align}\] where \(I(u,v)\) is the integral appearing in the Lorentzian correlator formula 7 . By equation 26 , \[\begin{align} |e^{-4b^2 g(\mathrm{i}(t-s), x-y)}| &\le |1-e^{-\pi \mathrm{i}(t-s+x-y)}|^{2b^2} |1-e^{-\pi \mathrm{i}(t-s-x+y)}|^{2b^2}. \end{align}\] Clearly, the upper bound tends to zero as \(u,v\to 0\). Thus, to prove the claim 30 , it suffices to show that \(I(u,v)\) is bounded in a neighborhood of zero.

Let \(\gamma\) be the contour for the Lorentzian correlator in this problem, so that \[\begin{align} I(u,v) &= \int_{\gamma} \exp\biggl( 4b^2\sum_{l=1}^2 g(\mathcal{T}(u-z_l)) \notag\\ &\qquad\qquad +4b^2\sum_{\ell=1}^2 g(\mathcal{T}(v-z_l)) -4b^2 g(\mathcal{T}(z_1-z_2)) \biggr) \,\mathrm{d}z_1\,\mathrm{d}z_2. \end{align}\] Applying the Cauchy–Schwarz inequality, we get \[|I(u,v)|\le \sqrt{J(u)J(v)},\] where \[\begin{align} J(u) &:= \int_{\gamma} \exp\biggl( 8b^2\sum_{l=1}^2 \operatorname{Re}(g(\mathcal{T}(u-z_l)))-4b^2 \operatorname{Re}(g(\mathcal{T}(z_1-z_2))) \biggr) \,|\mathrm{d}z_1|\,|\mathrm{d}z_2|, \end{align}\] and \(J(v)\) is defined similarly, replacing \(u\) by \(v\). We will now show that \(J(u)\) is uniformly bounded when \(u\) and \(v\) lie in a fixed compact neighborhood of zero. (Note that \(J(u)\) is not just a function of \(u\); it also depends on \(v\) through the definition of the contour \(\gamma\).) The uniform boundedness of \(J(v)\) will follow similarly.

Suppose that \(t\le s\). Then the contour \(\gamma\) can be written as \(\Gamma_-\cup \Gamma_0\cup \Gamma_+\), where \[\Gamma_-:=\{\mathrm{i}t-r:r>0\},\quad \Gamma_0:=\{\mathrm{i}\tau:t<\tau<s\},\quad \Gamma_+:=\{\mathrm{i}s+r:r>0\}.\] (If \(t>s\), we just have to redefine \(\Gamma_0 := \{\mathrm{i}\tau:t>\tau>s\}\). The proof will then proceed exactly the same as follows.) The integral defining \(J(u)\) splits up into parts depending on which of the above segments contain \(z_1\) and which contains \(z_2\). For example, if \(J_{++}(u)\) denotes the part where \(z_1\) and \(z_2\) are both in \(\Gamma_+\), then \[\begin{align} J_{++}(u) &= 2\int_{0< r_1< r_2<\infty}\int_{a_1, a_2\in [-1,1]}\exp\biggl( 8b^2\sum_{l=1}^2 \operatorname{Re}(g(\mathrm{i}(s - t) + r_l, a_l-x))\\ &\qquad -4b^2 \operatorname{Re}(g(r_2-r_1, a_2 - a_1)) \biggr) \,\mathrm{d}a_1 \,\mathrm{d}a_2\, \mathrm{d}r_2\, \mathrm{d}r_1. \end{align}\] From the formula 4 for \(g\), it follows that the integrand is bounded above by a constant times \[\begin{align} e^{-6\pi b^2r_1-2\pi b^2r_2}\prod_{l=1}^2 \prod_{\sigma\in \{-1,1\}}|1-e^{-\pi (\mathrm{i}(s - t) + r_l+ \mathrm{i}\sigma(a_l-x))}|^{-4b^2}. \end{align}\] Now, if \(r,\theta \in \mathbb{R}\), then \[\begin{align} \label{eq:rtheta} |1-re^{\mathrm{i}\theta}|^2 = 1-2r\cos \theta + r^2 \ge 1-2r+r^2=(1-r)^2. \end{align}\tag{31}\] This shows that the quantity from the previous display is bounded above by \[\begin{align} e^{-6\pi b^2r_1-2\pi b^2r_2}\prod_{l=1}^2(1-e^{-\pi r_l})^{-8b^2}. \end{align}\] Note that this bound has no dependence on \(u\) or \(v\). Moreover, since \(8b^2 < 1\), its integral over \(0<r_1<r_2<\infty\) is finite. Thus, \(J_{++}(u)\) is bounded above by a quantity that does not depend on \(u,v\).

Next, let \(J_{0+}(u)\) denote the part where \(z_1\in \Gamma_0\) and \(z_2\in \Gamma_+\). Then \[\begin{align} J_{0+}(u) &= 2\int_t^s\int_0^\infty\int_{a_1,a_2\in[-1,1]} \exp\biggl( 8b^2\operatorname{Re}(g(\mathrm{i}(\tau-t),a_1-x)) \notag\\ &\qquad\qquad\qquad +8b^2\operatorname{Re}(g(\mathrm{i}(s-t)+r,a_2-x))\\ &\qquad \qquad \qquad -4b^2\operatorname{Re}(g(r+\mathrm{i}(s-\tau),a_2-a_1)) \biggr) \,\mathrm{d}a_1\,\mathrm{d}a_2\,\mathrm{d}r\,\mathrm{d}\tau. \end{align}\] By equations 4 , 5 , and 31 , the integrand is bounded above by a constant times \[\begin{align} &e^{-2\pi b^2r}(1-e^{-\pi r})^{-8b^2}\prod_{\sigma\in\{-1,1\}}|1-e^{-\pi\mathrm{i}((\tau-t)+\sigma(a_1-x))}|^{-4b^2}. \end{align}\] The integral of the above with respect to \((r,\tau,a_1)\) factorizes into the product of an integral over \(r\) and an integral over \((\tau, a_1)\). The integral over \(r\) is finite and has no dependence on \((u,v)\). For the integral over \((\tau, a_1)\), a simple change of variable gives \[\begin{align} &\int_{t}^s \int_{-1}^1 \prod_{\sigma\in\{-1,1\}}|1-e^{-\pi\mathrm{i}((\tau-t)+\sigma(a_1-x))}|^{-4b^2} \, \mathrm{d}a_1 \, \mathrm{d}\tau\\ &= \int_0^{s-t} \int_{-1}^1 \prod_{\sigma\in\{-1,1\}}|1-e^{-\pi\mathrm{i}(\xi+\sigma c)}|^{-4b^2} \, \mathrm{d}c \, \mathrm{d}\xi\\ &\le \int_0^{1} \int_{-1}^1 \prod_{\sigma\in\{-1,1\}}|1-e^{-\pi\mathrm{i}(\xi+\sigma c)}|^{-4b^2} \, \mathrm{d}c \, \mathrm{d}\xi, \end{align}\] assuming that \(s-t\le 1\). Again, since \(8b^2 <1\), the above integral is finite; and it has no dependence on \((u,v)\). This shows that \(J_{0+}(u)\) is uniformly bounded in the region where \(0\le s-t\le 1\).

Next, let \(J_{00}(u)\) denote the part where \(z_1,z_2\in \Gamma_0\). Then \[\begin{align} J_{00}(u) &= 2\int_{t<\tau_1<\tau_2<s}\int_{a_1,a_2\in[-1,1]} \exp\biggl( 8b^2\sum_{l=1}^2 \operatorname{Re}(g(\mathrm{i}(\tau_l-t),a_l-x)) \notag\\ &\qquad\qquad\qquad -4b^2\operatorname{Re}(g(\mathrm{i}(\tau_2-\tau_1),a_2-a_1)) \biggr) \,\mathrm{d}a_1\,\mathrm{d}a_2\,\mathrm{d}\tau_2\,\mathrm{d}\tau_1. \end{align}\] By equation 5 , the interaction factor is bounded above by a constant, because each term \[|1-e^{-\pi\mathrm{i}((\tau_2-\tau_1)\pm(a_2-a_1))}|^{2b^2}\] is at most \(2^{2b^2}\). Therefore the integrand is bounded by a constant times \[\prod_{l=1}^2\prod_{\sigma\in\{-1,1\}} |1 - e^{-\pi \mathrm{i}((\tau_l-t)+\sigma(a_l-x))}|^{-4b^2}.\] Dropping the ordering constraint \(\tau_1<\tau_2\) and using the same change of variables as above for each pair \((\tau_l,a_l)\), we get the uniform boundedness of \(J_{00}(u)\).

The remaining pieces are estimated similarly. The term \(J_{--}(u)\) is the same as \(J_{++}(u)\) by symmetry, \(J_{-0}(u)\) is treated exactly like \(J_{0+}(u)\), and \(J_{-+}(u)\) is even easier because it has exponential decay in both tail variables. Therefore every piece in the decomposition of \(J(u)\) is uniformly bounded on a compact neighborhood of zero, and hence so is \(J(u)\) itself. The same argument applies to \(J(v)\). Consequently \(I(u,v)\) is bounded in a neighborhood of zero. Combined with the vanishing of the prefactor \(e^{-4b^2g(\mathrm{i}(t-s),x-y)}\), this proves 30 , and hence also 29 .

Now note that since \(A_\epsilon^*=A_\epsilon\), Theorem 7 gives \[(u_\epsilon,u_\epsilon) = \omega((A_\epsilon-\omega(A_\epsilon)1)^* (A_\epsilon-\omega(A_\epsilon)1)).\] Using \(\omega(A_\epsilon)=c_b\in\mathbb{R}\), this becomes \[(u_\epsilon,u_\epsilon) = \omega(A_\epsilon^*A_\epsilon)-|c_b|^2 .\] By the estimate just proved, \(\omega(A_\epsilon^*A_\epsilon)\to 0\). Therefore, for all sufficiently small \(\epsilon\), \[(u_\epsilon,u_\epsilon)<0.\] Taking \(u_-:=u_\epsilon\) for such an \(\epsilon\) completes the proof. ◻

Proof of Theorem 9. It is easy to see that \(\pi(C)\) is linear and continuous, as a map from \(\mathcal{H}_0\) into \(\mathbb{C}^\mathcal{A}\) (since we have not yet shown that it maps into \(\mathcal{H}\)). If \(A,B,C\in\mathcal{A}\), then \[(\pi(C)\Phi(A))_B =\Phi(A)_{C^*B} =\langle (C^*B)^*A \rangle =\langle B^*CA \rangle =\Phi(CA)_B.\] Thus, \[\pi(C)\Phi(A)=\Phi(CA)\in\mathcal{H}_0.\] In particular, this shows that \(\pi(C)(\mathcal{H}_0)\subseteq\mathcal{H}_0\). Continuity implies \[\pi(C)(\mathcal{H})=\pi(C)(\overline{\mathcal{H}_0})\subseteq\overline{\mathcal{H}_0}=\mathcal{H}.\] Thus \(\pi(C)\) is a well-defined continuous linear map on \(\mathcal{H}\), and item 1 follows. Next, for \(C,D\in\mathcal{A}\) and \(z\in\mathcal{H}\), we have \[(\pi(C+D)z)_A=z_{(C+D)^*A}=z_{C^*A}+z_{D^*A}=((\pi(C)+\pi(D))z)_A,\] and similarly \(\pi(\lambda C)=\lambda\pi(C)\) for every \(\lambda\in\mathbb{C}\). Also, \[(\pi(C)\pi(D)z)_A=(\pi(D)z)_{C^*A}=z_{D^*C^*A}=z_{(CD)^*A}=(\pi(CD)z)_A,\] while clearly \((\pi(1)z)_A=z_A\). Therefore \(C\mapsto\pi(C)\) is a unital algebra representation.

Now let \(u=\Phi(A)\in\mathcal{H}_0\) and \(z\in\mathcal{H}\). Then by item 1, \[(\pi(C)u,z) =(\Phi(CA),z) =z_{CA} =(\pi(C^*)z)_A =(u,\pi(C^*)z).\] This proves item 2. Next, set \(\Omega:=\Phi(1)\). Then for every \(C\in\mathcal{A}\), item 1 gives \[\pi(C)\Omega=\Phi(C).\] Hence \[\operatorname{span}\{\pi(C)\Omega:C\in\mathcal{A}\}=\Phi(\mathcal{A})=\mathcal{H}_0.\] Thus \(\Omega\) is cyclic, proving item 3. Finally, \[(\Omega,\pi(A)\Omega) =(\Phi(1),\Phi(A)) =\langle 1,A\rangle_{\mathcal{A}} =\omega(A),\] which proves item 4. ◻

Proof of Lemma 10. Uniqueness of the adjoint follows from Theorem 7: if \(S_1,S_2\) both satisfy \((Tu,z)=(u,S_i z)\) for all \(u\in \mathcal{H}_0,z\in \mathcal{H}\), then \((u,(S_1-S_2)z)=0\) for all \(u\in \mathcal{H}_0\), hence \((S_1-S_2)z=0\) for every \(z\in \mathcal{H}\).

It is immediate from the definition that \((T^\dagger)^\dagger=T\), and that the identity operator belongs to \(\operatorname{End}_{\dagger}(\mathcal{H})\). If \(T,R\in \operatorname{End}_{\dagger}(\mathcal{H})\), then \(TR\) preserves \(\mathcal{H}_0\), and for \(u\in \mathcal{H}_0,z\in \mathcal{H}\), \[(TRu,z)=(Ru,T^\dagger z)=(u,R^\dagger T^\dagger z).\] Similarly, \[(R^\dagger T^\dagger u,z)=(u,TRz).\] Thus \(TR\in\operatorname{End}_{\dagger}(\mathcal{H})\) and \[(TR)^\dagger=R^\dagger T^\dagger .\] Linearity is checked in the same way. Hence \(\operatorname{End}_{\dagger}(\mathcal{H})\) is a unital \(*\)-algebra, with \(*\) denoted by \(\dagger\).

The definition of \(\mathcal{T}_\dagger\) makes \(T\mapsto T^\dagger\) continuous. Composition with a fixed continuous linear map is continuous for \(\mathcal{T}_b\): on the right because continuous linear maps send bounded sets to bounded sets, and on the left because the pullback of a continuous seminorm by a continuous linear map is again a continuous seminorm. Applying this both to operators and to their adjoints proves separate continuity of multiplication for \(\mathcal{T}_\dagger\).

Finally, Theorem 9 gives \[(\pi(C)u,z)=(u,\pi(C^*)z)\] for all \(u\in \mathcal{H}_0,z\in \mathcal{H}\). Applying the same identity with \(C^*\) in place of \(C\) gives \[(\pi(C^*)u,z)=(u,\pi(C)z).\] Also \(\pi(C)\) and \(\pi(C^*)\) preserve \(\mathcal{H}_0\). Hence \(\pi(C)\in\operatorname{End}_{\dagger}(\mathcal{H})\) and \(\pi(C)^\dagger=\pi(C^*)\). ◻

Proof of Proposition 6. Cylinder translations preserve smooth compact support, and \[\eta_{a,\theta}\overline{f}=\overline{\eta_{a,\theta}f}, \qquad f\in\mathcal{F}.\] Since \(\eta_{a,\theta}\) was extended multiplicatively from the generators, each \(\eta_{a,\theta}\) is a \(*\)-automorphism of \(\mathcal{A}\), and the pullback definition immediately gives \[\eta_{a,\theta}\eta_{a',\theta'}=\eta_{a+a',\,\theta+\theta'}, \qquad \eta_{0,0}=\mathrm{Id}_{\mathcal{A}},\] with the second coordinate understood modulo \(2\).

