Proof. Using the growth assumptions on \(g\), \(g'\) and \(g''\), we see that the weight \(e^{mg}\) and its products with \(g'\), \(g''\) and \((g')^2\) grow at most exponentially and are thus integrable against the Gaussian law of \(B_{2t}\), locally uniformly in \(x\). We may therefore differentiate all the expectations below under the integral sign without further comment.

We first treat the case \(m>0\), and abbreviate \(\mathcal{L}=\mathcal{L}_{mg,t}\). Set \[H(t,x)=\mathbb{E}\,e^{mg(x+B_{2t})}, \qquad\text{so that}\qquad u=\tfrac1m\log H, \quad\text{i.e.}\quad H=e^{mu}.\] Since \(B_{2t}\) has variance \(2t\), the function \(H\) is the convolution of \(e^{mg}\) with the Gaussian heat kernel, and therefore solves the heat equation \(\partial_t H=\partial_x^2 H\).

Space derivative. Differentiating \(H\) in \(x\) and recalling the definition 20 of \(\mathcal{L}\), \[\partial_x H=m\,\mathbb{E}\bigl[g'(x+B_{2t})\,e^{mg(x+B_{2t})}\bigr]=mH\,\mathcal{L} g'.\] On the other hand, \(H=e^{mu}\) gives \(\partial_x H=mH\,\partial_x u\). Comparing the two expressions yields \(\partial_x u=\mathcal{L} g'\), which is 25 .

Time derivative. By the heat equation and \(u=\tfrac1m\log H\), we have \[\partial_t u=\frac{\partial_t H}{mH}=\frac{\partial_x^2 H}{mH},\] so it remains to compute \(\partial_x^2 H\). Differentiating the relation \(\partial_x H=mH\,\partial_x u\) once more in \(x\) gives \[\partial_x^2 H=mH\bigl(\partial_x^2 u+m(\partial_x u)^2\bigr),\] whereas differentiating the expression \(\partial_x H=m\,\mathbb{E}[g'(x+B_{2t})\,e^{mg(x+B_{2t})}]\) gives \[\partial_x^2 H=m\,\mathbb{E}\bigl[\bigl(g''+m(g')^2\bigr)(x+B_{2t})\,e^{mg(x+B_{2t})}\bigr]=mH\,\mathcal{L}\bigl(g''+m(g')^2\bigr).\] Dividing each of these by \(mH\) and recalling that \(\partial_t u=\partial_x^2 H/(mH)\), we obtain \[\partial_t u=\partial_x^2 u+m(\partial_x u)^2=\mathcal{L}\bigl(g''+m(g')^2\bigr),\] which is 26 .

Variation formula. Differentiating \(\varepsilon\mapsto\tfrac1m\log\mathbb{E}\,e^{m(g+\varepsilon h)(x+B_{2t})}\) at \(\varepsilon=0\) gives \[\left.\frac{d}{d\varepsilon}T_{m,t}(g+\varepsilon h)(x)\right|_{\varepsilon=0} =\frac{\mathbb{E}\bigl[h(x+B_{2t})\,e^{mg(x+B_{2t})}\bigr]}{\mathbb{E}\,e^{mg(x+B_{2t})}} =\mathcal{L} h(x),\] which is 27 .

The case \(m=0\). Here \(u(t,x)=\mathbb{E}\,g(x+B_{2t})\), and \(\mathcal{L}_{0,t}\) is the plain Gaussian average \(h\mapsto\mathbb{E}\,h(\,\cdot\,+B_{2t})\). Differentiating under the expectation gives \(\partial_x u=\mathbb{E}\,g'(x+B_{2t})=\mathcal{L}_{0,t}g'\), which is 25 . Moreover, the function \(u\) is now the convolution of \(g\) with the heat kernel, so it solves the heat equation; this gives \(\partial_t u=\partial_x^2 u=\mathbb{E}\,g''(x+B_{2t})=\mathcal{L}_{0,t}g''\), which is 26 . Finally, \(\frac{d}{d\varepsilon}\big|_{\varepsilon=0}\mathbb{E}\,(g+\varepsilon h)(x+B_{2t})=\mathbb{E}\,h(x+B_{2t})=\mathcal{L}_{0,t}h(x)\), which is 27 . ◻

Proof. We fix \(j\in\{0,\dots,k\}\) and apply Lemma 1 with \(u=f_j=T_{m_j,t_j}f_{j+1}\), \(g=f_{j+1}\), \(m=m_j\), \(t=t_j\). The spatial-derivative formula 25 gives \[f'_j=\partial_x u=\mathcal{L}_{m_j f_{j+1},\,t_j}f'_{j+1}\stackrel{\eqref{e46L95i61}}{=}\mathcal{L}_jf'_{j+1},\] which is 28 . Iterating 28 and using the composition convention 23 gives \(f'_j=\mathcal{L}_{j:k}f'_{k+1}\), with \(f'_{k+1}=\phi'\) by 17 .

For the bound, recall from 17 that \(f'_{k+1}=\phi'=\tanh\), so \(|f'_{k+1}|\leqslant 1\). Each \(\mathcal{L}_j\) is a positive linear operator with \(\mathcal{L}_j1=1\) by 21 ; hence it is order-preserving and \(|\mathcal{L}_jh|\leqslant\mathcal{L}_j|h|\leqslant\sup|h|\) for any bounded \(h\). Therefore \(|f'_{j+1}|\leqslant 1\) implies \(|f'_j|=|\mathcal{L}_jf'_{j+1}|\leqslant 1\), and downward induction gives \(|f'_j|\leqslant 1\) for all \(j\). ◻

Proof of Proposition 5. We decompose the proof into two steps.