To prove invariance of the point correlator, fix \(\boldsymbol{n}=(n_1,\ldots,n_k)\in\mathcal{I}_b^k\). If \(-\sum_{j=1}^k n_j<0\), then \(C_{b\boldsymbol{n}}^{\mathrm L}\equiv0\) and invariance is immediate. Otherwise, let \(u_j=(t_j,x_j)\) and write \(v_j=(\mathrm{i}t_j,x_j)\). Let \(\gamma\) be the contour used in the definition 7 of \(C_{b\boldsymbol{n}}^{\mathrm L}(u_1,\ldots,u_k)\), and let \[\widetilde{v}_j:=(\mathrm{i}(t_j+a),x_j+\theta), \qquad \widetilde{\gamma}:=\gamma+\mathrm{i}a.\] Then \(\widetilde{\gamma}\) is exactly the contour appearing in the definition of \(C_{b\boldsymbol{n}}^{\mathrm L}(T_{a,\theta}u_1,\ldots,T_{a,\theta}u_k)\). If \(\mathcal{T}_\gamma\) and \(\mathcal{T}_{\widetilde{\gamma}}\) denote the corresponding time-ordering operations, then shifting every screening variable by \((\mathrm{i}a,\theta)\) preserves the order along the contour and leaves all time-ordered differences unchanged: \[\begin{align} \mathcal{T}_{\widetilde{\gamma}}(\widetilde{v}_j-\widetilde{v}_{j'}) &=\mathcal{T}_\gamma(v_j-v_{j'}),\\ \mathcal{T}_{\widetilde{\gamma}}(\widetilde{v}_j-\widetilde{u}_l) &=\mathcal{T}_\gamma(v_j-u_l),\\ \mathcal{T}_{\widetilde{\gamma}}(\widetilde{u}_l-\widetilde{u}_{l'}) &=\mathcal{T}_\gamma(u_l-u_{l'}), \end{align}\] where \(u_l=(\gamma(s_l),b_l)\) and \(\widetilde{u}_l=(\gamma(s_l)+\mathrm{i}a,b_l+\theta)\). Since every \(g\)-factor in equation 7 depends only on these differences, the integrand is unchanged. Also, \(\widetilde{\gamma}'(s)=\gamma'(s)\), and the spatial integral is unchanged by translation modulo the \(2\)-periodicity in the second coordinate. Therefore the change of variables \(u_l\mapsto \widetilde{u}_l\) in equation 7 gives \[C_{b\boldsymbol{n}}^{\mathrm{L}}(T_{a,\theta}u_1,\ldots,T_{a,\theta}u_k) = C_{b\boldsymbol{n}}^{\mathrm{L}}(u_1,\ldots,u_k)\] for every admissible integer charge tuple \(\boldsymbol{n}\) and every configuration in the non-light-cone domain.

If \(C=\mathcal{O}_{f_1,n_1}\cdots\mathcal{O}_{f_k,n_k}\), then changing variables \(u_j\mapsto T_{a,\theta}u_j\) in the defining integral gives \[\omega(\eta_{a,\theta}C)=\omega(C).\] By linearity this holds for all \(C\in\mathcal{A}\). If \(O_1\subseteq O_2\), then every generator of \(\mathcal{A}(O_1)\) is also a generator of \(\mathcal{A}(O_2)\), so \(\mathcal{A}(O_1)\subseteq\mathcal{A}(O_2)\). Moreover, if \(\mathop{\mathrm{supp}}(f)\subseteq O\), then \(\mathop{\mathrm{supp}}(\eta_{a,\theta}f)\subseteq T_{a,\theta}O\), hence \(\eta_{a,\theta}\mathcal{A}(O)\subseteq\mathcal{A}(T_{a,\theta}O)\). Applying this inclusion to \((-a,-\theta)\) and then composing with \(\eta_{a,\theta}\) gives the reverse inclusion, so equality holds. Finally, \(\omega_O(1)=1\), and for every \(A\in\mathcal{A}(O)\), \[\omega_O(A^*)=\omega(A^*)=\overline{\omega(A)}=\overline{\omega_O(A)},\] where the middle identity follows from Theorem 5 with \(B=1\). Thus \(\omega_O\) is a normalized Hermitian functional on \(\mathcal{A}(O)\), and Proposition 6 follows. ◻

Proof of Theorem 13. A priori, it is easy to see that \(U_{\mathrm{cyl}}(a,\theta)\) is a linear and continuous map from \(\mathcal{H}\) into \(\mathbb{C}^\mathcal{A}\), although we do not know that it maps into \(\mathcal{H}\). For \(A,B\in\mathcal{A}\), \[\begin{align} (U_{\mathrm{cyl}}(a,\theta)\Phi(B))_A &=\Phi(B)_{\eta_{-a,-\theta}A} =\omega((\eta_{-a,-\theta}A)^*B) \\ &=\omega(\eta_{-a,-\theta}(A^*\eta_{a,\theta}B))) =\omega(A^*\eta_{a,\theta}B) =\Phi(\eta_{a,\theta}B)_A, \end{align}\] where we used Proposition 6. Hence \(U_{\mathrm{cyl}}(a,\theta)(\mathcal{H}_0)\subseteq\mathcal{H}_0\), and therefore \[U_{\mathrm{cyl}}(a,\theta)(\mathcal{H}) =U_{\mathrm{cyl}}(a,\theta)(\overline{\mathcal{H}_0}) \subseteq\overline{\mathcal{H}_0}=\mathcal{H}.\] This proves that \(U_{\mathrm{cyl}}(a,\theta)\) is a well-defined continuous linear map from \(\mathcal{H}\) into \(\mathcal{H}\), and the previous display proves item 1 of Theorem 13.

Next, for \(z\in\mathcal{H}\) and \(A\in\mathcal{A}\), \[\begin{align} (U_{\mathrm{cyl}}(a,\theta)U_{\mathrm{cyl}}(a',\theta')z)_A &=(U_{\mathrm{cyl}}(a',\theta')z)_{\eta_{-a,-\theta}A} =z_{\eta_{-a',-\theta'}\eta_{-a,-\theta}A} \\ &=z_{\eta_{-(a+a'),-(\theta+\theta')}A} =(U_{\mathrm{cyl}}(a+a',\theta+\theta')z)_A. \end{align}\] Hence \((U_{\mathrm{cyl}}(a,\theta))_{(a,\theta)}\) is an action of \(\mathbb{R}\times\mathbb{S}^1\) on \(\mathcal{H}\), and \[U_{\mathrm{cyl}}(a,\theta)^{-1}=U_{\mathrm{cyl}}(-a,-\theta).\] Since both maps are continuous, each \(U_{\mathrm{cyl}}(a,\theta)\) is a homeomorphism.

Now let \(u=\Phi(A)\in\mathcal{H}_0\) and \(z\in\mathcal{H}\). Then by item 1, \[(U_{\mathrm{cyl}}(a,\theta)u,z) =(\Phi(\eta_{a,\theta}A),z) =z_{\eta_{a,\theta}A} =(U_{\mathrm{cyl}}(-a,-\theta)z)_A =(u,U_{\mathrm{cyl}}(-a,-\theta)z).\] Applying this with \(z=U_{\mathrm{cyl}}(a,\theta)v\) and using the inverse relation gives \[(U_{\mathrm{cyl}}(a,\theta)u,U_{\mathrm{cyl}}(a,\theta)v)=(u,v), \qquad u,v\in\mathcal{H}_0.\] This proves item 2. Since \(\Omega=\Phi(1)\), item 1 gives \[U_{\mathrm{cyl}}(a,\theta)\Omega=\Phi(\eta_{a,\theta}1)=\Phi(1)=\Omega.\] Now let \(C\in\mathcal{A}\), \(z\in\mathcal{H}\), and \(A\in\mathcal{A}\). Then \[\begin{align} (U_{\mathrm{cyl}}(a,\theta)\pi(C)U_{\mathrm{cyl}}(a,\theta)^{-1}z)_A &=(\pi(C)U_{\mathrm{cyl}}(-a,-\theta)z)_{\eta_{-a,-\theta}A} \\ &=(U_{\mathrm{cyl}}(-a,-\theta)z)_{C^*\eta_{-a,-\theta}A} =z_{\eta_{a,\theta}(C^*\eta_{-a,-\theta}A)} \\ &=z_{\eta_{a,\theta}(C)^*A} =(\pi(\eta_{a,\theta}C)z)_A, \end{align}\] because \(\eta_{a,\theta}\) is a \(*\)-automorphism. This proves item 3.

To prove item 4, first note that by item 2, \(U_{\mathrm{cyl}}(a,\theta)\) is adjointable and \[U_{\mathrm{cyl}}(a,\theta)^\dagger=U_{\mathrm{cyl}}(a,\theta)^{-1} =U_{\mathrm{cyl}}(-a,-\theta).\] Thus conjugation by \(U_{\mathrm{cyl}}(a,\theta)\), \[\Gamma_{a,\theta}(T) := U_{\mathrm{cyl}}(a,\theta)T U_{\mathrm{cyl}}(a,\theta)^{-1},\] maps \(\operatorname{End}_{\dagger}(\mathcal{H})\) onto itself and satisfies \[\Gamma_{a,\theta}(T)^\dagger = \Gamma_{a,\theta}(T^\dagger).\] Since \(U_{\mathrm{cyl}}(a,\theta)\) and its inverse send bounded sets to bounded sets, \(\Gamma_{a,\theta}\) is a homeomorphism of \(\operatorname{End}_{\dagger}(\mathcal{H})\) for the graph-bounded topology \(\mathcal{T}_\dagger\).

If \(O_1\subseteq O_2\), then Proposition 6 gives \(\mathcal{A}(O_1)\subseteq\mathcal{A}(O_2)\), hence \(\pi(\mathcal{A}(O_1))\subseteq\pi(\mathcal{A}(O_2))\). Therefore the closed unital \(*\)-subalgebra generated by \(\pi(\mathcal{A}(O_1))\) is contained in the corresponding closed unital \(*\)-subalgebra generated by \(\pi(\mathcal{A}(O_2))\), that is, \[\mathfrak A_{\mathrm{loc}}(O_1)\subseteq\mathfrak A_{\mathrm{loc}}(O_2).\] Likewise, item 3 gives \[\Gamma_{U_{\mathrm{cyl}}(a,\theta)}\bigl(\pi(\mathcal{A}(O))\bigr) =\pi\bigl(\eta_{a,\theta}\mathcal{A}(O)\bigr) =\pi\bigl(\mathcal{A}(T_{a,\theta}O)\bigr),\] again by Proposition 6. Since \(\Gamma_{U_{\mathrm{cyl}}(a,\theta)}\) is a homeomorphism and an algebra automorphism of \(\mathrm{End}_\dagger(\mathcal{H})\), it carries the closed unital \(\dagger\)-subalgebra generated by \(\pi(\mathcal{A}(O))\) onto the closed unital \(\dagger\)-subalgebra generated by \(\pi(\mathcal{A}(T_{a,\theta}O))\). Hence \[U_{\mathrm{cyl}}(a,\theta)\,\mathfrak A_{\mathrm{loc}}(O) \,U_{\mathrm{cyl}}(a,\theta)^{-1} =\mathfrak A_{\mathrm{loc}}(T_{a,\theta}O).\] Finally, by definition \(U(t)=U_{\mathrm{cyl}}(t,0)\) and \(\tau_t=\eta_{-t,0}\), so item 1 gives \[U(t)\Phi(A)=\Phi(\eta_{t,0}A)=\Phi(\tau_{-t}A),\qquad A\in\mathcal{A},\] and the one-parameter-group statement in item 5 is just the \(\theta=0\) special case of the algebraic action already proved. No continuity in the group parameter is asserted here. This completes the proof of Theorem 13. ◻

Proof. Choose a lift of \(x\) with \(|t|<|x|\le 1\); the general case follows by the \(2\)-periodicity in \(x\). It is enough to consider \(0<x\le 1\), since the case \(x<0\) is obtained by replacing \((t,x)\) by \((-t,-x)\). Let \[L(y):=\log(1-e^{-\pi \mathrm{i}y}),\] with the principal branch of \(\log\). For \(y\notin2\mathbb{Z}\), let \(r(y)\in(0,2)\) be the representative of \(y\) modulo \(2\). Write \(y=r(y)+2m\) for some \(m\in\mathbb{Z}\). Since \(1-e^{\pi\mathrm{i}y}\) and \(1-e^{-\pi\mathrm{i}y}\) are both in the open right half-plane and \(\log a - \log b = \log (a/b)\) for \(a,b\) in the open right half-plane, \[\begin{align} L(-y)-L(y) &= \log\biggl(\frac{1-e^{\pi \mathrm{i}y}}{1-e^{-\pi \mathrm{i}y}}\biggr) \\ &= \log(-e^{\pi \mathrm{i}y}) = \log(-e^{\pi \mathrm{i}r(y)}) = \log(e^{\pi \mathrm{i}(r(y)-1)}). \end{align}\] Since \(r(y)-1\in(-1,1)\), the number \(e^{\pi \mathrm{i}(r(y)-1)}\) has principal argument \(\pi(r(y)-1)\in(-\pi,\pi)\), so the principal logarithm gives \[\label{eq:log-branch-difference} L(-y)-L(y)=\mathrm{i}\pi(r(y)-1).\tag{32}\] Since \(|t|<x\le 1\), we have \[t+x\in(0,2),\qquad t-x\in(-2,0).\] Thus \[r(t+x)=t+x,\qquad r(t-x)=t-x+2.\] Using the definition of \(g\), equation 32 , and the last two identities gives \[\begin{align} g(\mathrm{i}t,x)-g(-\mathrm{i}t,-x) &= -\pi \mathrm{i}t +\frac{\mathrm{i}\pi}{2}\{r(t+x)+r(t-x)-2\} \\ &= -\pi \mathrm{i}t+\frac{\mathrm{i}\pi}{2}(2t)=0. \end{align}\] This proves the claim. ◻

Proof. The screening number \(w\) remains unchanged after adding the extra charge. In the Euclidean formula, every free factor involving the inserted charge has exponent zero, and every screening-external factor involving the inserted charge also has exponent zero. Thus the Euclidean correlator with the inserted zero-charge point is identically the original Euclidean correlator: \[\begin{align} & C_{\alpha_1,\ldots,\alpha_r,0,\alpha_{r+1},\ldots,\alpha_k} (q_1,\ldots,q_r,q_*,q_{r+1},\ldots,q_k) \\ &\qquad = C_{\alpha_1,\ldots,\alpha_k}(q_1,\ldots,q_k), \qquad q_1,\ldots,q_k,q_*\in M. \end{align}\] Let us write \(u_j = (t_j,x_j)\) and \(u_* = (t_*,x_*)\). Then by the above observation, we see that on the enlarged ordered tube \(\Omega_{k+1}\), the function \[(\tau_1,\ldots,\tau_r,\tau_*,\tau_{r+1},\ldots,\tau_k) \longmapsto C_{\alpha_1,\ldots,\alpha_k}(\tau_1,x_1;\ldots;\tau_k,x_k)\] is analytic and has the same Euclidean boundary values as the analytic continuation of the enlarged zero-charge correlator \[\begin{align} &(\tau_1,\ldots,\tau_r,\tau_*,\tau_{r+1},\ldots,\tau_k)\\ &\longmapsto C_{\alpha_1,\ldots,\alpha_r,0,\alpha_{r+1},\ldots,\alpha_k}(\tau_1,x_1;\ldots;\tau_r,x_r; \tau_*,x_*; \tau_{r+1},x_{r+1};\ldots;\tau_k,x_k). \end{align}\] By uniqueness of analytic continuation to the ordered tube (Lemma 48), the two analytic functions coincide. Taking Lorentzian boundary values gives the claim. ◻