Step 1. For convenience, we change variables and use \[\label{e46def46ti} t_i=q_i-q_{i-1}, \qquad i\in\{0,\dots,k\}.\tag{29}\] Again abusing notation, we write \(\partial_{t_i} \psi_\circ\) to denote the derivative of the mapping in ?? , but seen as a function of the \((t_j)\) rather than of the \((q_j)\). We also set \[\label{e46A95i61} A_i=f_i''+m_i(f_i')^2, \qquad i\in\{0,\dots,k+1\}.\tag{30}\] In this first step, we show that for every \(i \in \{0, \ldots, k\}\), we have \[\label{e46drt46psi} \frac{\partial f_0(0)}{\partial t_i}=\bigl(\mathcal{L}_{0:i-1}A_i\bigr)(0).\tag{31}\] In order to compute this derivative, we denote, for any sufficiently small \(\varepsilon\), \[t_i^\varepsilon=t_i+\varepsilon \qquad\text{and}\qquad t_r^\varepsilon=t_r\quad \text{for r\neq i},\] and we let \(f_r^\varepsilon\) be the recursion obtained from the perturbed increments: \[f_{k+1}^\varepsilon=\phi\qquad\text{and}\qquad f_r^\varepsilon = T_{m_r,t_r^\varepsilon}f_{r+1}^\varepsilon \quad\text{for r\in\{0,\dots,k\}}.\] Define \[\begin{align} \label{e46D95r61} D_r(x) = \left. \frac{d}{d\varepsilon}f_r^\varepsilon(x) \right|_{\varepsilon=0}\qquad\text{for r\in\{0,\dots,k\}}. \end{align}\tag{32}\]

First, for \(r\geqslant i+1\), the function \(f_r^\varepsilon\) is unchanged. Indeed, the recursion defining \(f_r\) uses only the interval lengths \(t_r,t_{r+1},\ldots,t_k\), and none of these equals \(t_i\). Hence, we have \(f_r^\varepsilon=f_r\) for \(r\geqslant i+1\). At level \(i\), the input function \(f_{i+1}\) is therefore fixed, so \(f_i^\varepsilon =T_{m_i,t_i+\varepsilon}f_{i+1}\), and thus \[\begin{align} \label{e46D95i61A95i} D_i = \left. \frac{d}{d\varepsilon} T_{m_i,t_i+\varepsilon}f_{i+1} \right|_{\varepsilon=0} \stackrel{\eqref{eq:HC-time-derivative}}{=} f_i''+m_i(f_i')^2 \stackrel{\eqref{e46A95i61}}{=} A_i. \end{align}\tag{33}\]

Now take \(r<i\). At this level, the time parameter \(t_r\) is fixed. The only dependence on \(\varepsilon\) comes through the input function \(f_{r+1}^\varepsilon\): \[f_r^\varepsilon = T_{m_r,t_r}f_{r+1}^\varepsilon.\] Since \(f_{r+1}^\varepsilon=f_{r+1}+\varepsilon D_{r+1}+o(\varepsilon)\), we may apply the variation formula 27 with \(g=f_{r+1}\), \(h=D_{r+1}\), \(m=m_r\), \(t=t_r\) to obtain \[D_r = \mathcal{L}_{m_r f_{r+1},\,t_r}D_{r+1} = \mathcal{L}_rD_{r+1}, \qquad r=i-1,i-2,\ldots,0.\] Iterating this and using 33 , we get \(D_0=\mathcal{L}_0\mathcal{L}_1\cdots\mathcal{L}_{i-1}A_i\stackrel{\eqref{eq:L-composition-convention}}{=}\mathcal{L}_{0:i-1}A_i\). Since \(\frac{\partial f_0(0)}{\partial t_i}=D_0(0)\) due to 32 , we can conclude \(\frac{\partial f_0(0)}{\partial t_i}=\bigl(\mathcal{L}_{0:i-1}A_i\bigr)(0)\) as desired.

Step 2. We now proceed to complete the proof of the proposition. To lighten notation, we set \[\delta_i=m_{i+1}-m_i, \qquad i\in\{0,\dots,k\}.\] By 30 and the second relation in 26 , we have \[A_i = \mathcal{L}_i\bigl(f_{i+1}''+m_i(f_{i+1}')^2\bigr),\] and thus \[f_{i+1}''+m_i(f_{i+1}')^2 = A_{i+1}-\delta_i(f_{i+1}')^2.\] Combining the above two displays, we get \[\label{eq:A-recursion} A_i = \mathcal{L}_iA_{i+1} - \delta_i\mathcal{L}_i[(f_{i+1}')^2].\tag{34}\]

The terminal condition \(\phi(x)=\log\cosh x\) satisfies \(\phi'(x)=\tanh x\) and \(\phi''(x)=\cosh^{-2} (x)\). Since \(m_{k+1}=1\) as in 15 , we have \[\label{e46A95k431611} A_{k+1} = \phi''+(\phi')^2 = \cosh^{-2}(x) +\tanh^2(x) =1.\tag{35}\] Since every \(\mathcal{L}_i\) preserves constants by 21 , iteration of 34 gives \[\label{eq:A-expansion} A_i = 1- \sum_{r=i}^k \delta_r \bigl(\mathcal{L}_{i:r}[(f_{r+1}')^2]\bigr).\tag{36}\] Using the result 31 from the previous step, we write \[\frac{\partial f_0(0)}{\partial t_i} = \bigl(\mathcal{L}_{0:i-1}A_i\bigr)(0)\stackrel{\eqref{eq:A-expansion}}{=} 1- \sum_{r=i}^k\delta_r \bigl(\mathcal{L}_{0:i-1}\mathcal{L}_{i:r}[(f_{r+1}')^2]\bigr)(0) \stackrel{\eqref{eq:Ui-def-Lab}}{=} 1- \sum_{r=i}^k\delta_r U_r(q).\] Using this and 19 , we get \[\label{eq:t-gradient-formula} \frac{\partial \psi_\circ}{\partial t_i} = \sum_{r=i}^k \delta_r U_r(q).\tag{37}\] Since \(t_i=q_i-q_{i-1}\), we have \(\frac{\partial}{\partial q_k}=\frac{\partial}{\partial t_k}\) and \(\frac{\partial}{\partial q_i}=\frac{\partial}{\partial t_i}-\frac{\partial}{\partial t_{i+1}}\) for \(i\in\{0,\dots,k-1\}\). By 37 , we get \[\frac{\partial \psi_\circ}{\partial q_k}=\frac{\partial \psi_\circ}{\partial t_k}=\delta_kU_k \qquad \text{and}\qquad \frac{\partial \psi_\circ}{\partial q_i}=\sum_{r=i}^k \delta_rU_r-\sum_{r=i+1}^k \delta_rU_r=\delta_iU_i \quad\text{for i\in\{0,\dots,k-1\}}\] as announced. ◻

Proof of Proposition 2. Let \(q,q'\in\mathcal{Q}\) take finitely many values. As explained in the paragraph below 19 , we may choose a common set of discretization points \(0=m_0<m_1<\cdots<m_{k+1}=1\) such that \[q=\sum_{i=0}^kq_i\mathbb{1}_{[m_i,m_{i+1})}, \qquad q'=\sum_{i=0}^kq'_i\mathbb{1}_{[m_i,m_{i+1})},\] with \(0\leqslant q_0\leqslant\cdots\leqslant q_k\) and \(0\leqslant q'_0\leqslant\cdots\leqslant q'_k\). We have \[\label{e46lip46psi46L1} |q-q'|_{L^1}=\sum_{i=0}^k(m_{i+1} - m_i)\,|q_i-q'_i|.\tag{38}\] Identifying a finite-valued path with the vector of its values, we regard \(\psi_\circ\) as a function on the cone \(\{0\leqslant q_0\leqslant\cdots\leqslant q_k\}\) via the recursion 18 –19 .

We first prove the bound when \(q\) and \(q'\) both lie in the open cone \(\{0<q_0<\cdots<q_k\}\). There, by Proposition 5, the function \(\psi_\circ\) is differentiable with \[\partial_{q_i} \psi_\circ=(m_{i+1} - m_i)\,U_i(q), \qquad i\in\{0,\dots,k\},\] where \(U_i(q)=\bigl(\mathcal{L}_{0:i}[(f'_{i+1})^2]\bigr)(0)\) as in 24 . By Lemma 2, we have \(|f'_j|\leqslant 1\) for all \(j\), so \(0\leqslant(f'_{i+1})^2\leqslant 1\). Since each \(\mathcal{L}_j\) is a positive operator with \(\mathcal{L}_j1=1\), applying \(\mathcal{L}_{0:i}\) and evaluating at \(0\) gives \[0\leqslant U_i(q)\leqslant 1, \qquad\text{so that}\qquad 0\leqslant\partial_{q_i} \psi_\circ \leqslant m_{i+1}-m_i.\] The open cone is convex, so the segment \(q_\theta:=(1-\theta)q'+\theta q\), \(\theta\in[0,1]\), remains in it. By the fundamental theorem of calculus, \[\psi_\circ(q)-\psi_\circ(q') =\int_0^1\sum_{i=0}^k\partial_{q_i} \psi_\circ(q_\theta)\,(q_i-q'_i)\,\mathrm{d}\theta,\] and therefore, using \(0\leqslant\partial_{q_i} \psi_\circ \leqslant m_{i+1}-m_i\) together with 38 , we obtain that \[|\psi_\circ(q)-\psi_\circ(q')| \leqslant\sum_{i=0}^k(m_{i+1} - m_i)\,|q_i-q'_i| =|q-q'|_{L^1}.\] Both sides above are continuous in \((q_0,\dots,q_k)\) on the closed cone; for the left-hand side this is clear from 19 and the continuity of \((q_i)\mapsto f_0(0)\), recalling that \(T_{m,0}\) is the identity map. Since the open cone is dense in the closed cone, the inequality extends to all \(q,q'\) taking finitely many values. Finally, paths taking finitely many values are dense in \(\mathcal{Q}_1\) for the \(L^1\) norm, so \(\psi_\circ\) admits a (unique, \(1\)-Lipschitz) extension to \(\mathcal{Q}_1\) satisfying the same bound. ◻

Proof of Lemma 3. Theorem 2 is normalized so that the largest endpoint is \(1\). To apply it here, we fix \(Q>q_k\) and set \[p_0=0,\qquad p_{r+1}=q_r\;\text{ for }r\in\{0,\dots,k\},\qquad \text{and}\quad p_{k+2}=Q.\] We take \(K=k+1\), set \(\widehat q_r=p_r/Q\) for \(r\in\{0,\ldots,k+2\}\), keep the masses \(m_0,\ldots,m_{k+1}\), choose \(\xi(s)=Qs^2\), and take the terminal function in Theorem 2 to be \(\Phi=\phi\). Then, by 39 , \[\label{e46compar95var} \xi'(\widehat q_{r+1})-\xi'(\widehat q_r)=2(p_{r+1}-p_r),\tag{45}\] so that, for \(r\leqslant k\), the Gaussian increment \(z_r\) has the same law as \(B_{2(q_r-q_{r-1})}\).

The only additional interval is \([q_k,Q]\), with mass \(m_{k+1}=1\). It does not change the quantities of interest, since \[\label{eq:T1-terminal-constant} T_{1,Q-q_k}\phi(x)=\log\mathbb{E}\cosh(x+B_{2(Q-q_k)})=\phi(x)+(Q-q_k).\tag{46}\] Thus the functions \(\Phi_i\) generated by 40 satisfy \[\Phi_i=f_i+Q-q_k,\qquad i\in\{0,\ldots,k+1\}.\] In particular, \(\Phi_i'=f_i'\) for these indices. The convexity and parity assertions for the \(f_i\)’s therefore follow directly from Theorem 2.

For the tilted operators, the additive constant cancels from 41 . Using 45 , we get, for \(i\in\{0,\ldots,k\}\), \[\mathbb{E}_{z_i}V_i(x,z_i)g(x+z_i) =\frac{\mathbb{E}\left[g(x+B_{2(q_i-q_{i-1})})e^{m_i f_{i+1}(x+B_{2(q_i-q_{i-1})})}\right]}{\mathbb{E}e^{m_i f_{i+1}(x+B_{2(q_i-q_{i-1})})}} =\mathcal{L}_i g(x).\] Thus each operator \(\mathcal{L}_i\), for \(i\in\{0,\ldots,k\}\), sends \(\mathcal{C}\) into \(\mathcal{C}\) by Theorem 2.

It remains to identify our \(U_i\) with the quantity \(U_i^{\mathrm P}\) from Theorem 2. Differentiating 40 gives \[\Phi_r'(x)=\mathbb{E}_{z_r}\left[V_r(x,z_r)\Phi_{r+1}'(x+z_r)\right].\] Iterating this identity from \(r=i+1\) to \(K\) gives \[\Phi_{i+1}'(Z_{i+1}) = \mathbb{E}\left[W_{i+1}\cdots W_K\Phi'(Z)\mid z_0,\ldots,z_i\right].\] Using the preceding identification of the tilted operators for the steps \(0,\ldots,i\), and recalling that \(W_0=1\) because \(m_0=0\), we obtain \[U_i(q) =\mathbb{E}\left[W_0\cdots W_i\left(\Phi_{i+1}'(Z_{i+1})\right)^2\right] =U^{\mathrm P}_{i+1}.\] The monotonicity of \(U_i(q)\) in each mass coordinate now follows from the monotonicity of \(U^{\mathrm P}_{i+1}\) in Theorem 2. ◻

Proof. Being a Hessian, \(H(q)\) is symmetric, so it suffices to treat the case \(i<j\). Since \(m_{i+1}-m_i>0\), identity 47 reduces the bound \(H_{ij}\leqslant 0\) to \[\label{e46dU47dq600} \partial_{q_j} U_i\leqslant 0,\qquad i<j.\tag{48}\] Fix such a pair \(i<j\), and write \(e_j\) for the \(j\)-th standard basis vector of \(\mathbb{R}^{k+1}\). It suffices to show that \(U_i(q+\varepsilon e_j)\leqslant U_i(q)\) for all sufficiently small \(\varepsilon>0\): dividing by \(\varepsilon\) and letting \(\varepsilon\downarrow0\) then gives 48 . The mechanism is that displacing the single breakpoint \(q_j\) to the right amounts, once both paths are recorded on a common refinement, to lowering one mass coordinate, to which the monotonicity of Lemma 3 [lem:panchenko40341] applies.