Proof. For any \(r_1,\ldots, r_n \in \mathbb{R}\) and \(y_1,\ldots, y_n\in [-1,1]\), let us insert phantom charges at \((r_1,y_1),\ldots,(r_n, y_n)\) in between \((t_j,x_j)\) and \((t_{j+1}, x_{j+1})\). Given \(r_1,\ldots,r_n\), we choose \(y_1,\ldots,y_n\) such that these new insertions preserve the no-light-cone condition. By Lemma 27, the insertion of these phantom charges does not alter the value of the correlation. Moreover, the only effect of inserting these charges in the formula for the correlation is that it alters the contour \(\gamma\), which now goes from \(\mathrm{i}t_j\) to \(\mathrm{i}r_1\), then to \(\mathrm{i}r_2,\ldots,\mathrm{i}r_n\), and finally to \(\mathrm{i}t_{j+1}\). This proves the claim. ◻

Proof. First, we substitute \[a_i^+=a_i^-+\delta_i^A,\quad b_i^+=b_i^-+\delta_i^B,\quad S_{ij}^+=S_{ij}^-+\Delta_{ij},\] and write \(D\) as a polynomial in \(a_i^-\), \(b_i^-\), \(S_{ij}^+\), \(\delta_i^A\), \(\delta_i^B\), and \(\Delta_{ij}\), for \(1\le i<j\le n\). To write this out carefully, we proceed as follows. Introduce, for \(c\in\{1,2,3\}\), \[r_c:=1_{\{c=3\}}, \quad \ell_c:=1_{\{c=1\}}.\] Then we have \[W_{c,i} = (a_i^-+r_c\delta_i^A) (b_i^-+(1-\ell_c)\delta_i^B),\quad S_{ij}^{cd} = S_{ij}^+ -(1-\ell_c)(1-r_d)\Delta_{ij}.\] With these substitutions, we have \[\begin{align} D_2 &:= \sum_{c_1,\ldots,c_n\in\{1,2,3\}} \prod_{i=1}^n \sigma_{c_i} \prod_{i=1}^n W_{c_i,i} \prod_{1\le i<j\le n} S_{ij}^{c_i c_j}\\ &= \sum_{c_1,\ldots,c_n\in\{1,2,3\}} \sum_{P,Q,E} Y_{P,Q,E} \prod_{i=1}^n \sigma_{c_i} \prod_{i\in P} r_{c_i} \delta_i^A \prod_{i\in Q} (1-\ell_{c_i}) \delta_i^B\\ &\hskip1in \cdot \prod_{(i,j)\in E}(1-\ell_{c_i})(1-r_{c_j}) \Delta_{ij}, \end{align}\] where \(Y_{P,Q,E}\) is a polynomial in \(a_i^-\), \(b_i^-\), and \(S_{ij}^+\), for \(1\le i<j\le n\); \(P,Q\) are subsets of \(\{1,\ldots,n\}\); and \(E\) is a subset of \(\{(i,j):1\le i<j\le n\}\).

Since \(Y_{P,Q,E}\) has no dependence on \(c_1,\ldots,c_n\), we can interchange the order of summation and write \[D_2 = \sum_{P,Q,E} Y_{P,Q,E}Z_{P,Q,E},\] where \[\begin{align} Z_{P,Q,E} &:= \sum_{c_1,\ldots,c_n\in\{1,2,3\}} \prod_{i=1}^n \sigma_{c_i} \prod_{i\in P} r_{c_i} \delta_i^A \prod_{i\in Q} (1-\ell_{c_i}) \delta_i^B\\ &\hskip1in \cdot \prod_{(i,j)\in E}(1-\ell_{c_i})(1-r_{c_j}) \Delta_{ij}\\ &= C_{P,Q,E} \prod_{i\in P} \delta_i^A \prod_{i\in Q} \delta_i^B\prod_{(i,j)\in E} \Delta_{ij}, \end{align}\] with \[C_{P,Q,E} := \sum_{c_1,\ldots,c_n\in\{1,2,3\}} \prod_{i=1}^n \sigma_{c_i} \prod_{i\in P} r_{c_i} \prod_{i\in Q} (1-\ell_{c_i}) \prod_{(i,j)\in E} (1-\ell_{c_i})(1-r_{c_j}).\] Suppose that \(Q\ne \varnothing\) and that the coefficient \(C_{P,Q,E}\) is nonzero. Then, we claim there is some \(i_0\in P\) and \(i_l\in Q\) such that either \(i_0=i_l\), or \(i_0<i_l\) and there is a directed path from \(i_0\) to \(i_l\) through the directed graph defined by \(E\).

To see this, first note that since \(r_c\) takes values in \(\{0,1\}\), we have \(r_c^m = r_c\) for any positive integer \(m\). The same is true for \(\ell_c\) and \(1-\ell_c\). Thus, \[\begin{align} \prod_{i\in P} r_{c_i} \prod_{i\in Q} (1-\ell_{c_i}) \prod_{(i,j)\in E}(1-\ell_{c_i})(1-r_{c_j}) &= \prod_{i=1}^n r_{c_i}^{p_i} (1-\ell_{c_i})^{L_i} (1-r_{c_i})^{R_i}, \end{align}\] where \[\begin{align} p_i &:= 1_{\{i\in P\}}, \\ L_i &:= 1_{\{i\in Q \text{ or } i \text{ has an outgoing edge in } E\}},\\ R_i &:= 1_{\{i \text{ has an incoming edge in } E\}}. \end{align}\] From this, we get \[\begin{align} C_{P,Q,E} &= \sum_{c_1,\ldots,c_n\in\{1,2,3\}} \prod_{i=1}^n \sigma_{c_i}r_{c_i}^{p_i} (1-\ell_{c_i})^{L_i} (1-r_{c_i})^{R_i}= \prod_{i=1}^n M(p_i,L_i,R_i), \end{align}\] where \[M(p,L,R) := \sum_{c=1}^3 \sigma_c r_{c}^{p} (1-\ell_{c})^{L} (1-r_{c})^{R}.\] A simple calculation yields the following table for \(M\): \[\begin{align} \label{eq:table1} \begin{array}{c|c|c|c} p & L & R & M(p,L,R)\\ \hline 0&0&0&1\\ 0&1&0&0\\ 0&0&1&0\\ 0&1&1&-1\\ 1&0&0&1\\ 1&1&0&1\\ 1&0&1&0\\ 1&1&1&0 . \end{array} \end{align}\tag{33}\] Now, if there is some \(i\) such that \(i\) is both in \(P\) and \(Q\), then the claim is already proved. So let us assume that \(P,Q\) are disjoint. Take any \(i\in Q\). Then \(L_i=1\) and (since \(i\notin P\)) \(p_i = 0\). If \(i\) has no incoming edge in \(E\), then \(R_i=0\). But then by table 33 , \(M(p_i,L_i,R_i) =0\), and this is impossible since \(C_{P,Q,E}\ne 0\). Thus, \(i\) must have an incoming edge in \(E\). Let \((i_1,i)\) be such an edge. If \(i_1\notin P\), then the same argument shows that \(i_1\) must have an incoming edge \((i_2,i_1)\) in \(E\). The process cannot continue indefinitely. This yields a directed path from \(P\) to \(Q\) in \(E\), proving the claim.

Thus, we have proved the claim that if \(Q\ne \varnothing\) and \(C_{P,Q,E}\ne 0\), then there is a path from \(P\) to \(Q\) in \(E\). Consequently, for any \(Q\ne \varnothing\), \(Y_{P,Q,E}Z_{P,Q,E} \in \mathcal{I}\).

Next, suppose that \(Q = \varnothing\) but \(E \ne \varnothing\), and \(C_{P,Q,E}\ne 0\). Take any \(i\in P\). Then \(p_i=1\). So, for \(C_{P,Q,E}\) to be nonzero, \(R_i\) must be zero (by table 33 ). In other words, \(i\) cannot have an incoming edge in \(E\). On the other hand, if \(i\notin P\), then \(p_i=0\), and therefore by table 33 , we must have \(L_i=R_i\). But since \(Q=\varnothing\) the condition \(L_i=R_i\) implies that \(i\) has both an incoming edge in \(E\) and an outgoing edge in \(E\). To summarize, we have shown that every \(i\) has either no incoming edge in \(E\), or both an incoming and an outgoing edge in \(E\). We claim that this is impossible. To see this, take the largest \(i\) that has at least one edge from \(E\) that is incident to it. This is well-defined, since \(E\) is nonempty. Since \(i\) is the largest such vertex, there can be no outgoing edge incident to it. So there must be an incoming edge but no outgoing edge, which contradicts our previous deduction. This proves that if \(Q=\varnothing\) and \(E\ne \varnothing\), then \(C_{P,Q,E}\) must be zero.

Thus, the only part of \(D_2\) that we have not yet shown to be in \(\mathcal{I}\) is \[D_2' := \sum_P Y_{P,\varnothing, \varnothing} Z_{P, \varnothing, \varnothing}.\] We will take care of this part later. Next, writing \(U_i = (a_i^-+\delta_i^A)b_i^-\), we get the expansion \[\begin{align} D_1 &:= \sum_{c_1,\ldots,c_n\in\{1,2,3\}}\prod_{i=1}^n \sigma_{c_i}\prod_{i=1}^n U_i\prod_{1\le i<j\le n} S_{ij}^{c_i c_j}\\ &= \sum_{c_1,\ldots,c_n\in\{1,2,3\}}\sum_{P,E}W_{P,E}\prod_{i=1}^n \sigma_{c_i}\prod_{i\in P}\delta_i^A \prod_{(i,j)\in E} (1-\ell_{c_i})(1-r_{c_j})\Delta_{ij}, \end{align}\] where \(W_{P,E}\) is a polynomial in \(a_i^-\), \(b_i^-\), and \(S_{ij}^+\), for \(1\le i<j\le n\); \(P\) is a subset of \(\{1,\ldots,n\}\); and \(E\) is a subset of \(\{(i,j):1\le i<j\le n\}\).

Since \(W_{P,E}\) has no dependence on \(c_1,\ldots,c_n\), we can interchange the order of summation and write \[D_1 = \sum_{P,E} W_{P,E}V_{P,E},\] where \[\begin{align} V_{P,E} &:= \sum_{c_1,\ldots,c_n\in\{1,2,3\}} \prod_{i=1}^n \sigma_{c_i} \prod_{i\in P} \delta_i^A \prod_{(i,j)\in E} (1-\ell_{c_i})(1-r_{c_j}) \Delta_{ij}\\ &= U_{P,E} \prod_{i\in P} \delta_i^A \prod_{(i,j)\in E} \Delta_{ij}, \end{align}\] with \[U_{P,E} := \sum_{c_1,\ldots,c_n\in\{1,2,3\}} \prod_{i=1}^n \sigma_{c_i} \prod_{(i,j)\in E} (1-\ell_{c_i})(1-r_{c_j}).\] Define \[\begin{align} L_i &:= 1_{\{i \text{ has an outgoing edge in } E\}},\quad R_i := 1_{\{i \text{ has an incoming edge in } E\}}. \end{align}\] Then \[\begin{align} U_{P,E} &= \sum_{c_1,\ldots,c_n\in\{1,2,3\}} \prod_{i=1}^n \sigma_{c_i}(1-\ell_{c_i})^{L_i} (1-r_{c_i})^{R_i}\\ &= \prod_{i=1}^n \biggl(\sum_{c=1}^3 \sigma_c (1-\ell_c)^{L_i}(1-r_c)^{R_i}\biggr) = \prod_{i=1}^n M(0, L_i, R_i). \end{align}\] Suppose that \(E\ne \varnothing\). Then for \(U_{P,E}\) to be nonzero, the above formula shows that each \(M(0,L_i,R_i)\) has to be nonzero. By table 33 , this means that each \(i\) has either no incoming or outgoing edges in \(E\), or has an incoming and an outgoing edge in \(E\). But this is impossible, as can be seen by considering the largest \(i\) which has an edge in \(E\) incident to it. Thus, \(U_{P,E} = 0\) whenever \(E\ne \varnothing\). We conclude that \[D_1 = \sum_P W_{P,\varnothing} V_{P,\varnothing}.\] Our final claim is that \(D_1=D_2'\). Indeed, when \(Q=\varnothing\) and \(E=\varnothing\), no \(\delta_i^B\)-factor and no \(\Delta_{ij}\)-factor has been selected. Therefore \[Y_{P,\varnothing,\varnothing} = \biggl(\prod_{i\notin P} a_i^-\biggr) \biggl(\prod_{i=1}^n b_i^-\biggr) \biggl(\prod_{1\le i<j\le n} S_{ij}^+\biggr).\] Also, \[\begin{align} Z_{P,\varnothing,\varnothing} &= \sum_{c_1,\ldots,c_n\in\{1,2,3\}} \prod_{i=1}^n\sigma_{c_i} \prod_{i\in P} r_{c_i}\delta_i^A \\ &= \biggl(\prod_{i\in P}\delta_i^A\biggr) \biggl(\prod_{i\notin P}\sum_{c=1}^3\sigma_c\biggr) \biggl(\prod_{i\in P}\sum_{c=1}^3\sigma_c r_c\biggr) = \prod_{i\in P}\delta_i^A, \end{align}\] since \(\sum_{c=1}^3\sigma_c=1\) and \(\sum_{c=1}^3\sigma_c r_c=1\). Similarly, in the expansion of \(D_1\), \[W_{P,\varnothing} = \biggl(\prod_{i\notin P} a_i^-\biggr) \biggl(\prod_{i=1}^n b_i^-\biggr) \biggl(\prod_{1\le i<j\le n} S_{ij}^+\biggr), \qquad V_{P,\varnothing} = \prod_{i\in P}\delta_i^A .\] Thus \[W_{P,\varnothing}V_{P,\varnothing} = Y_{P,\varnothing,\varnothing}Z_{P,\varnothing,\varnothing}\] for every \(P\), and hence \(D_1=D_2'\). Since the preceding argument showed that \(D_2-D_2'\in\mathcal{I}\), we conclude that \[D=D_1-D_2=(D_1-D_2')-(D_2-D_2')\in\mathcal{I} .\] This completes the proof. ◻

Proof. Write \(u = (t,x)\), \(v=(s,y)\), and \(w = (r,z)\). Then \(t<s<r\), and \[s - t > d_1(x-y), \quad r-s > d_1(y-z).\] We claim that for any \(a,b\), \(d_1(a+b)\le d_1(a)+d_1(b)\). To see this, find integers \(m,n\) that minimize \(|a+2m|\) and \(|b+2n|\). Then \[d_1(a+b) \le |a+b+2(m+n)|\le |a+2m|+|b+2n|=d_1(a)+d_1(b).\] Having proved the claim, we may now conclude that \[\begin{align} r - t > d_1(x-y)+d_1(y-z) \ge d_1(x-z). \end{align}\] This shows that \(u\prec w\). ◻