Step 1: a common refinement. The recursion 18 defining the functions \(f_l\), and hence the quantities \(U_i\), depends on a path only through the mass attached to each spatial interval, and not on the particular breakpoints used to record it. Indeed, subdividing an interval into two adjacent pieces carrying the same mass \(m\) replaces a single step \(T_{m,a+b}\) of the recursion by two consecutive steps \(T_{m,a}T_{m,b}\), and these coincide: \[\label{e46semigroup46T} T_{m,a}T_{m,b}=T_{m,a+b}\qquad(m\geqslant 0),\tag{49}\] as one checks directly from 13 . We may therefore evaluate \(U_i\) at \(q\) and at \(q^\varepsilon:=q+\varepsilon e_j\) from any single list of breakpoints that represents both paths.

Assume first that \(j<k\), and take \(\varepsilon>0\) small enough that \(q_j+\varepsilon<q_{j+1}\), so that \(q^\varepsilon\) still lies in the open cone. Inserting the point \(q_j+\varepsilon\) among the breakpoints of \(q\) (or, equivalently, the point \(q_j\) among those of \(q^\varepsilon\)) produces the common list \[(q_0,\ldots,q_{j-1},\,q_j,\,q_j+\varepsilon,\,q_{j+1},\ldots,q_k),\] which represents both paths. On this list the two paths attach the same mass to every interval except \([q_j,q_j+\varepsilon]\): for \(q\) this interval is part of \([q_j,q_{j+1}]\) and carries mass \(m_{j+1}\), whereas for \(q^\varepsilon\) it is part of \([q_{j-1},q_j+\varepsilon]\) and carries mass \(m_j\). Since \(m_j<m_{j+1}\), passing from \(q\) to \(q^\varepsilon\) lowers exactly one mass coordinate—the mass on \([q_j,q_j+\varepsilon]\), from \(m_{j+1}\) down to \(m_j\)—and leaves all the others unchanged. The two representations are shown below.

Step 2: monotonicity in the mass. Because \(i<j\), the breakpoint \(q_i\) lies strictly to the left of the modified interval \([q_j,q_j+\varepsilon]\). For a single fixed path, subdividing intervals that lie to the right of \(q_i\) leaves \(U_i=\mathcal{L}_{0:i}[(f'_{i+1})^2](0)\) unchanged: by the semigroup identity 49 it alters neither the operators \(\mathcal{L}_0,\ldots,\mathcal{L}_i\) nor the function \(f_{i+1}\) entering this expression. Consequently, we can view \(U_i(q)\) and \(U_i(q^\varepsilon)\) as two values of one and the same function, namely the quantity \(U_i\) attached to the common refined list and regarded as a function of its mass coordinates, but evaluated at the masses of \(q\) and of \(q^\varepsilon\) respectively. By Step 1 these two assignments share every coordinate but one: the mass on \([q_j,q_j+\varepsilon]\), which equals \(m_{j+1}\) for \(q\) and \(m_j\) for \(q^\varepsilon\).

Let us write \(g(\mu)\) for the value of this function when the mass on \([q_j,q_j+\varepsilon]\) is set to \(\mu\) and all the other coordinates are held fixed, so that \(g(m_{j+1})=U_i(q)\) and \(g(m_j)=U_i(q^\varepsilon)\). For \(\mu\in(m_j,m_{j+1})\) the masses of the refined list are strictly ordered, so Lemma 3 [lem:panchenko40341] applies and shows that \(g\) is nondecreasing on this interval; since \(g\) is moreover continuous (the operators \(T_{m,t}\) depend continuously on \(m\)), it follows that \(g(m_j)\leqslant g(m_{j+1})\), that is, \(U_i(q^\varepsilon)\leqslant U_i(q)\). This proves 48 when \(j<k\).

It remains to treat \(j=k\). Here \(q_k\) is the last breakpoint, so we first make room to its right. Fix \(Q>q_k+\varepsilon\) and append the terminal interval \([q_k,Q]\) with mass \(m_{k+1}=1\), exactly as in the proof of Lemma 3. By 46 this adds the constant \(Q-q_k\) to every \(f_l\), hence leaves all the derivatives \(f_l'\), and with them every \(U_i\), unchanged. With the terminal interval in place, \(q_k\) is no longer last, and Steps 1 and 2 apply verbatim to the common list \[(q_0,\ldots,q_{k-1},\,q_k,\,q_k+\varepsilon,\,Q):\] on it the interval \([q_k,q_k+\varepsilon]\) carries mass \(1\) for \(q\) and mass \(m_k\) for \(q^\varepsilon\), so moving \(q_k\) to \(q_k+\varepsilon\) again lowers a single mass coordinate, this time from \(1\) to \(m_k\). Arguing as in Step 2 then gives \(U_i(q+\varepsilon e_k)\leqslant U_i(q)\) for \(i<k\), which is 48 for \(j=k\). The proof is thus complete. ◻

Proof. For convenience, we denote by \(P_t:=T_{0,t} = \mathcal{L}_{0,t}\) the heat semigroup with Brownian variance \(2t\), so that \[P_tf(x)=\mathbb{E}f(x+B_{2t}).\] We set \(\mathbb{1}=(1,\ldots,1)\in\mathbb{R}^{k+1}\). The \(i\)-th row sum is the directional derivative of \(\partial_{q_i} \psi_\circ\) in the direction \(\mathbb{1}\). By Proposition 5, we have \[\label{eq:row-sum-directional} \sum_{j=0}^k H_{ij} = (m_{i+1} - m_i) \left. \frac{\mathrm{d}}{\mathrm{d}\varepsilon} U_i(q+\varepsilon\mathbb{1}) \right|_{\varepsilon=0}.\tag{50}\] We set \(q^\varepsilon:=q+\varepsilon\mathbb{1}\). In the notation \((t_i)\) from 29 , we see that \[t_0^\varepsilon=t_0+\varepsilon, \qquad \text{and}\qquad t_r^\varepsilon=t_r\quad \text{for r\geqslant 1}.\] Therefore the functions \(f_r\) for \(r\geqslant 1\) are unchanged, because their recursion uses only \(t_r,t_{r+1},\ldots,t_k\). Likewise the operators \(\mathcal{L}_1,\ldots,\mathcal{L}_i\) are unchanged. Only \(\mathcal{L}_0\) changes, and since \(m_0=0\) it changes from \(P_{t_0}\) to \(P_{t_0+\varepsilon}\).

For the original value of \(q\), set \(R_i= \mathcal{L}_{1:i}[(f_{i+1}')^2]\) for \(i\in\{0,\dots,k\}\). By our convention, we have \(R_0 = (f_1')^2\). Then, in view of 24 , we have \(U_i(q)=P_{t_0}R_i(0)\). By the preceding paragraph, the same \(R_i\) is used for \(q^\varepsilon\), and thus \[U_i(q+\varepsilon\mathbb{1}) = P_{t_0+\varepsilon}R_i(0) = P_\varepsilon G_i(0), \qquad \text{where}\quad G_i:=P_{t_0}R_i.\] Here the last equality uses the heat semigroup property \(P_{t_0+\varepsilon}=P_\varepsilon P_{t_0}\).

By Lemma 3 [lem:panchenko40141], we have \(f_{i+1}'\in\mathcal{C}'\) and thus \((f_{i+1}')^2\in\mathcal{C}\). Since \(\mathcal{L}_r\) preserve \(\mathcal{C}\) by Lemma 3 [lem:panchenko40241] and the heat semigroup \(P_{t_0}\) also preserves \(\mathcal{C}\), we have \(G_i\in\mathcal{C}\). Hence \(G_i\) is even and nondecreasing in \(|x|\).

Let \(Z\) be a standard Gaussian random variable. We have \[\label{eq:heat-monotone-epsilon} P_\varepsilon G_i(0) = \mathbb{E}G_i(\sqrt{2\varepsilon}\,Z) = \mathbb{E}G_i(\sqrt{2\varepsilon}\,|Z|).\tag{51}\] Since \(G_i(r)\) is nondecreasing for \(r\geqslant 0\), the right-hand side of 51 is nondecreasing in \(\varepsilon\). Therefore, we conclude \(\left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}U_i(q+\varepsilon\mathbb{1})\right|_{\varepsilon=0}\geqslant 0\). Using also 50 , we obtain that \(\sum_{j=0}^k H_{ij}\geqslant 0\), as desired. ◻

Proof. For every vector \(a=(a_i)\), we have \[a^\top Ha = \sum_i \left(\sum_jH_{ij}\right)a_i^2 + \sum_{i<j} (-H_{ij})(a_i-a_j)^2.\] Since both terms in the display are nonnegative under the assumptions, the matrix \(H\) is positive semidefinite. ◻

Proof of Proposition 4. The map in ?? is smooth, and its Hessian \(H\) is given by 47 . This Hessian is symmetric, with nonpositive off-diagonal entries by Lemma 4 and nonnegative row sums by Lemma 5. By Lemma 6, it is positive semidefinite, and thus the map is convex. ◻

Proof of Proposition 3. We first observe that, for any fixed discretization points \(0=m_0<m_1<\cdots<m_{k+1}=1\), the map \((q_i)\mapsto\psi_\circ(q)\) is convex on the closed cone \(\{0\leqslant q_0\leqslant\cdots\leqslant q_k\}\). Indeed, it is convex on the open cone from Proposition 4; it is continuous on the closed cone; and the open cone is dense in the closed cone.

Now let \(q_0,q_1\in\mathcal{Q}_1\) and \(\lambda\in[0,1]\). By Proposition 2, \(\psi_\circ\) is continuous on \(\mathcal{Q}_1\), and paths taking finitely many values are dense in \(\mathcal{Q}_1\) for the \(L^1\) norm. It therefore suffices to prove the claimed inequality when \(q_0\) and \(q_1\) take finitely many values. As in the proof of Proposition 2, we may then choose a common set of discretization points \(0=m_0<m_1<\cdots<m_{k+1}=1\) and write \[q_0=\sum_{i=0}^k a_i\mathbb{1}_{[m_i,m_{i+1})} \qquad\text{and}\qquad q_1=\sum_{i=0}^k b_i\mathbb{1}_{[m_i,m_{i+1})},\] with \((a_i)\) and \((b_i)\) both lying in the closed cone \(\{0\leqslant q_0\leqslant\cdots\leqslant q_k\}\). The path \((1-\lambda)q_0+\lambda q_1\) has values \(\bigl((1-\lambda)a_i+\lambda b_i\bigr)_i\), which again lie in this cone. By the convexity on the closed cone established in the previous paragraph, \[\psi_\circ\bigl((1-\lambda)q_0+\lambda q_1\bigr)\leqslant(1-\lambda)\psi_\circ(q_0)+\lambda\psi_\circ(q_1),\] as desired. ◻

Proof of Corollary 1. By the decomposition 7 , the map \(\psi\) is a nonnegative linear combination of the maps \(q\mapsto\psi_\circ(q_s)\), \(s\in\mathscr{S}\), each of which is convex on \(\mathcal{Q}_1\) by Proposition 3. ◻

Proof. This follows from the local Lipschitzness of \(\nabla\xi\) and \(\theta\) and results from Lemma 9 together with 63 . ◻

Proof. The relations 87 and 88 follow directly from the definitions in 80 , 73 , and 86 . Indeed, \[\begin{align} a^{\mathrm{vec}}\cdot b = \sum_{s\in\mathscr{S}}\sum_{d\in\mathsf{D}_s}a_s b_d = {D}\sum_{s\in\mathscr{S}}a_s b_s^{\mathrm{sum}} = {D}a\cdot b^{\mathrm{sum}}. \end{align}\]

We next prove 89 . By 74 , \(\xi^{\mathrm{vec}}(b)={D}\xi(b^{\mathrm{sum}})\). Hence, for \(d\in\mathsf{D}_s\), the chain rule gives \[\begin{align} \nabla_d\xi^{\mathrm{vec}}(b) = {D}\sum_{s'\in\mathscr{S}}\nabla_{s'}\xi(b^{\mathrm{sum}})\, \frac{\partial b_{s'}^{\mathrm{sum}}}{\partial b_d} = {D}\nabla_s\xi(b^{\mathrm{sum}})\frac{1}{{D}} = \nabla_s\xi(b^{\mathrm{sum}}). \end{align}\] This proves the first relation in 89 , and the second follows from the definition of the vectorization map \(a\mapsto a^{\mathrm{vec}}\).