Proof. Move the insertions of the \(J\)-block leftward past the insertions of the \(I\)-block by finitely many adjacent transpositions. Each transposition is allowed by the spacelike separation hypothesis and Theorem 3. ◻

Proof of Theorem 15. We first prove the algebraic commutativity identity. Let \(A\in\mathcal{A}(O_1)\), \(B\in\mathcal{A}(O_2)\), and let \(X,Y\in\mathcal{A}\). By linearity in all four variables, it is enough to treat the case where \(A\), \(B\), \(X\), and \(Y\) are monomials. If either \(A\) or \(B\) is the unit, the identity is immediate, so assume that \(A\) and \(B\) have positive lengths. Write \[A=\mathcal{O}_{f_1,n_1}\cdots \mathcal{O}_{f_r,n_r},\qquad B=\mathcal{O}_{g_1,m_1}\cdots \mathcal{O}_{g_s,m_s},\] with \(n_i,m_j\in\mathcal{I}_b\), \(\mathop{\mathrm{supp}}(f_i)\subseteq O_1\), and \(\mathop{\mathrm{supp}}(g_j)\subseteq O_2\). Because \(X\) and \(Y\) are monomials in \(\mathcal{A}\), every point correlator that appears in the smeared integrals for \(\omega(XABY)\) and \(\omega(XBAY)\) still has integer charges. In those integrals, the insertions belonging to \(A\) and the insertions belonging to \(B\) occur as adjacent blocks, possibly with spectator insertions coming from \(X\) and \(Y\). The total screening number is the same for the two orders. If it is negative, both ordered point correlators vanish. If it is nonnegative, then the Lorentzian correlator formula applies. Because \(O_1\) and \(O_2\) are spacelike separated, every point in the \(A\)-block is spacelike separated from every point in the \(B\)-block. Corollary 31 therefore gives pointwise equality of the two Lorentzian point correlators corresponding to the orders \(XABY\) and \(XBAY\), away from the light-cone exceptional set. That exceptional set has Lebesgue measure zero, and Theorem 4 supplies absolute integrability of the relevant smeared correlators. Therefore \[\label{eq:algebraic-locality} \omega(XABY)=\omega(XBAY)\tag{35}\] for all \(X,Y\in\mathcal{A}\).

Now let \(E,F\in\mathcal{A}\). Since \(\pi(C)\Phi(E)=\Phi(CE)\) for every \(C\in\mathcal{A}\), we have \[\pi(A)\pi(B)\Phi(E)=\Phi(ABE), \qquad \pi(B)\pi(A)\Phi(E)=\Phi(BAE).\] For each coordinate \(F\in\mathcal{A}\), \[\begin{align} \bigl(\Phi(ABE)-\Phi(BAE)\bigr)_F &= \omega(F^*ABE)-\omega(F^*BAE). \end{align}\] Since \(A\in\mathcal{A}(O_1)\) and \(B\in\mathcal{A}(O_2)\), equation 35 , applied with \(X=F^*\) and \(Y=E\), shows that the right side above is zero. Thus \[\Phi(ABE)=\Phi(BAE),\] and therefore \[\pi(A)\pi(B)\Phi(E)=\pi(B)\pi(A)\Phi(E)\] for every \(E\in\mathcal{A}\). Since \(\mathcal{H}_0=\Phi(\mathcal{A})\) is dense in \(\mathcal{H}\), and \(\pi(A)\) and \(\pi(B)\) are continuous, we obtain \[[\pi(A),\pi(B)]=0\] on all of \(\mathcal{H}\). By linearity, the same conclusion holds for all \(A\in\mathcal{A}(O_1)\) and \(B\in\mathcal{A}(O_2)\). Hence every element of the unital \(*\)-subalgebra generated by \(\pi(\mathcal{A}(O_1))\) commutes with every element of the corresponding unital \(*\)-subalgebra generated by \(\pi(\mathcal{A}(O_2))\).

It remains to pass to the closed represented local algebras. Let \[S\in\mathfrak A_{\mathrm{loc}}(O_1), \qquad T\in\mathfrak A_{\mathrm{loc}}(O_2).\] Choose nets \(S_i\) and \(T_j\) in the unital \(\dagger\)-subalgebras \(\pi(\mathcal{A}(O_1))\) and \(\pi(\mathcal{A}(O_2))\), respectively, such that \[S_i\to S,\qquad T_j\to T\] in the graph-bounded topology \(\mathcal{T}_\dagger\). In particular, these convergences hold in the underlying topology \(\mathcal{T}_b\) of uniform convergence on bounded subsets of \(\mathcal{H}\). By the algebraic part just proved, we have \[S_iT_j=T_jS_i \qquad\text{for all }i,j.\] Composition with a fixed continuous linear map is continuous in this topology. Indeed, if \(R_\nu\to R\), then \(R_\nu U\to RU\) because a continuous linear map \(U\) sends bounded sets to bounded sets; and \(UR_\nu\to UR\) because for every continuous seminorm \(p\) on \(\mathcal{H}\), the seminorm \(p\circ U\) is again continuous. Therefore, first letting \(i\) tend to the limit gives \[ST_j=T_jS\] for every \(j\), and then letting \(j\) tend to the limit gives \[ST=TS.\] This proves the theorem. ◻

Proof. Let \(u:= t-s\) and \(v:= x-y\). Then note that \[\begin{align} &e_{m,n}^{\mathrm{c}}(t,x) e_{m,n}^{\mathrm{c}}(s,y)+ e_{m,n}^{\mathrm{s}}(t,x) e_{m,n}^{\mathrm{s}}(s,y)\\ &= \frac{1}{2T}\biggl\{\cos\biggl(\pi\frac{m}{T}t+\pi nx\biggr)\cos\biggl(\pi\frac{m}{T}s+\pi ny\biggr) \\ &\qquad \qquad + \sin\biggl(\pi\frac{m}{T}t+\pi nx\biggr)\sin\biggl(\pi\frac{m}{T}s+\pi ny\biggr)\biggr\}\\ &= \frac{1}{2T}\cos\biggl(\pi \frac{m}{T}u + \pi nv\biggr)\\ &= \frac{1}{4T}\biggl\{\exp\biggl(\pi \mathrm{i}\frac{m}{T}u + \pi\mathrm{i}n v\biggr)+ \exp\biggl(-\pi \mathrm{i}\frac{m}{T}u - \pi\mathrm{i}nv\biggr)\biggr\}. \end{align}\] By the definitions of \(G_{T,N}\) and \(\mathbb{Z}^2_*\), this shows that \[\begin{align} &G_{T,N}((t,x),(s,y)) \\ &= \frac{\pi}{2T} \sum_{\substack{|m|\le N, \;|n|\le N\\ (m,n)\in\mathbb{Z}^2_*\setminus\{(0,0)\}}}\frac{1}{\lambda_{m,n}} \Biggl\{\exp\biggl(\pi \mathrm{i}\frac{m}{T}u + \pi\mathrm{i}nv\biggr)+ \exp\biggl(-\pi \mathrm{i}\frac{m}{T}u - \pi\mathrm{i}nv\biggr)\Biggr\} \\ &= \frac{\pi}{2T} \sum_{\substack{|m|\le N, \;|n|\le N\\ (m,n)\ne (0,0)}}\frac{1}{\lambda_{m,n}}\exp\biggl(\pi \mathrm{i}\frac{m}{T}u + \pi\mathrm{i}nv\biggr) = g_{T,N}(u,v). \end{align}\] This completes the proof. ◻

Proof. Let \(f(x) := (\pi/a)e^{-2\pi a|x|}\). The Fourier transform of \(f\) is \[\begin{align} \widehat{f}(\xi) &= \int_{\mathbb{R}} f(x) e^{-2\pi \mathrm{i}x\xi} \mathrm{d}x\\ &= \frac{\pi}{a}\int_0^\infty e^{-(2\pi a+2\pi \mathrm{i}\xi)x}\mathrm{d}x + \frac{\pi}{a}\int_0^\infty e^{-(2\pi a-2\pi \mathrm{i}\xi) x} \mathrm{d}x\\ &= \frac{\pi}{a(2\pi a+2\pi \mathrm{i}\xi)} + \frac{\pi}{a(2\pi a-2\pi \mathrm{i}\xi)}\\ &= \frac{1}{a^2+\xi^2}. \end{align}\] Then by the Poisson summation formula [45], \[\begin{align} \sum_{k\in \mathbb{Z}} \frac{e^{\mathrm{i}k\theta}}{k^2+a^2} &= \sum_{k\in \mathbb{Z}} e^{\mathrm{i}k \theta} \widehat{f}(k)\notag \\ &= \sum_{k\in \mathbb{Z}} f \biggl(k + \frac{\theta}{2\pi}\biggr) = \frac{\pi}{a}\sum_{k\in \mathbb{Z}} e^{-a |2\pi k + \theta|}. \end{align}\] This proves the first claim. Next, suppose that \(\theta \in [-\pi,\pi]\) and \(a\ge 1\). Then \(2\pi \pm \theta \ge |\theta|\), and hence, \[\begin{align} \sum_{k\in \mathbb{Z}} e^{-a |2\pi k + \theta|} &= e^{-a|\theta|} + \sum_{k=1}^\infty e^{-a (2\pi k + \theta)} + \sum_{k=1}^\infty e^{-a(2\pi k - \theta)}\\ &= e^{-a|\theta|} + \frac{e^{-a(2\pi + \theta)}}{1-e^{-2\pi a}} + \frac{e^{-a(2\pi - \theta)}}{1-e^{-2\pi a}}\le Ce^{-a|\theta|}, \end{align}\] where \(C\) is universal. This completes the proof. ◻

Proof. Let \(A_n:= a_1+\cdots +a_n\) and \(A_0:=0\). Then for any \(M\le N\), \[\begin{align} \sum_{n=M}^N a_nb_n &= \sum_{n=M}^N (A_n - A_{n-1})b_n\\ &= \sum_{n=M}^N A_n b_n - \sum_{n=M-1}^{N-1} A_n b_{n+1}\\ &= A_N b_N -A_{M-1} b_M+ \sum_{n=M}^{N-1}A_n(b_n - b_{n+1}). \end{align}\] From the given conditions, this gives \[\begin{align} \biggl|\sum_{n=M}^N a_nb_n \biggr| &\le Cb_N + C b_M + C\sum_{n=M}^{N-1}|b_n - b_{n+1}|\\ &= Cb_N + Cb_M + C(b_M - b_N) = 2Cb_M. \end{align}\] It is clear that this suffices to prove both assertions. ◻

Proof. Since \(g_{T,N}(t,x)\) is periodic with period \(2T\) in the first coordinate and period \(2\) in the second coordinate, we may assume without loss that \((t,x)\in M_T\setminus\{(0,0)\}\). Define \(\alpha := \pi t/T\) and \(\beta := \pi x\). First, suppose that \(\alpha\neq 0\). Define, for \(n\neq 0\), \[B_n:=\sum_{m\in\mathbb{Z}}\frac{e^{\mathrm{i}\alpha m}}{\lambda_{m,n}}, \qquad B_{n,N}:=\sum_{|m|\le N}\frac{e^{\mathrm{i}\alpha m}}{\lambda_{m,n}},\] and for \(n=0\) define \[B_0^\ast:=\sum_{m\in\mathbb{Z}\setminus\{0\}}\frac{e^{\mathrm{i}\alpha m}}{\lambda_{m,0}}, \qquad B_{0,N}^\ast:=\sum_{0<|m|\le N}\frac{e^{\mathrm{i}\alpha m}}{\lambda_{m,0}}.\] (Note that all of the above series are absolutely convergent.) By definition, \[\begin{align} \label{eq:gnexp} g_{T,N}(t,x)&=\frac{\pi}{2T}\biggl(B_{0,N}^\ast+\sum_{0<|n|\le N}e^{\mathrm{i}\beta n}B_{n,N}\biggr). \end{align}\tag{37}\] We claim that \[\label{eq:blimit} \lim_{N\to \infty}\sum_{0<|n|\le N}|B_n-B_{n,N}| = 0, \qquad \lim_{N\to \infty}|B_0^\ast-B_{0,N}^\ast|=0.\tag{38}\] To prove this, first note that \[\begin{align} B_n - B_{n,N}&=\sum_{|m|> N}\frac{e^{\mathrm{i}\alpha m}}{\lambda_{m,n}}, \end{align}\] that \(\lambda_{m,n}\) is positive, \(1/\lambda_{m,n}\) decreases to zero as \(m\to\infty\), and that for any \(0\le M_1\le M_2\), \[\sum_{m=M_1}^{M_2}e^{\pm\mathrm{i}\alpha m} = \frac{e^{\pm\mathrm{i}\alpha M_1} - e^{\pm\mathrm{i}\alpha (M_2+1)}}{1-e^{\pm \mathrm{i}\alpha}},\] whose absolute value is clearly bounded by a constant \(C_1(\alpha)\) that depends only on \(\alpha\). Thus, by Lemma 34, \[\begin{align} |B_n - B_{n,N}| &\le \frac{2C_1(\alpha)}{\lambda_{N,n}} = \frac{C_2(\alpha)T^2}{N^2+n^2T^2}. \end{align}\] This gives \[\begin{align} \sum_{0<|n|\le N}|B_n-B_{n,N}| &\le C_2(\alpha)T^2\sum_{0<|n|\le N} \frac{1}{N^2+n^2T^2}\le \frac{C_3(\alpha)T^2}{N}. \end{align}\] Also, since \(\lambda_{m,0}\) grows quadratically in \(m\), \(B_0^* - B_{0,N}^*\to 0\) as \(N\to\infty\). This completes the proof of the claim 38 . So, by equation 37 , it only remains to show that the limit \[\begin{align} \label{eq:bnlim} \lim_{N\to\infty}\sum_{0<|n|\le N}e^{\mathrm{i}\beta n}B_{n} \end{align}\tag{39}\] exists and is finite. Since \(\alpha \in [-\pi,\pi]\) and \(\alpha \ne 0\), Lemma 33 implies that for \(|n|\) so large that \(|n|T\ge1\), \[0\le B_n\le \frac{CT}{|n|}e^{-|\alpha| |n| T},\] for some universal constant \(C\). Clearly, this proves that the limit displayed in equation 39 exists and is finite. This completes the proof when \(\alpha \ne 0\). The same argument works when \(\beta \ne 0\), upon swapping \(m,n\) everywhere. ◻

Proof. Fix \(t>0\) and set \[\begin{align} p_t^{(1)}(x)&:=\frac{1}{2T}\sum_{m\in\mathbb{Z}}\exp\biggl( -\frac{t\pi^2 m^2}{T^2} + \pi\mathrm{i}\frac{m}{T}x\biggr),\\ p_t^{(2)}(y)&:=\frac{1}{2}\sum_{n\in\mathbb{Z}}\exp(-t\pi^2 n^2+\pi\mathrm{i}ny). \end{align}\] Then \(p_t(x,y)=p_t^{(1)}(x)\,p_t^{(2)}(y)\). We will prove a formula for \(p_t^{(1)}(x)\), and the corresponding formula for \(p_t^{(2)}(y)\) is identical with \(T\) replaced by \(1\). To that end, consider on \(\mathbb{R}\) the function \(f(x):=\exp(-t\pi^2 x^2/T^2)\), whose Fourier transform is \[\widehat f(\xi)=\int_{\mathbb{R}} f(x)e^{-2\pi \mathrm{i}x\xi}\,\mathrm{d}x = \frac{T}{\sqrt{\pi t}}\exp\biggl(-\frac{T^2\xi^2}{t}\biggr).\] Applying the Poisson summation formula [45] to the function \(g(x):=f(x)e^{2\pi\mathrm{i}x\xi}\), whose Fourier transform is \(\widehat g(\eta)=\widehat f(\eta-\xi)\), we get \[\sum_{m\in\mathbb{Z}}f(m)e^{2\pi\mathrm{i}m\xi}=\sum_{k\in\mathbb{Z}}\widehat f(k-\xi).\] Taking \(\xi=x/2T\) and dividing both sides by \(2T\) gives \[p_t^{(1)}(x) = \frac{1}{2T}\sum_{m\in\mathbb{Z}}\exp\biggl( -\frac{t\pi^2 m^2}{T^2}+\pi\mathrm{i}\frac{m}{T}x\biggr) =\frac{1}{2\sqrt{\pi t}}\sum_{k\in\mathbb{Z}}\exp\biggl(-\frac{(x-2Tk)^2}{4t}\biggr),\] which completes the proof. ◻

Proof. Clearly, it suffices to work with \((t,x)\in M_T\). Define \[\begin{align} \label{eq:alphabetadef} \alpha:=\pi\frac{t}{T},\quad \beta:=\pi x. \end{align}\tag{41}\] By pairing \((m,n)\) with \((-m,-n)\), we get \[\label{eq:cos-form} g_{T,N}(t,x)=\frac{\pi}{2T}\sum_{\substack{|m|\le N,\;|n|\le N\\(m,n)\neq(0,0)}} \frac{1}{\lambda_{m,n}}\cos(\alpha m+\beta n).\tag{42}\] For \(\epsilon>0\), define the Abel-regularized kernel \[g^\epsilon(t,x):=\frac{\pi}{2T}\sum_{(m,n)\in\mathbb{Z}^2\setminus\{(0,0)\}} \frac{e^{-\epsilon\lambda_{m,n}}}{\lambda_{m,n}}\, e^{\mathrm{i}(\alpha m+\beta n)}.\] This series converges absolutely for each \(\epsilon>0\). Let \(p_s\) be the heat kernel defined in Lemma 36. By the identity \(\lambda^{-1}e^{-\epsilon \lambda}=\int_\epsilon^\infty e^{-s\lambda}\,ds\) and Fubini (justified by absolute convergence when \(\epsilon>0\)), we have \[\begin{align} \label{eq:gepsilonform} g^\epsilon(t,x) &=\frac{\pi}{2T}\sum_{(m,n)\neq(0,0)}\frac{e^{-\epsilon\lambda_{m,n}}}{\lambda_{m,n}}\, e^{\mathrm{i}(\alpha m+\beta n)} \notag \\ &=\frac{\pi}{2T}\sum_{(m,n)\neq(0,0)}e^{\mathrm{i}(\alpha m+\beta n)} \int_\epsilon^\infty e^{-s\lambda_{m,n}}\,\mathrm{d}s \notag \\ &=\int_\epsilon^\infty \biggl(\frac{\pi}{2T}\sum_{(m,n)\neq(0,0)} e^{-s\lambda_{m,n}}e^{\mathrm{i}(\alpha m+\beta n)}\biggr)\, \mathrm{d}s \notag \\ &=2\pi \int_\epsilon^\infty\biggl(p_s(t,x)-\frac{1}{4T}\biggr)\,\mathrm{d}s. \end{align}\tag{43}\] Now recall that by Lemma 36, \(p_s\) is a nonnegative function. Thus, \[\int_\epsilon^1\biggl(p_s-\frac{1}{4T}\biggr)\,\mathrm{d}s\ge -\frac{1-\epsilon}{4T}\ge -\frac{1}{4T}.\] Next, let \(\lambda_\star:=\min\{\pi^2/T^2,\pi^2\}\). Then note that for \(s\ge1\), \[\begin{align} \label{eq:pttail} \biggl|p_s(t,x)-\frac{1}{4T}\biggr| &=\frac{1}{4T}\biggl|\sum_{(m,n)\neq(0,0)}e^{-s\lambda_{m,n}}e^{\mathrm{i}(\alpha m+\beta n)}\biggr|\notag\\ &\le \frac{1}{4T}\sum_{(m,n)\neq(0,0)}e^{-s\lambda_{m,n}} \le C_1 e^{-(s-1)\lambda_\star}, \end{align}\tag{44}\] where \[C_1:=\frac{1}{4T}\sum_{(m,n)\neq(0,0)}e^{-\lambda_{m,n}}<\infty.\] Therefore, \[\int_1^\infty\biggl(p_s-\frac{1}{4T}\biggr)\,\mathrm{d}s\ge -\int_1^\infty C_1 e^{-(s-1)\lambda_\star}\,\mathrm{d}s = -\frac{C_1}{\lambda_\star}.\] Combining with the above estimates yields the uniform lower bound \[\label{eq:abel-lower} g^\epsilon(t,x)\ge -2\pi \biggl(\frac{1}{4T}+\frac{C_1}{\lambda_\star}\biggr)=:-C_{\mathrm{Abel}} \quad\text{for all }\epsilon>0,\;(t,x)\in M_T.\tag{45}\] The next step is to compare \(g_{T,N}\) with \(g^{\epsilon_N}\), where \(\epsilon_N:=(N+1)^{-2}\). Note that \[g_{T,N}(t,x)-g^{\epsilon_N}(t,x) =\frac{\pi}{2T}\sum_{(m,n)\neq(0,0)} \frac{1_{\{|m|\le N,|n|\le N\}}-e^{-\epsilon_N\lambda_{m,n}}}{\lambda_{m,n}} e^{\mathrm{i}(\alpha m+\beta n)},\] where \(1_{\{|m|\le N,|n|\le N\}}\) is \(1\) if \(|m|\le N\) and \(|n|\le N\), and \(0\) otherwise. Thus, \[|g_{T,N}(t,x)-g^{\epsilon_N}(t,x)| \le \frac{\pi}{2T}\sum_{(m,n)\neq(0,0)}\frac{\big|1_{\{|m|\le N,|n|\le N\}}-e^{-\epsilon_N\lambda_{m,n}}\big|}{\lambda_{m,n}} =:R_N.\] Split \(R_N=R_N^{\mathrm{in}}+R_N^{\mathrm{out}}\) into contributions from the square \(\{|m|\le N,\, |n|\le N\}\) and its complement. Take any \((m,n)\ne 0\) inside the square. Using \(|1-e^{-x}|\le x\), we get \[\frac{|1-e^{-\epsilon_N\lambda_{m,n}}|}{\lambda_{m,n}}\le \epsilon_N.\] There are at most \((2N+1)^2\) such terms. Hence, \[R_N^{\mathrm{in}}\le \frac{\pi(2N+1)^2}{2T}\epsilon_N\le \frac{2\pi}{T}.\] Next, take \((m,n)\) outside the square. Then \(m^2+n^2\ge (N+1)^2\). Moreover, \[\lambda_{m,n}\ge \kappa(m^2+n^2),\qquad \kappa:=\frac{\pi^2}{\max\{T^2,1\}}.\] This gives \[R_N^{\mathrm{out}} \le \frac{\pi}{2T\kappa}\sum_{\text{outside}}\frac{e^{-\epsilon_N\kappa(m^2+n^2)}}{m^2+n^2}.\] Comparing the lattice sum to a polar-coordinate integral, we get \[\sum_{\text{outside}}\frac{e^{-\epsilon_N\kappa(m^2+n^2)}}{m^2+n^2} \lesssim \int_{r\ge N+1} \frac{e^{-\epsilon_N\kappa r^2}}{r^2}r\,\mathrm{d}r =\int_{s\ge 1} e^{-\kappa s^2}\, \frac{\mathrm{d}s}{s}<\infty.\] This gives \(R_N^{\mathrm{out}}\lesssim 1\). Since we have already established that \(R_N^{\mathrm{in}}\lesssim 1\), we get \[\begin{align} \label{eq:rnbound} R_N \le C_{\mathrm{compare}} \end{align}\tag{46}\] for some positive constant \(C_{\mathrm{compare}}\) that depends only on \(T\). Combining this with equation 45 , we obtain \[g_{T,N}(t,x)\ge g^{\epsilon_N}(t,x)-C_{\mathrm{compare}}\ge -C_{\mathrm{Abel}}-C_{\mathrm{compare}}=:-C,\] which proves the claim. ◻

Proof. In this proof, \(C_0, C_1,\ldots\) will denote constants depending only on \(T\), whose values may change from line to line. First, note that since \(g_{T,N}(t,x)\) and \(r\) are invariant under adding an integer multiple of \(2T\) to \(t\) and an even integer to \(x\), we may assume without loss that \(|t|\le T\) and \(|x|\le 1\). As in the proof of Lemma 37, write \[g_{T,N}(t,x)=g^{\epsilon_N}(t,x)+(g_{T,N}(t,x)-g^{\epsilon_N}(t,x)), \qquad \epsilon_N:=(N+1)^{-2}.\] By the inequality 46 , the comparison term \(|g_{T,N}-g^{\epsilon_N}|\) is uniformly bounded in \(N\). Thus, it suffices to bound \(|g^{\epsilon_N}(t,x)|\) from above. Using the representation 43 for \(g^\epsilon(t,x)\) and the alternative expression for \(p_s(t,x)\) from Lemma 36, we have \[g^\epsilon(t,x)=2\pi \int_{\epsilon}^\infty \biggl\{\frac{1}{4\pi s}\sum_{k,l\in\mathbb{Z}} \exp\biggl(-\frac{(t-2Tk)^2+(x-2l)^2}{4s}\biggr) - \frac{1}{4T}\biggr\} \, \mathrm{d}s.\] First, we get an upper bound on the absolute value of the integral from \(\epsilon\) to \(1\). The \((k,l)=(0,0)\) term yields \[\begin{align} \int_\epsilon^1 \frac{1}{4\pi s}\exp\biggl(-\frac{r^2}{4s}\biggr)\, \mathrm{d}s &=\int_{\epsilon/r^2}^{1/r^2}\frac{1}{4\pi q} e^{-1/(4q)}\, \mathrm{d}q\\ &\le \int_{0}^{1/r^2}\frac{1}{4\pi q} e^{-1/(4q)}\, \mathrm{d}q\\ &\le \int_0^1 \frac{1}{4\pi q} e^{-1/(4q)} \, \mathrm{d}q + \int_1^{\max\{1,1/r^2\}} \frac{1}{4\pi q} \, \mathrm{d}q\\ &\le C+ \frac{|\log r|}{2\pi}. \end{align}\] Next, note that since \(t\in [-T,T]\) and \(x\in [-1,1]\), we have \[(t-2Tk)^2+(x-2l)^2\gtrsim k^2+l^2\] when \((k,l)\ne(0,0)\). Therefore, for \(s\in(0,1]\), \[\begin{align} \sum_{(k,l)\ne(0,0)}\exp\biggl(-\frac{(t-2Tk)^2+(x-2l)^2}{4s}\biggr) &\le \sum_{(k,l)\ne(0,0)} e^{-C_0(k^2+l^2)/s}\\ &\lesssim \int_1^\infty e^{-C_1\rho^2/s}\rho \,\mathrm{d}\rho = \frac{s}{2C_1} e^{-C_1/s} \lesssim s. \end{align}\] This gives a uniform bound on the contribution of the nonzero lattice terms to the integral over \(s\in(0,1]\). Lastly, by the inequality 44 , \(\int_1^\infty|p_s(t,x)-1/(4T)|\, \mathrm{d}s\lesssim 1\). This completes the proof. ◻

Proof. Take any \(f\in C^\infty(M_T)\). Define \[G_{T,N} f(u):=\int_{M_T} G_{T,N}(u,v)f(v)\,\mathrm{d}v.\] It is easy to see that \(G_{T,N}f\) is well-defined and smooth. By the bound from Lemma 38 and dominated convergence, \[\lim_{N\to \infty} G_{T,N} f(u)=\int_{M_T} G_T(u,v)f(v)\,\mathrm{d}v =: G_T f(u).\] This shows, in particular, that \(G_T\) is well-defined as an integral operator on smooth functions. (We have not yet shown that \(G_T f\) is smooth; this will be shown later in the proof.) Now, let \(h:=\Delta f\). We will first show that \[\begin{align} \label{eq:gtshow1} -\frac{1}{2\pi}G_T h = f-c(f). \end{align}\tag{47}\] To prove this, take \(u=(t,x)\in M_T\). First note that \[\begin{align} G_{T,N} h(u) &= \frac{\pi}{2T}\sum_{\substack{|m|\le N,\;|n|\le N\\ (m,n)\neq(0,0)}} \frac{1}{\lambda_{m,n}}\int_{M_T}\exp\biggl(\pi \mathrm{i} \frac{m}{T}(t-s)+\pi \mathrm{i} n (x-y)\biggr) h (s,y)\, \mathrm{d}s \, \mathrm{d}y. \end{align}\] Integration by parts gives \[\begin{align} &\int_{M_T}\exp\biggl(\pi \mathrm{i} \frac{m}{T}(t-s)+\pi \mathrm{i} n (x-y)\biggr) h (s,y)\, \mathrm{d}s \, \mathrm{d}y\\ &= \int_{M_T}\Delta\biggl\{\exp\biggl(\pi \mathrm{i} \frac{m}{T}(t-s)+\pi \mathrm{i} n (x-y)\biggr)\biggr\}f (s,y)\, \mathrm{d}s \, \mathrm{d}y\\ &= -\lambda_{m,n} \int_{M_T}\exp\biggl(\pi \mathrm{i} \frac{m}{T}(t-s)+\pi \mathrm{i} n (x-y)\biggr)f (s,y)\, \mathrm{d}s \, \mathrm{d}y. \end{align}\] Combining the last two displays, we get \[\begin{align} \label{eq:gtexpansion} G_{T,N} h(u) &= -\frac{\pi}{2T}\sum_{\substack{|m|\le N,\;|n|\le N\\ (m,n)\neq(0,0)}} \int_{M_T}\exp\biggl(\pi \mathrm{i} \frac{m}{T}(t-s)+\pi \mathrm{i} n (x-y)\biggr) f (s,y)\, \mathrm{d}s \, \mathrm{d}y. \end{align}\tag{48}\] Let \(\mathbb{T}^2\) denote the unit torus \([0,1]^2\), and define \(f_0:\mathbb{T}^2\to \mathbb{R}\) as \[\begin{align} f_0(t,x) := f(2Tt - T, 2x-1). \end{align}\] For \(m,n\in \mathbb{Z}\), \[\widehat{f}_0(m,n) := \int_{\mathbb{T}^2} f_0(t,x) e^{-2\pi \mathrm{i}(mt + nx)} \, \mathrm{d}t\, \mathrm{d}x\] denote the Fourier coefficients of \(f_0\). Since \(f_0\) is smooth, standard results [45] imply that \(\widehat{f}_0(m,n)\) is rapidly decaying in \(|m|+|n|\). From this, it follows [45] that for almost every \((t,x)\in \mathbb{T}^2\), \[f_0(t,x) = \sum_{m,n\in \mathbb{Z}} \widehat{f}_0(m,n) e^{2\pi \mathrm{i}(mt+nx)}\] But the rapid decay of the Fourier coefficients implies that the right side above is a continuous function of \((t,x)\), and the left side is continuous anyway. Thus, the identity must hold for all \((t,x)\).