For 90 , using 89 and 87 , we have \[\begin{align} b'\cdot \nabla\xi^{\mathrm{vec}}(b) = b'\cdot \left(\nabla\xi(b^{\mathrm{sum}})\right)^{\mathrm{vec}} = {D}b'^{\mathrm{sum}}\cdot \nabla\xi(b^{\mathrm{sum}}). \end{align}\] Taking \(b'=b\) and using \(\xi^{\mathrm{vec}}(b)={D}\xi(b^{\mathrm{sum}})\) gives \[\begin{align} \theta^{\mathrm{vec}}(b) = b\cdot \nabla\xi^{\mathrm{vec}}(b)-\xi^{\mathrm{vec}}(b) = {D}b^{\mathrm{sum}}\cdot\nabla\xi(b^{\mathrm{sum}})-{D}\xi(b^{\mathrm{sum}}) = {D}\theta(b^{\mathrm{sum}}). \end{align}\]

We now prove 91 . Since \(q^{\mathrm{vec}}=\underline q\), \(p=\underline p^{\mathrm{sum}}\), and by 89 , \[\begin{align} \underline q+t\nabla\xi^{\mathrm{vec}}(\underline p) = q^{\mathrm{vec}}+t\left(\nabla\xi(\underline p^{\mathrm{sum}})\right)^{\mathrm{vec}} = \left(q+t\nabla\xi(p)\right)^{\mathrm{vec}}. \end{align}\] Therefore, by 85 , \[\begin{align} \psi^{\mathrm{vec}}\left(\underline q+t\nabla\xi^{\mathrm{vec}}(\underline p)\right) = {D}\psi\left(q+t\nabla\xi(p)\right). \end{align}\] Combining this identity with \(\theta^{\mathrm{vec}}(\underline p)={D}\theta(\underline p^{\mathrm{sum}})={D}\theta(p)\) pointwise gives \[\begin{align} \mathscr{P}^{\mathrm{vec}}_{t,\underline q}(\underline p) &= \psi^{\mathrm{vec}}\left(\underline q+t\nabla\xi^{\mathrm{vec}}(\underline p)\right) - t\int_0^1\theta^{\mathrm{vec}}(\underline p(r))\,\mathrm{d}r \\ &= {D}\psi\left(q+t\nabla\xi(p)\right) - t{D}\int_0^1\theta(p(r))\,\mathrm{d}r = {D}\mathscr{P}_{t,q}(p). \end{align}\]

It remains to prove 92 . Set \[\begin{align} q':=q+t\nabla\xi(p), \qquad \underline q':=\underline q+t\nabla\xi^{\mathrm{vec}}(\underline p). \end{align}\] By the previous computation, \(\underline q'=q'^{\mathrm{vec}}\). Assume that \[\begin{align} \underline p=\partial_{\underline q}\psi^{\mathrm{vec}}(\underline q'). \end{align}\] Let \(h\in \mathrm{Adm}(\mathcal{Q}_2^\mathscr{S},q')\). Then \(h^{\mathrm{vec}}\in \mathrm{Adm}(\mathcal{Q}_2^{D},\underline q')\) and \(\underline q'+\varepsilon h^{\mathrm{vec}}=(q'+\varepsilon h)^{\mathrm{vec}}\) for all sufficiently small \(\varepsilon\geqslant 0\). Using the Gateaux derivative of \(\psi^{\mathrm{vec}}\) at \(\underline q'\), the identity 85 , and then the Gateaux derivative of \(\psi\) at \(q'\), we obtain \[\begin{align} \left\langle\underline p,h^{\mathrm{vec}}\right\rangle_{L^2([0,1];\mathbb{R}^{D})} &= \lim_{\varepsilon\searrow0} \frac{ \psi^{\mathrm{vec}}(\underline q'+\varepsilon h^{\mathrm{vec}})-\psi^{\mathrm{vec}}(\underline q') }{\varepsilon} \\ &= \lim_{\varepsilon\searrow0} \frac{ {D}\psi(q'+\varepsilon h)-{D}\psi(q') }{\varepsilon} = {D}\left\langle\partial_q\psi(q'),h\right\rangle_{L^2([0,1];\mathbb{R}^\mathscr{S})}. \end{align}\] On the other hand, applying 87 pointwise and integrating gives \[\begin{align} \left\langle\underline p,h^{\mathrm{vec}}\right\rangle_{L^2([0,1];\mathbb{R}^{D})} = {D}\left\langle\underline p^{\mathrm{sum}},h\right\rangle_{L^2([0,1];\mathbb{R}^\mathscr{S})} = {D}\left\langle p,h\right\rangle_{L^2([0,1];\mathbb{R}^\mathscr{S})}. \end{align}\] Hence, for every \(h\in \mathrm{Adm}(\mathcal{Q}_2^\mathscr{S},q')\), \[\begin{align} \left\langle p,h\right\rangle_{L^2([0,1];\mathbb{R}^\mathscr{S})} = \left\langle\partial_q\psi(q'),h\right\rangle_{L^2([0,1];\mathbb{R}^\mathscr{S})}. \end{align}\] By the uniqueness in the definition of the Gateaux derivative, this implies \[\begin{align} p=\partial_q\psi(q') = \partial_q\psi(q+t\nabla\xi(p)). \end{align}\] This proves 92 . ◻

Proof. This is essentially the uniform-in-\(q\) version of [3], which was stated for a fixed \(q\). Inspecting the proof shows that the estimates do not depend on \(q\). One can also see this directly from the \(q\)-dependent part of \(H^{t,q}_{N+1}(\rho,\alpha)\) in 78 appearing in \((N+1)\overline{F}^x_{N+1}(t,q)\). This part is given by \(\sqrt{2}W^q_{N+1}(\alpha)\cdot\rho-(N+1)q(1)\cdot \vec{\mathbf{1}}\), and it decomposes into \(\sqrt{2}W^q_N(\alpha)\cdot\sigma-Nq(1)\cdot \vec{\mathbf{1}}\) plus \(\sqrt{2}w^q(\alpha)\cdot \tau-q(1)\cdot \vec{\mathbf{1}}\), with the two terms taken to be independent. The first term is exactly the corresponding contribution in \(H^{t,q}_N(\sigma,\alpha)\) inside \(N\overline{F}_N^x(t,q)\), while the second term is included in \(U(\sigma,\alpha,\tau)\) inside \(A_N(x,q)\). Therefore the argument of [3] applies with estimates that are independent of \(q\). ◻