Translating this identity to the torus \(M_T\), we get, for \((t,x)\in M_T\), \[\begin{align} \label{eq:ff0} f(t,x) &= f_0\biggl(\frac{t+T}{2T}, \frac{x+1}{2}\biggr)\notag \\ &= \sum_{m,n\in \mathbb{Z}} \widehat{f}_0(m,n) \exp\biggl\{2\pi \mathrm{i}\biggl(m \frac{t+T}{2T} +n \frac{x+1}{2}\biggr)\biggr\}\notag \\ &= \sum_{m,n\in \mathbb{Z}} \widehat{f}_0(m,n)e^{\pi \mathrm{i}(m+n)} \exp\biggl(\pi \mathrm{i}m \frac{t}{T} +\pi \mathrm{i}n x\biggr). \end{align}\tag{49}\] This, together with another application of the rapid decay of the Fourier coefficients, gives \[\begin{align} &\int_{M_T}\exp\biggl(\pi \mathrm{i} \frac{m}{T}(t-s)+\pi \mathrm{i} n (x-y)\biggr) f (s,y)\, \mathrm{d}s \, \mathrm{d}y\\ &= \sum_{m',n'\in \mathbb{Z}}\widehat{f}_0(m',n')e^{\pi\mathrm{i}(m'+n')}\exp\biggl(\pi \mathrm{i} \frac{m'}{T}t+\pi \mathrm{i} n' x\biggr)\\ &\qquad \qquad \cdot \int_{M_T}\exp\biggl(\pi \mathrm{i}\frac{m-m'}{T} s + \pi \mathrm{i}(n-n')y\biggr) \, \mathrm{d}s\, \mathrm{d}y\\ &= 4T\widehat{f}_0(m,n)e^{\pi\mathrm{i}(m+n)}\exp\biggl(\pi \mathrm{i} \frac{m}{T}t+\pi \mathrm{i} n x\biggr). \end{align}\] Substituting this in equation 48 , we get \[\begin{align} G_{T,N}h(u) &=-2\pi\sum_{\substack{|m|\le N,\;|n|\le N\\ (m,n)\neq(0,0)}}\widehat{f}_0(m,n)e^{\pi\mathrm{i}(m+n)}\exp\biggl(\pi \mathrm{i} \frac{m}{T}t+\pi \mathrm{i} n x\biggr). \end{align}\] Again, the rapid decay of the Fourier coefficients allows us to take \(N\to\infty\) on the right. Using equation 49 , this gives \[\begin{align} G_{T} h(u) &=-2\pi\sum_{(m,n)\neq(0,0)}\widehat{f}_0(m,n)e^{\pi\mathrm{i}(m+n)}\exp\biggl(\pi \mathrm{i} \frac{m}{T}t+\pi \mathrm{i} n x\biggr)\\ &= -2\pi f(u) + 2\pi \widehat{f}_0(0,0) = -2\pi(f(u)-c(f)). \end{align}\] This proves the claim 47 .

Next, let \(g := G_T f\). We claim that \(g\) is smooth, and \[\begin{align} \label{eq:gtshow2} -\frac{1}{2\pi}\Delta g = f - c(f). \end{align}\tag{50}\] To see that \(g\) is smooth, note that we can write \[g(u) = \int_{M_T} f(u-v) g_T(v) \, \mathrm{d}v.\] By the bound from Lemma 38 and dominated convergence, we can differentiate the right side as many times as we want, by moving the derivatives inside the integral. This shows that \(g\) is smooth. Next, take any \(\varphi\in C^\infty(M_T)\). Again using Lemma 38 to apply Fubini, and using the observation that \(g_T(-u)=g_T(u)\) (which follows from the same property for \(g_{T,N}\)), we get \[\begin{align} \int_{M_T}\varphi(u)\Delta g(u)\,\mathrm{d}u &=\int_{M_T}g(u)\Delta \varphi(u)\,\mathrm{d}u \\ &= \int_{M_T}\int_{M_T} f(v) g_T(u-v) \Delta \varphi(u) \,\mathrm{d}v \, \mathrm{d}u \\ &= \int_{M_T}\int_{M_T} f(v) g_T(v-u) \Delta \varphi(u) \,\mathrm{d}u \, \mathrm{d}v \\ &= \int_{M_T} f(v) G_T \Delta \varphi(v) \, \mathrm{d}v. \end{align}\] But by equation 47 , \(G_T\Delta \varphi(v) = -2\pi(\varphi(v)-c(\varphi))\). Thus, \[\int_{M_T}\varphi(u)\Delta g(u)\,\mathrm{d}u = -2\pi \int_{M_T} f(v)(\varphi(v)-c(\varphi)) \, \mathrm{d}v = -2\pi \int_{M_T} \varphi(v)(f(v)-c(f)) \, \mathrm{d}v.\] Rearranging this, we get \[\int_{M_T}\varphi(u) \biggl(\frac{1}{2\pi}\Delta g(u)+f(u)-c(f) \biggr)\, \mathrm{d}u =0.\] Since this holds for every smooth \(\varphi\), we can take \[\varphi(u) = \frac{1}{2\pi}\Delta g(u)+f(u)-c(f).\] This choice gives \[\int_{M_T}\biggl(\frac{1}{2\pi}\Delta g(u)+f(u)-c(f) \biggr)^2\, \mathrm{d}u =0.\] Since \(f\) and \(\Delta g\) are smooth, this proves the claim 50 . ◻

Proof. Let \(g(\theta)\) denote the left side. Pairing \(m\) with \(-m\), we get \[\begin{align} \label{eq:ktuform} g(\theta)=\frac{2}{\pi}\sum_{m=1}^\infty\frac{\cos(m\theta)}{m^2}. \end{align}\tag{51}\] Let \(f(x):= x^2\) for \(x\in [-\pi,\pi]\). The Fourier coefficients of \(f\) are given by \(\widehat{f}(0)=\pi^2/3\) and for \(m\ne 0\), \[\begin{align} \widehat{f}(m) &= \frac{1}{2\pi}\int_{-\pi}^\pi f(x) e^{\mathrm{i}m x} \mathrm{d}x\\ &= \frac{1}{2\pi}\biggl[\frac{x^2e^{\mathrm{i}mx}}{\mathrm{i}m}\biggr]^\pi_{-\pi} - \frac{1}{\mathrm{i}\pi m}\int_{-\pi}^\pi x e^{\mathrm{i}m x} \mathrm{d}x\\ &= -\frac{1}{\mathrm{i}\pi m}\biggl[\frac{xe^{\mathrm{i}m x}}{\mathrm{i}m}\biggr]^\pi_{-\pi} + \frac{1}{\mathrm{i}^2 \pi m^2}\int_{-\pi}^\pi e^{\mathrm{i}m x} \mathrm{d}x\\ &= \frac{2\cos (m \pi)}{m^2} = \frac{2(-1)^m}{m^2}. \end{align}\] Since \(f\in L^1\) and \(\sum_{m\in \mathbb{Z}} |\widehat{f}(m)|<\infty\), the Fourier inversion formula [46] shows that for almost all \(x\in [-\pi,\pi]\), \[\begin{align} \label{eq:xexpan} x^2 = f(x) &= \sum_{m\in \mathbb{Z}} \widehat{f}(m)e^{\mathrm{i}m x} = \frac{\pi^2}{3} + 4\sum_{m=1}^\infty \frac{(-1)^m \cos (mx)}{m^2}. \end{align}\tag{52}\] The right side is absolutely convergent and is a continuous function of \(x\). This shows that the above identity holds for all \(x\in [-\pi,\pi]\).

Replacing \(x\) by \(\theta-\pi\) in equation 52 , and using \(\cos(m(\theta-\pi))=(-1)^m\cos(m\theta)\), we get \[\sum_{m=1}^\infty\frac{\cos(m\theta)}{m^2} =\frac{\pi^2}{6}-\frac{\pi\theta}{2}+\frac{\theta^2}{4}, \qquad \theta\in[0,2\pi].\] By evenness of the left side, this gives \[\sum_{m=1}^\infty\frac{\cos(m\theta)}{m^2} =\frac{\pi^2}{6}-\frac{\pi|\theta|}{2}+\frac{\theta^2}{4}, \qquad \theta\in[-2\pi,2\pi].\] Substituting this into equation 51 completes the proof. ◻

Proof. In this proof, \(C,C_1,\ldots\) will denote positive universal constants, whose values may change from line to line. Since \(g_T\) is \(2\)-periodic in the second argument, it suffices to prove the claim for \((t,x)\in M\setminus\{(0,0)\}\). Let us fix such a \((t,x)\). As in the proof of Lemma 35, let \(\alpha:=\pi t/T\) and \(\beta:=\pi x\), and define \[B_0^*:=\sum_{m\in\mathbb{Z}\setminus\{0\}}\frac{e^{\mathrm{i}\alpha m}}{\lambda_{m,0}}, \qquad B_n:=\sum_{m\in\mathbb{Z}}\frac{e^{\mathrm{i}\alpha m}}{\lambda_{m,n}},\quad n\neq 0.\] The proof of Lemma 35 shows that \[\label{eq:gtlimit-g-fourier} g_T(t,x)=\frac{\pi}{2T}\biggl(B_0^*+\sum_{n\in\mathbb{Z}\setminus\{0\}}e^{\mathrm{i}\beta n}B_n\biggr),\tag{53}\] and also that \[\begin{align} \label{eq:bnbound} 0\le B_n\le \frac{CT}{|n|}e^{-|n||\alpha|T}=\frac{CT}{|n|}e^{-\pi|n||t|}, \qquad n\neq 0. \end{align}\tag{54}\] Since \(t\ne 0\), this shows that the series in the preceding display is absolutely convergent. We first compute \(B_0^*\). Since \(\lambda_{m,0}=\pi^2m^2/T^2\), we have \[\label{eq:gtlimit-b0-pre} \frac{\pi}{2T}B_0^* = \frac{T}{2\pi}\sum_{m\in\mathbb{Z}\setminus\{0\}}\frac{e^{\mathrm{i}m\pi t/T}}{m^2}.\tag{55}\] Whenever \(|t|\le T\), we have \(\pi t/T\in[-\pi,\pi]\), and therefore Lemma 40 gives \[\label{eq:gtlimit-b0-eval} \frac{\pi}{2T}B_0^* = \frac{\pi T}{6}-\frac{\pi |t|}{2}+\frac{\pi t^2}{4T}.\tag{56}\] Next, for \(n\neq 0\), note that \[\frac{\pi}{2T}B_n=\frac{T}{2\pi}\sum_{m\in\mathbb{Z}}\frac{e^{\mathrm{i}\alpha m}}{m^2+n^2T^2}.\] Applying Lemma 33 with \(a=|n|T\) and \(\theta=\alpha=\pi t/T\), we get \[\label{eq:gtlimit-bn-eval} \frac{\pi}{2T}B_n = \frac{1}{2|n|}\sum_{k\in\mathbb{Z}}e^{-\pi |n||t+2kT|}.\tag{57}\] Since \(t\ne 0\), the bound 54 allows us to apply the dominated convergence theorem and get \[\begin{align} \lim_{T\to\infty}\frac{\pi}{2T}\sum_{n\in\mathbb{Z}\setminus\{0\}}e^{\mathrm{i}\beta n}B_n &= \sum_{n\in\mathbb{Z}\setminus\{0\}}e^{\mathrm{i}\beta n}\biggl(\lim_{T\to\infty}\frac{\pi}{2T}B_n\biggr)\notag \\ &= \frac{1}{2}\sum_{n\in\mathbb{Z}\setminus\{0\}}\frac{e^{\pi \mathrm{i}nx}}{|n|}e^{-\pi |n||t|}, \end{align}\] and the series on the right is absolutely convergent. By the power series expansion of logarithm at \(1\), \[\begin{align} \frac{1}{2}\sum_{n\in\mathbb{Z}\setminus\{0\}}\frac{e^{\pi \mathrm{i}nx}}{|n|}e^{-\pi |n||t|} &= \frac{1}{2}\sum_{n=1}^\infty\frac{e^{\pi \mathrm{i}nx}}{n}e^{-\pi n|t|} + \frac{1}{2}\sum_{n=1}^\infty\frac{e^{-\pi \mathrm{i}nx}}{n}e^{-\pi n|t|}\\ &= -\frac{1}{2}\log(1-e^{\pi \mathrm{i}x - \pi |t|}) -\frac{1}{2}\log(1-e^{-\pi \mathrm{i}x - \pi |t|})\\ &= -\log|1-e^{-\pi |t|+\pi \mathrm{i}x}|. \end{align}\] Combining the steps, this completes the proof. ◻

Proof. Applying Lemma 33 with \(\theta=0\) and \(a=nT\), we get \[\sum_{m\in\mathbb{Z}}\frac{1}{m^2+n^2T^2} = \frac{\pi}{nT}\sum_{k\in\mathbb{Z}}e^{-2\pi nT|k|}.\] Therefore, \[\begin{align} a_n(T) &= \frac{T}{\pi}\cdot \frac{\pi}{nT}\sum_{k\in\mathbb{Z}} e^{-2\pi nT|k|}\\ &= \frac{1}{n} + \frac{2}{n}\sum_{k=1}^\infty e^{-2\pi Tnk} = \frac{1}{n}+ \frac{2e^{-2\pi Tn}}{n(1-e^{-2\pi Tn})}. \end{align}\] Rearranging the last term completes the proof. ◻

Proof. Since \(g_T(0,\cdot)\) and \(x\mapsto -\log|1-e^{\pi\mathrm{i}x}|\) are both \(2\)-periodic, it suffices to prove the claim for \(x\in[-1,1)\setminus\{0\}\). For fixed \(T\), write \[g_{T,N}(0,x)=A_{T,N}+B_{T,N}(x),\] where \[A_{T,N} = \frac{T}{2\pi}\sum_{0<|m|\le N}\frac{1}{m^2}, \qquad B_{T,N}(x) = \frac{T}{2\pi} \sum_{\substack{|m|\le N\\0< |n|\le N}} \frac{e^{\pi \mathrm{i}n x}}{m^2+n^2T^2}.\] Since the denominator is even in \(n\), pairing \(n\) and \(-n\) gives \[B_{T,N}(x) = \frac{T}{\pi} \sum_{n=1}^N \cos(\pi n x) \biggl(\sum_{|m|\le N}\frac{1}{m^2+n^2T^2}\biggr).\] Clearly, \[\lim_{N\to \infty} A_{T,N}= \frac{T}{\pi}\sum_{m=1}^\infty \frac{1}{m^2} = \frac{\pi T}{6}.\] Thus it remains to understand \(B_{T,N}(x)\). For \(k\ge 1\), set \[C_k(x):=\sum_{n=1}^k \cos(\pi n x).\] Note that for any \(k\), \[\begin{align} C_k(x) &= \frac{1}{2}\sum_{n=1}^k e^{\pi\mathrm{i}n x} + \frac{1}{2}\sum_{n=1}^k e^{-\pi\mathrm{i}n x}\\ &= \frac{e^{\pi\mathrm{i}x} - e^{\pi\mathrm{i}(k+1)x}}{2(1-e^{\pi\mathrm{i}x})} + \frac{e^{-\pi\mathrm{i}x} - e^{-\pi\mathrm{i}(k+1)x}}{2(1-e^{-\pi\mathrm{i}x})}. \end{align}\] Since \(x\in [-1,1)\setminus\{0\}\), we have \(e^{\pm \pi\mathrm{i}x}\ne 1\). Therefore the above identity shows that \[M(x) := \sup_{k\ge 1} |C_k(x)|<\infty.\] Let \(a_n(T)\) be as in Lemma 42. Since \(a_n(T)\) is positive and decreases to zero as \(n\to \infty\), and \(|C_k(x)|\le M(x)\) for all \(k\), Lemma 34 shows that the following function is well-defined: \[F_T(x)=\sum_{n=1}^\infty \cos(\pi n x)\,a_n(T) := \lim_{N\to\infty}\sum_{n=1}^N \cos(\pi n x)\,a_n(T).\] Now set \[\begin{align} &b_{n,N}(T) := \frac{T}{\pi}\sum_{|m|\le N}\frac{1}{m^2+n^2T^2},\\ &d_{n,N}(T):=a_n(T)-b_{n,N}(T) = \frac{T}{\pi}\sum_{|m|>N}\frac{1}{m^2+n^2T^2}. \end{align}\] Then \(d_{n,N}(T)\ge 0\), and \(d_{n,N}(T)\) decreases in \(n\). Moreover, \[0\le d_{1,N}(T) = \frac{T}{\pi}\sum_{|m|>N}\frac{1}{m^2+T^2} \le \frac{T}{\pi}\sum_{|m|>N}\frac{1}{m^2} \xlongrightarrow{N\to \infty} 0.\] Since \[B_{T,N}(x)=\sum_{n=1}^N \cos(\pi n x)\,b_{n,N}(T),\] we have \[F_T(x)-B_{T,N}(x) = \sum_{n=1}^N \cos(\pi n x)\,d_{n,N}(T) + \sum_{n=N+1}^\infty \cos(\pi n x)\,a_n(T).\] Applying Lemma 34 to each sum yields \[|F_T(x)-B_{T,N}(x)| \le 2M(x)\,d_{1,N}(T)+2M(x)\,a_{N+1}(T).\] Now \(d_{1,N}(T)\to 0\) as \(N\to\infty\), and Lemma 42 shows that \(a_{N+1}(T)\to 0\) as \(N\to\infty\). Hence, \[\lim_{N\to\infty} B_{T,N}(x)= F_T(x).\] We conclude that \[g_T(0,x) = \lim_{N\to\infty} g_{T,N}(0,x)=\frac{\pi T}{6}+F_T(x).\] Now define \[\begin{align} \label{eq:sxdef} S(x):=\sum_{n=1}^\infty \frac{\cos(\pi n x)}{n}, \end{align}\tag{58}\] where the series on the right is conditionally convergent by Lemma 34 and the uniform boundedness of \(|C_k(x)|\). For fixed \(T\), set \[c_n(T) := a_n(T) - \frac{1}{n} = \frac{2}{n(e^{2\pi nT}-1)}.\] Then \(c_n(T)\ge 0\) and decreases to zero as \(n\to\infty\). Therefore, again by Lemma 34, \[\biggl| \biggl(\lim_{N\to\infty} g_{T,N}(0,x)-\frac{\pi T}{6}\biggr)-S(x) \biggr| = \biggl|\sum_{n=1}^\infty \cos(\pi n x)\,c_n(T)\biggr| \le 2M(x)\,c_1(T).\] But \[c_1(T)=\frac{2}{e^{2\pi T}-1}\xlongrightarrow{T\to \infty} 0.\] Hence, \[\begin{align} \label{eq:prevlim} \lim_{T\to\infty} \biggl(g_{T}(0,x)-\frac{\pi T}{6} \biggr) = S(x). \end{align}\tag{59}\] It remains to evaluate \(S(x)\). For \(r\in (0,1)\), define \[S_r(x) := \sum_{n=1}^\infty \frac{r^n\cos(\pi n x)}{n},\] which is an absolutely convergent series. Note that \[\begin{align} S_r(x) &= \operatorname{Re}\biggl(\sum_{n=1}^\infty \frac{r^n e^{\pi \mathrm{i}nx}}{n}\biggr)\\ &= \operatorname{Re}(-\log(1-r e^{\pi \mathrm{i}x})) = -\log |1-re^{\pi \mathrm{i}x}|. \end{align}\] Now, define the partial sum \[\begin{align} U_n(x) := \sum_{j=1}^n \frac{\cos (\pi j x)}{j}. \end{align}\] By Lemma 34, \(U_n(x)\to S(x)\) as \(n\to\infty\). In particular, the sequence is bounded by some constant \(C\) in absolute value. Thus, summation by parts gives \[\begin{align} S_r(x) &= \sum_{n=1}^\infty r^n(U_n(x)-U_{n-1}(x)) \\ &= \sum_{n=1}^\infty U_n(x)(r^n- r^{n+1}) = (1-r)\sum_{n=1}^\infty U_n(x) r^n. \end{align}\] Take any \(\epsilon>0\) and find \(N\) so large that \(|U_n(x)-S(x)|\le \epsilon\) for all \(n\ge N\). Then \[\begin{align} |S_r(x)-S(x)| &\le (1-r)\sum_{n=1}^N 2C r^n + (1-r)\sum_{n=N+1}^\infty \epsilon r^n\\ &\le 2CN(1-r) + \epsilon. \end{align}\] This shows that \[\limsup_{r\to 1}|S_r(x)-S(x)| \le \epsilon.\] Since \(\epsilon\) is arbitrary, this shows that \(S_r(x)\to S(x)\) as \(r\to 1\). Since \(x\in[-1,1)\setminus\{0\}\), we have \(e^{\pi\mathrm{i}x}\neq 1\). Hence \[\begin{align} \label{eq:sxform} S(x)=\lim_{r\uparrow 1}S_r(x)= -\lim_{r\uparrow 1}\log |1-re^{\pi \mathrm{i}x}| = -\log|1-e^{\pi \mathrm{i}x}|. \end{align}\tag{60}\] Combining with equation 59 gives the claim. ◻

Proof. To see that \(f_T\) is concave on \([0,\infty)\), simply observe that its second derivative is nonpositive. Nonnegativity on \([0,T]\) is simple to check (in fact, it is nonnegative on \([0,2T]\)), and \(f_T\) is increasing on \([0,T]\) because \(f_T'\) is nonnegative on \([0,T]\). The bound on the Lipschitz constant follows from the maximum value of \(|f_T'|\) in \([0,T]\). The upper and lower bounds on \(f_T(r)\) are straightforward from the formula for \(f_T\). Lastly, for the subadditive inequality, take any \(a,b\in [0,T]\). Let \(c:= \min\{a+b,T\}-a\). Note that \(c\le b\). Thus, if we can show that \[\begin{align} \label{eq:fttoshow} f_T(a+c)\le f_T(a)+f_T(c), \end{align}\tag{62}\] then by the increasing nature of \(f_T\) on \([0,T]\), it will follow that \(f_T(a+c)\le f_T(a)+f_T(b)\), which is the required inequality. So, let us prove the inequality 62 . If \(a+c=0\), then the inequality is trivial since \(f_T(0)=0\). So, let us assume that \(a+c>0\). Let \[\lambda := \frac{a}{a+c},\] so that \[a = \lambda (a+c), \quad c = (1-\lambda)(a+c).\] By the concavity of \(f_T\), we get \[\begin{align} &f_T(a)\ge \lambda f_T(a+c) + (1-\lambda)f_T(0) = \frac{a}{a+c}f_T(a+c),\\ &f_T(c)\ge (1-\lambda)f_T(a+c)+\lambda f_T(0) = \frac{c}{a+c} f_T(a+c). \end{align}\] Adding up the above inequalities gives 62 . ◻

Proof. Throughout this proof, \(c_1,c_2,\ldots\) will denote positive universal constants whose values may change from line to line. Using \[|1-e^{-u+\mathrm{i}v}|^2=(1-e^{-u})^2+2e^{-u}(1-\cos v),\] consider two cases. If \(u^2+v^2\le 1\), then \(u\le 1\) and \(|v|\le 1\). Hence \[1-e^{-u}\ge c_1u,\qquad 1-\cos v\ge c_2v^2,\qquad e^{-u}\ge e^{-1},\] which implies \[|1-e^{-u+\mathrm{i}v}|^2\ge c_3(u^2+v^2).\] If \(u^2+v^2>1\), then either \(u\ge \frac{1}{2}\) or \(u<\frac{1}{2}\). If \(u\ge \frac{1}{2}\), then \[|1-e^{-u+\mathrm{i}v}|\ge 1-e^{-u}\ge 1-e^{-1/2}.\] If \(u<\frac{1}{2}\), then \(|v|\ge \sqrt{1-u^2}>\frac{\sqrt3}{2}\) and \(|v|\le \pi\), and so, \[|1-e^{-u+\mathrm{i}v}|^2\ge 2e^{-u}(1-\cos v) \ge 2e^{-1/2}(1-\cos(\sqrt3/2)).\] Thus in either subcase, \(|1-e^{-u+\mathrm{i}v}|\ge c_4\). This proves the lower bound. For the upper bound, first use \(|e^{-u}|\le1\) to get \[|1-e^{-u+\mathrm{i}v}|\le 1+e^{-u}\le2.\] Also, \[|1-e^{-u+\mathrm{i}v}| \le |1-e^{-u}|+e^{-u}|1-e^{\mathrm{i}v}| \le u+|v|,\] where we used \(1-e^{-u}\le u\) and \(|1-e^{\mathrm{i}v}|\le |v|\). Since \(u+|v|\le\sqrt2\sqrt{u^2+v^2}\), the upper bound follows after increasing the constant \(C\). ◻

Proof. Throughout this proof, \(C,C_1,\ldots\) will denote positive universal constants, whose values may change from line to line. Since both \(h_T\) and \(r_T\) are \(2T\)-periodic in the first coordinate and \(2\)-periodic in the second coordinate, we may assume without loss that \((t,x)\in M_T\). Write \[h_T(t,x)=h_T^{(0)}(t,x)+R_T(t,x),\] where \[h_T^{(0)}(t,x):=\sum_{n=1}^\infty \frac{\cos(\pi n x)}{n}e^{-\pi n|t|} =-\log|1-e^{-\pi |t|+\pi \mathrm{i}x}|,\] and \[R_T(t,x):=\sum_{n=1}^\infty \frac{\cos(\pi n x)}{n} \biggl(\sum_{m\in\mathbb{Z}\setminus\{0\}}e^{-\pi n|t+2mT|}\biggr).\] Since \(|t|\le T\), we have \[|t+2mT|\ge (2|m|-1)T,\qquad m\neq 0.\] Since \(T\ge 1\), this implies that \[\begin{align} \label{eq:rtbound} |R_T(t,x)| \le \sum_{n=1}^\infty \frac{1}{n}\biggl(\sum_{m\neq 0}e^{-\pi n|t+2mT|}\biggr) \le 2\sum_{n=1}^\infty \frac{1}{n}\biggl(\sum_{m=1}^\infty e^{-\pi n(2m-1)T}\biggr) \le C_1. \end{align}\tag{64}\] Since \(|1-z|\le 2\) whenever \(|z|\le 1\), we also have \[h_T^{(0)}(t,x)\ge -\log 2.\] This gives the lower bound \(h_T(t,x)\ge -C_1\). Next, let \(a:=\pi|t|\) and \(b:=\pi x\). Since \((t,x)\in M_T\), we have \(d_T(t)=|t|\) and \(d_1(x)=|x|\), and so, \[r_T(t,x)=\sqrt{t^2+x^2}=\frac{1}{\pi}\sqrt{a^2+b^2}.\] Now note that \[|1-e^{-a+\mathrm{i}b}|\ge C^{-1}\min\{1,\sqrt{a^2+b^2}\},\] by Lemma 45 (applied with \(u=a\) and \(v=b\in[-\pi,\pi]\)). Therefore, for \((t,x)\neq (0,0)\), \[\begin{align} h_T^{(0)}(t,x) &=-\log|1-e^{-a+\mathrm{i}b}|\\ &\le \log C -\log(\min\{1,\sqrt{a^2+b^2}\})\\ &\le C_1 + |\log\sqrt{a^2+b^2}|\\ &\le C_2 + |\log r_T(t,x)|. \end{align}\] At \((t,x)=(0,0)\), the right side is \(+\infty\), so the same inequality is immediate. Combining this with the bound 64 on \(R_T\) completes the proof. ◻

Proof. If \(w=0\), then all sums and products in the definitions are empty, so \(\Phi_T\equiv 1\) and the claim is immediate after taking any fixed \(\delta\in(0,2)\) and increasing \(C\) if needed. So assume \(w\ge 1\).