Proof. By a standard Gaussian interpolation argument, there exists a constant \(C_1>0\) such that \[\begin{align} \label{e46124F-F94x12460} \sup_{x,q}\left|\overline{F}^x_N(t,{q})-\overline{F}_N(t,{q})\right|\leqslant C_1N^{-1/16},\qquad \sup_{x,q}\left|\widetilde{F}^x_N(t,{q})-\overline{F}_N(t,{q})\right|\leqslant C_1N^{-1/16} \end{align}\tag{108}\] for every \(N\in\mathbb{N}\). For the details, we refer to the proof of [3]. That proof shows that the above differences vanish as \(N\to\infty\). The stated rate is not written explicitly there, but it follows from the same estimates. By [3], there exists a constant \(C_2\) such that, for every \(N\in\mathbb{N}\), \(t\geqslant 0\), and \(q\in\mathcal{Q}_1^{D}\), \[\begin{align} \left|N\overline{F}_N(t,{q})-(N+1)\overline{F}_{N+1}(t,{q})\right|\leqslant C_2\left(t+|{q}|_{L^1}\right). \end{align}\] The Lipschitz estimate for \(\overline{F}_N\) in Proposition 7 and Remark 9 also gives a constant \(C_3>0\) such that \(|\overline{F}_N(t,q)|\leqslant C_3(1+|t|+|q|_{L^1})\). Combining these estimates yields 107 . ◻

Proof. We first prove the uniform version for each fixed test triple \((\mathbf{f}_j,n_j,h_j)\). The proof of [3] is uniform over the background Hamiltonian. In the present notation, the path \(q\) only enters through the unperturbed part of the Gibbs weight, namely through the cascade field in \(H_N^{t,q}\). The perturbative Hamiltonians indexed by \(x\) are independent of this field, and the estimates in the proof of [3] depend only on the bounded test function, on \(n_j,h_j\), and on the uniform bounds on the overlaps, but not on the particular choice of \(q\) in a bounded subset of \(\mathcal{Q}_\infty^{D}\). Therefore, for every \(R<\infty\) and every fixed \(j\), \[\begin{align} \label{e46uniform95Delta95j} \lim_{N\to\infty}\sup_{|q|_{L^\infty}\leqslant R}\mathbb{E}_x\Delta^{x,q}_N(\mathbf{f}_j,n_j,h_j)=0. \end{align}\tag{114}\] Using 112 , for every \(J\in\mathbb{N}\) we have \[\begin{align} \sup_{|q|_{L^\infty}\leqslant R}\mathbb{E}_x\Delta_N(x,q) &\leqslant\sum_{j=1}^J2^{-j}\sup_{|q|_{L^\infty}\leqslant R}\mathbb{E}_x\Delta^{x,q}_N(\mathbf{f}_j,n_j,h_j)+\sum_{j>J}2^{-j}. \end{align}\] Taking \(\limsup_{N\to\infty}\) and using 114 and then letting \(J\to\infty\), we obtain ?? . ◻

Proof. Let \(\eta:[0,3]\to[0,1]\) be smooth and satisfy \(\eta=1\) on \([1,2]\) and \(\eta(0)=\eta(3)=0\). Using the properties of \(\eta\) and integrating by parts, we have \[\begin{align} \int_1^2\left|f'-g'\right|^2 \leqslant\int_0^3\eta \left(f'-g'\right)^2= - \int_0^3\eta'(f-g)(f'-g')-\int_0^3 \eta(f-g)(f''-g''). \end{align}\] Next, we estimate each term on the right. We write \(L^\infty = L^\infty[0,3]\). We start with \[\begin{align} \left|\int_0^3\eta'(f-g)(f'-g')\right|\leqslant 6a\|\eta'\|_{L^\infty} \|f-g\|_{L^\infty} \end{align}\] to bound the first term. For the second term, we have \[\begin{align} \left|\int_0^3 \eta(f-g)(f''-g'')\right|\leqslant\|\eta\|_{L^\infty}\|f-g\|_{L^\infty}\int_0^3\left|f''-g''\right|. \end{align}\] Since \(b-f'',\, b-g''\geqslant 0\), we have \[\begin{align} \int_0^3\left|f''-g''\right|\leqslant\int_0^3 \left|b-f''\right|+\left|b-g''\right| = \int_0^3 2b-f''-g''=6b-f'(3)-g'(3)+f'(0)+g'(0) \\ \leqslant 6b + 4a. \end{align}\] Combining the above displays, we can deduce 123 . ◻

Proof. Since \(r, a_j\geqslant 0\) and \(q_i, q_j\) are increasing paths, we can deduce from the definition of \(\mathcal{Q}_{\uparrow,c}\) in 54 that \[\begin{align} q+a_i r q_i+N^{-\gamma/2}\sum_{j\neq i}a_j y_jq_j \in \mathcal{Q}_{\uparrow,c}^{D},\qquad\forall i \in\mathbb{N},\;N\in\mathbb{N},\;r\in[0,1). \end{align}\] Fix any \(i\in\mathbb{N}\) and write \(q_* = q+N^{-\gamma/2}\sum_{j\neq i}a_j y_jq_j\) and \(\kappa = a_iq_i\). Then, the above display ensures that \(q_*+r\kappa\in\mathcal{Q}_{\uparrow,c}^{D}\) for every \(N\) and \(r\). We also write \(F(r)=\overline{F}_N(t,q_*+r\kappa)\). For \(\varepsilon>0\) small, applying Proposition 8¹ (see also Remark 9) with \(\frac{1}{2}, q_*+r\kappa, q_*+(r+\varepsilon)\kappa\) substituted for \(r, q,q'\) therein, we get \[\begin{align} \tfrac{1}{2}F(r)+\tfrac{1}{2}F(r+\varepsilon)- F\left(r+\tfrac{\varepsilon}{2}\right)\leqslant\tfrac{C}{4} c^{-2}\varepsilon^2\left|\accentset{\large\bfseries .}{\kappa}\right|_{L^2}^2 \end{align}\] for some absolute constant \(C>0\). Dividing both sides by \(\varepsilon^2\), sending \(\varepsilon\to0\), and using \(\kappa=a_iq_i\), we can get \(\frac{\mathrm{d}^2}{\mathrm{d}r^2}F(r)\leqslant 2Cc^{-2}\left|a_i\accentset{\large\bfseries .}{q}_i\right|_{L^2}^2 \leqslant 2Cc^{-2}\), where we also used 120 in the last inequality. This implies the desired result.