Fix \(T\) large enough so that all the points \(q_1,\dots,q_k\) belong to \(M_T\). Fix \(u_1,\ldots,u_w\in M_T\), with \(u_l=(t_l,x_l)\). Throughout this proof, \(C,C_1,\ldots\) will denote constants that do not depend on \(T\) or \(u_1,\ldots,u_w\), whose values may change from line to line. We will also use \(O(1)\) to denote any quantity whose absolute value is bounded above by such a constant. Write \[\widetilde{g}_T(t,x):=g_T(t,x)-\frac{\pi}{6}T,\] so that \(\widetilde{G}_T((t,x),(s,y))=\widetilde{g}_T(t-s,x-y)\). Take \((t,x)\in M_T\setminus\{(0,0)\}\). Since \(|t|\le T\), we may combine equations 53 , 56 , and 57 to get \[\begin{align} \widetilde{g}_T(t,x) &= -\frac{\pi|t|}{2}+\frac{\pi t^2}{4T}+\frac{1}{2}\sum_{n\in \mathbb{Z}\setminus\{0\}} \frac{e^{\pi \mathrm{i}n x}}{|n|}\biggl(\sum_{m\in \mathbb{Z}}e^{-\pi|n||t+2mT|}\biggr). \end{align}\] After pairing \(n\) and \(-n\), this gives \[\label{eq:gt-decomp-tail} \widetilde{g}_T(t,x)= -f_T(|t|)+h_T(t,x),\tag{65}\] where \(f_T\) is the function defined in equation 61 and \(h_T\) is the function defined in equation 63 . Next, let \[E_T^{(1)}:=-4b\sum_{j=1}^k\sum_{l=1}^w \alpha_j \widetilde{G}_T(q_j,u_l).\] Fix \(l\). Write \(q_j=(\tau_j,\xi_j)\), and let \[d_{jl}:=d_T(t_l-\tau_j).\] Since \(d_T\) is a \(1\)-Lipschitz map, we have \[|d_{jl}-|t_l||= |d_T(t_l - \tau_j) - d_T(t_l)| \le |\tau_j|.\] Using the Lipschitz bound \(|f_T(a)-f_T(b)|\le \frac{\pi}{2}|a-b|\) from Lemma 44, this implies \[f_T(d_{jl})=f_T(|t_l|)+O(1).\] Therefore, using the decomposition 65 , \[\begin{align} -4b\sum_{j=1}^k \alpha_j \widetilde{G}_T(q_j,u_l) &=4b\sum_{j=1}^k \alpha_j f_T(d_{jl}) -4b\sum_{j=1}^k \alpha_j h_T(q_j-u_l) \notag\\ &=4b f_T(|t_l|)\sum_{j=1}^k \alpha_j +O(1)-4b\sum_{j=1}^k \alpha_j h_T(q_j-u_l)\notag \\ &= -4b^2 w f_T(|t_l|)+O(1)-4b\sum_{j=1}^k \alpha_j h_T(q_j-u_l). \label{eq:insert-split} \end{align}\tag{66}\] Taking real parts in 66 , using the lower bound from Lemma 46 for those \(j\) with \(\operatorname{Re}(\alpha_j)\ge 0\), and the upper bound from Lemma 46 for those \(j\) with \(\operatorname{Re}(\alpha_j)<0\), we get \[\begin{align} \operatorname{Re}\biggl(-4b\sum_{j=1}^k \alpha_j \widetilde{G}_T(q_j,u_l)\biggr) &\le -4b^2w f_T(|t_l|) + C +(2-\delta)\sum_{j=1}^k |\log r_T(q_j,u_l)|, \label{eq:insert-final} \end{align}\tag{67}\] where \[\delta:=\min_{1\le j\le k}(2+4b\operatorname{Re}(\alpha_j))>0.\] Summing equation 67 over \(l=1,\dots,w\) gives \[\begin{align} \operatorname{Re}(E_T^{(1)}) \le -4b^2w\sum_{l=1}^w f_T(|t_l|) + C +(2-\delta)\sum_{j=1}^k\sum_{l=1}^w |\log r_T(q_j,u_l)|. \label{eq:E1-bound} \end{align}\tag{68}\] Next, let \[E_T^{(2)}:=-4b^2\sum_{1\le l<l'\le w}\widetilde{G}_T(u_l,u_{l'}).\] Using the decomposition 65 , we have \[E_T^{(2)} = 4b^2\sum_{1\le l<l'\le w} f_T(d_T(t_l-t_{l'})) -4b^2\sum_{1\le l<l'\le w} h_T(u_l-u_{l'}).\] By the lower bound from Lemma 46, the second term is bounded above by a constant: \[-4b^2\sum_{1\le l<l'\le w} h_T(u_l-u_{l'})\le C.\] For the first term, note that \[d_T(t_l-t_{l'})\le \min\{|t_l|+|t_{l'}|,T\}.\] Hence, by Lemma 44, \[f_T(d_T(t_l-t_{l'}))\le f_T(\min\{|t_l|+|t_{l'}|,T\}) \le f_T(|t_l|)+f_T(|t_{l'}|).\] Therefore, \[\begin{align} \operatorname{Re}(E_T^{(2)}) &\le 4b^2\sum_{1\le l<l'\le w}(f_T(|t_l|)+f_T(|t_{l'}|))+C\notag\\ &=4b^2(w-1)\sum_{l=1}^w f_T(|t_l|)+C. \label{eq:E2-bound} \end{align}\tag{69}\] From equations 68 and 69 , we get \[\begin{align} \operatorname{Re}(E_T^{(1)}+E_T^{(2)}) &\le -4b^2\sum_{l=1}^w f_T(|t_l|) + C +(2-\delta)\sum_{j=1}^k\sum_{l=1}^w |\log r_T(q_j,u_l)|. \end{align}\] But again by Lemma 44, \[f_T(|t_l|)\ge \frac{\pi}{4}|t_l|.\] Using this in the previous display, we get \[\begin{align} \label{eq:re1e2} \operatorname{Re}(E_T^{(1)}+E_T^{(2)}) \le -\pi b^2\sum_{l=1}^w |t_l| + C +(2-\delta)\sum_{j=1}^k\sum_{l=1}^w |\log r_T(q_j,u_l)|. \end{align}\tag{70}\] Set \[a:=2-\delta,\qquad \rho_{jl}:=r_T(q_j,u_l),\qquad D_{jl}:=d(q_j,u_l).\] Since \(q_j=(\tau_j,\xi_j)\) is fixed, by increasing \(T_0\) if needed we may assume \[T\ge 1+\max_{1\le j\le k}|\tau_j|.\] Fix any \(\epsilon>0\). We claim that there exists \(C_\epsilon\) (independent of \(T\) and \(u_1,\dots,u_w\)) such that \[\label{eq:r-to-d-bound} e^{a|\log \rho_{jl}|}\le C_\epsilon e^{\epsilon|t_l|} D_{jl}^{-a}, \qquad 1\le j\le k,\;1\le l\le w.\tag{71}\] To prove this, consider two cases. If \(D_{jl}\le 1\), then \(|t_l-\tau_j|\le D_{jl}\le 1<T\), and hence \(d_T(t_l-\tau_j)=|t_l-\tau_j|\). This means that \(\rho_{jl}=D_{jl}\). Thus \[e^{a|\log \rho_{jl}|}=D_{jl}^{-a}\le e^{\epsilon|t_l|}D_{jl}^{-a}.\] If \(D_{jl}>1\), then \(\rho_{jl}\ge 1\) and therefore \(e^{a|\log \rho_{jl}|}=\rho_{jl}^a\). (Indeed, if \(|t_l-\tau_j|\le T\), then \(\rho_{jl}=D_{jl}>1\); if \(|t_l-\tau_j|>T\), then \(|t_l-\tau_j|\le T+|\tau_j|\), so \(d_T(t_l-\tau_j)=2T-|t_l-\tau_j|\ge T-|\tau_j|\ge 1\).) Also, \(d_T(\cdot)\le |\cdot|\) gives \(\rho_{jl}\le D_{jl}\le |t_l-\tau_j|+1\le C(1+|t_l|)\), and so, \[e^{a|\log \rho_{jl}|}=\rho_{jl}^a\le D_{jl}^a =D_{jl}^{-a}D_{jl}^{2a} \le C(1+|t_l|)^{2a}D_{jl}^{-a} \le C_\epsilon e^{\epsilon|t_l|}D_{jl}^{-a}.\] This proves the claim 71 . Exponentiating the bound 70 on \(\operatorname{Re}(E_T^{(1)}+E_T^{(2)})\) and applying the inequality 71 to each pair \((j,l)\) yields \[|\Phi_T(u_1,\dots,u_w)| \le C_\epsilon\exp\biggl(-(\pi b^2-k\epsilon)\sum_{l=1}^w |t_l|\biggr) \prod_{j=1}^k\prod_{l=1}^w d(q_j,u_l)^{-(2-\delta)}.\] Choose \(\epsilon>0\) so small that \(\pi b^2-k\epsilon>0\), and define \(\eta:=\pi b^2-k\epsilon\). Also, \[\sum_{l=1}^w |t_l| \ge \sum_{l=1}^w (|u_l|-|x_l|) \ge \sum_{l=1}^w |u_l|-w,\] because \(|x_l|\le 1\) on \(M_T\). Therefore, \[C_\epsilon\exp\biggl(-\eta\sum_{l=1}^w |t_l|\biggr) \le C_\epsilon e^{\eta w}\exp\biggl(-\eta\sum_{l=1}^w |u_l|\biggr).\] Absorbing \(e^{\eta w}\) into the constant completes the proof. ◻

Proof. We prove the result by induction on \(m\). For \(m=1\), we have \(D_1=\mathbb{C}\) and \(B_1=\mathbb{R}\). Since \(F\) is continuous on \(\mathbb{C}\cup\mathbb{R}=\mathbb{C}\) and vanishes on \(\mathbb{R}\), the one-variable identity theorem gives \(F\equiv 0\) on \(\mathbb{C}\).

Assume the claim holds for \(m\), and let \(F\) be analytic on \(D_{m+1}\) and zero on \(B_{m+1}\). Fix any \(r>0\) and define \[G_r(z_1,\ldots,z_m):=F(z_1,\ldots,z_m,r), \qquad (z_1,\ldots,z_m)\in D_m.\] Then \(G_r\) is analytic on \(D_m\), and \(G_r=0\) on \(B_m\) because \(F=0\) on \(B_{m+1}\). By the induction hypothesis, \[G_r\equiv 0\quad\text{on }D_m\qquad\text{for every }r>0.\] Now fix any \((z_1,\ldots,z_m)\in D_m\), and consider the one-variable function \[h(\zeta):=F(z_1,\ldots,z_m,\zeta),\] which is analytic on \(\mathbb{H}_+\). For each \(r>0\), since \((z_1,\ldots,z_m)\in D_m\) and \(G_r\equiv 0\) on \(D_m\), we have \[h(r)=G_r(z_1,\ldots,z_m)=0.\] Thus \(h\) vanishes on \((0,\infty)\). Hence the one-variable identity theorem implies that \(h\equiv 0\) on \(\mathbb{H}_+\). Since \((z_1,\ldots,z_m)\) was arbitrary, \(F\equiv 0\) on \(D_{m+1}\). This completes the induction. ◻

Proof. If \(H_\nu=\varnothing\), then \(\|\ell_\nu\|\) is bounded above and below by positive constants on the compact set \(K\), and the corresponding factor is harmless. If \(H_\nu\ne\varnothing\), then \(\|\ell_\nu(x)\|\) is comparable to \(\operatorname{dist}(x,H_\nu)\). It therefore suffices to prove the assertion with \[\prod_\nu \operatorname{dist}(x,H_\nu)^{a_\nu}\] in place of the displayed integrand.

We prove the claim by induction on the ambient dimension \(d\), starting with \(d=1\). In one dimension, every nonempty proper affine subspace is a point. Fix \(x^0\in K\). The factors whose zero sets do not contain \(x^0\) are bounded above and below by positive constants near \(x^0\). If no zero set contains \(x^0\), there is nothing to prove locally. Otherwise, after discarding these bounded factors, the local integrand is comparable to \[|x-x^0|^A, \qquad A:=\sum_{\nu:\,x^0\in H_\nu}a_\nu .\] The flat \(\{x^0\}\) has codimension one, so the hypothesis gives \(A>-1\). Hence \(|x-x^0|^A\) is locally integrable, proving the base case.

Now assume \(d\ge2\) and that the result is known in dimensions \(1,\ldots,d-1\). Fix a point \(x^0\in K\). By compactness of \(K\), it is enough to prove local integrability near \(x^0\). Let \[I(x^0):=\{\nu:x^0\in H_\nu\}.\] If \(I(x^0)=\varnothing\), then no singular factor vanishes near \(x^0\), so local integrability is immediate. Factors with \(\nu\notin I(x^0)\) are bounded above and below by positive constants in a small neighborhood of \(x^0\). Translating \(x^0\) to the origin, write \(V_\nu:=H_\nu-x^0\) for \(\nu\in I(x^0)\), and let \[W:=\bigcap_{\nu\in I(x^0)}V_\nu .\] Note that \(W\) is a proper subspace of \(\mathbb{R}^d\). Indeed, \(I(x^0)\) is nonempty by assumption, and for each \(\nu\in I(x^0)\) the set \(H_\nu\) is a nonempty proper affine subspace; hence each \(V_\nu\) is a proper subspace of \(\mathbb{R}^d\), and therefore so is \(W\).

If \(W\ne\{0\}\), split \(\mathbb{R}^d=W\oplus W^\perp\) and write a point near the origin as \(w+y\), with \(w\in W\) and \(y\in W^\perp\). Since \(W\subseteq V_\nu\) for all \(\nu\in I(x^0)\), each singular subspace decomposes as \[V_\nu=W\oplus (V_\nu\cap W^\perp).\] Indeed, if \(v\in V_\nu\) and \(v=w+y\) is its orthogonal decomposition, then \(w\in W\subseteq V_\nu\), and hence \(y=v-w\in V_\nu\cap W^\perp\). Consequently translation in the \(W\) direction does not change any of the singular distances: \[\operatorname{dist}(w+y,V_\nu) =\operatorname{dist}(y,V_\nu\cap W^\perp), \qquad \nu\in I(x^0).\] Thus the local integrand depends on the transverse variable \(y\) through the arrangement of subspaces \[V_\nu^\perp:=V_\nu\cap W^\perp\subseteq W^\perp, \qquad \nu\in I(x^0),\] up to the bounded nonsingular factors already discarded.

The power-counting hypotheses are inherited by this transverse arrangement. To see this, let \(M\subseteq W^\perp\) be a flat obtained as an intersection of some of the subspaces \(V_\nu^\perp\). Then \(W\oplus M\) is the corresponding flat of the original arrangement. Moreover, \[\operatorname{codim}_{W^\perp} M =\operatorname{codim}_{\mathbb{R}^d}(W\oplus M),\] and \(M\subseteq V_\nu^\perp\) holds exactly when \(W\oplus M\subseteq V_\nu\). Therefore the exponent sum attached to \(M\) in the transverse arrangement is the same as the exponent sum attached to \(W\oplus M\) in the original arrangement, while the codimension is also the same. The original hypothesis therefore gives precisely the hypothesis needed for the arrangement in \(W^\perp\).

Since \(\dim W^\perp<d\), the induction hypothesis gives local integrability in the transverse variable \(y\). The remaining \(w\) variable ranges over a bounded set, so Fubini’s theorem gives local integrability in the full neighborhood.

It remains to treat the case \(W=\{0\}\). Then all singular subspaces pass through the origin and have total intersection \(\{0\}\). In polar coordinates \(x=r\theta\), \(0<r<\epsilon\), \(\theta\in\mathbb{S}^{d-1}\), the singular part is \[r^{A}\prod_{\nu\in I(x^0)}\operatorname{dist}(\theta,V_\nu)^{a_\nu}, \qquad A:=\sum_{\nu\in I(x^0)}a_\nu .\] The hypothesis applied to the flat \(\{0\}\) gives \(A>-d\), so the radial integral \(\int_0^\epsilon r^{A+d-1}\,\mathrm{d}r\) is finite. It remains to prove integrability of the angular factor on \(\mathbb{S}^{d-1}\).

Fix \(\theta^0\in\mathbb{S}^{d-1}\) and let \(J:=\{\nu\in I(x^0):\theta^0\in V_\nu\}\). The factors with \(\nu\notin J\) are bounded near \(\theta^0\). In a smooth coordinate chart centered at \(\theta^0\), the factors with \(\nu\in J\) are comparable to distances to the linear subspaces \(V_\nu\cap(\theta^0)^\perp\) of the tangent space \((\theta^0)^\perp\). Indeed, after rotating so that \(\theta^0\) is the last coordinate vector, each \(V_\nu\) with \(\theta^0\in V_\nu\) is defined by linear equations involving only the first \(d-1\) coordinates in this chart. For any flat \(M\) in this tangent arrangement, \(M\oplus\mathbb{R}\theta^0\) is a flat of the original arrangement, \[\operatorname{codim}_{(\theta^0)^\perp}M =\operatorname{codim}_{\mathbb{R}^d}(M\oplus\mathbb{R}\theta^0),\] and the factors containing \(M\) in the tangent arrangement are exactly the factors whose original subspaces contain \(M\oplus\mathbb{R}\theta^0\). Hence the power-counting hypothesis is inherited by the tangent arrangement. By the induction hypothesis in dimension \(d-1\), the angular factor is locally integrable near \(\theta^0\). Compactness of \(\mathbb{S}^{d-1}\) gives angular integrability, and hence local integrability near \(x^0\). This proves the lemma. ◻