To obtain the estimates for \(\widetilde{F}_N^x\) and \(\overline{F}^x_{N+1}\), we can again apply Proposition 8 together with Remark 10. ◻

Proof. Fix any \(i\in\mathbb{N}\) and fix any \(y_j\in[0,3]\) for every \(j\neq i\). We write \(\widetilde{f}(y_i) = \widetilde{F}_N^x(t, q+q_N(y))\) and \(\overline{f}(y_i) = \overline{F}_{N+1}^x(t, q+q_N(y))\) as functions of \(y_i\in[0,3]\) only. In view of the definition of \(q_N(y)\) in 121 , we can compute the derivatives \[\begin{align} \label{e46f3961} \widetilde{f}'(y_i) = N^{-\gamma/2}a_i\left\langle q_i,\partial_q \widetilde{F}_N^x(t, q+q_N(y))\right\rangle_{L^2}\quad\text{and}\quad \overline{f}'(y_i) = N^{-\gamma/2}a_i\left\langle q_i,\partial_q \overline{F}_{N+1}^x(t, q+q_N(y))\right\rangle_{L^2}. \end{align}\tag{125}\] By the boundedness of \(\partial_q \widetilde{F}_N^x\) and \(\partial_q \overline{F}_{N+1}^x\) as ensured by Proposition 7 and Remark 10, there is some constant \(C_1>0\) such that \[\begin{align} \label{e46124f3912460} \big|\widetilde{f}'(y_i)\big|,\;\big|\overline{f}'(y_i)\big|\leqslant C_1N^{-\gamma/2}a_i|q_i|_{L^2}\stackrel{\eqref{e46a95i61}}{\leqslant}C_1N^{-\gamma/2},\qquad\forall y_i\in [0,3]. \end{align}\tag{126}\] Applying Lemma 15 with \(N^{-\gamma/2}y_i\) substituted for \(r\) and using the chain rule (since \(y_i\in[0,3]\), we only need \(3N^{-\gamma/2}< 1\) to ensure \(N^{-\gamma/2}y_i<1\)), we get \[\begin{align} \label{e46f393960} \widetilde{f}''(y_i),\;\overline{f}''(y_i)\leqslant N^{-\gamma}b,\qquad\forall y_i\in[0,3],\;N\in\mathbb{N}. \end{align}\tag{127}\] By Lemma 13, there is some absolute constant \(C_2\) such that \(\|\widetilde{f}-\overline{f}\|_{L^\infty[0,3]}\leqslant C_2 N^{-\gamma}\). Inserting this, 126 and 127 into Lemma 14, we get \[\begin{align} \int_1^2 \left|\widetilde{f}' - \overline{f}'\right|^2\leqslant C_3\left(C_1N^{-\gamma/2}+N^{-\gamma}b\right)C_2 N^{-\gamma},\qquad\forall N\in\mathbb{N}, \end{align}\] for some constant \(C_3>0\). Comparing 125 with 124 , we have \(\int_1^2 \left|\widetilde{f}' - \overline{f}'\right|^2 = N^{-\gamma}a_i^2 \mathbb{E}_y \mathcal{D}_{N,i}(x,y)\), which together with the above display gives the desired result. ◻

Proof. The boundedness of \(\partial_q \widetilde{F}_N^x\) and \(\partial_q \overline{F}_{N+1}^x\) from ?? , together with 124 , implies that there is an absolute constant \(C>0\) such that \(\mathcal{D}_{N,i}(x,y)\leqslant C|q_i|_{L^2}^2\) uniformly in \(i,N,x,y\). By the choice of \(a_i\) in 120 , this gives \(\left|a_i^2\mathcal{D}_{N,i}(x,y)\right|\leqslant C\) uniformly in \(i,N,x,y\), and therefore justifies interchanging \(\sum_{i=1}^\infty\) with \(\mathbb{E}_{x,y}\) in \(\mathbb{E}_{x,y}\mathcal{D}_N(x,y)\). Combining this with Lemma 16, we obtain \(\mathbb{E}_{x,y} \mathcal{D}_N(x,y)\leqslant CN^{-\gamma/2}\), and hence \[\begin{align} \lim_{N\to\infty} \mathbb{E}_{x,y}\mathcal{D}_N(x,y)=0. \end{align}\] Recall from 113 that \[\begin{align} \Delta_N(x,q+q_N(y)) = \sum_{j=1}^\infty 2^{-j}\Delta_N^{x,q+q_N(y)}(\mathbf{f}_j,n_j,h_j). \end{align}\] By 111 and 112 , we have \(0\leqslant\Delta_N^{x,q+q_N(y)}(\mathbf{f}_j,n_j,h_j)\leqslant 1\) uniformly in \(x,y,j\). By 122 , we have \(|q+q_N(y)|_{L^\infty}\leqslant|q|_{L^\infty}+1\) uniformly in \(N\) and \(y\). Applying Proposition 12, we get \(\lim_{N\to\infty}\mathbb{E}_x\Delta_N(x,y)=0\) for every \(y\). The same bounded-convergence argument as above then yields \[\begin{align} \lim_{N\to\infty}\mathbb{E}_{x,y}\Delta_N(x,y)=0. \end{align}\] Setting \(\mathcal{E}_N(x,y)=\mathcal{D}_N(x,y)+\Delta_N(x,y)\), we have \[\begin{align} \label{e46limEE95N610} \lim_{N\to\infty}\mathbb{E}_{x,y} \mathcal{E}_N(x,y) = 0. \end{align}\tag{134}\] It remains to choose a sequence along which 130 holds and \(\mathcal{E}_N\) vanishes. We use the argument of [5], which we recall in the present notation. From the expression of \(A_N\) in 104 and Jensen’s inequality, there exists a constant \(c>0\) such that \(|A_N(x,y)|\leqslant c\) uniformly in \(N,x,y\). For any \(\varepsilon>0\), define \[\begin{align} \Omega_{\varepsilon,N} = \left\{(x,y):\:A_N(x,y)\leqslant\mathbb{E}_{x,y}A_N(x,y)+\varepsilon\right\}. \end{align}\] Let \(\mathbb{P}_{x,y}\) be the probability measure associated with \(\mathbb{E}_{x,y}\). Then \[\begin{align} \mathbb{E}_{x,y} A_N(x,y) \geqslant\left(\mathbb{E}_{x,y}A_N(x,y)+\varepsilon\right)\mathbb{P}_{x,y}\left(\Omega_{\varepsilon,N}^{\complement}\right)-c \mathbb{P}_{x,y}(\Omega_{\varepsilon,N}), \end{align}\] and therefore \[\begin{align} \mathbb{P}_{x,y}(\Omega_{\varepsilon,N})\geqslant\frac{\varepsilon}{\mathbb{E}_{x,y}A_N(x,y)+\varepsilon+c}>\frac{\varepsilon}{3c},\qquad\forall \varepsilon\in(0,c). \end{align}\] On the other hand, Markov’s inequality gives \[\begin{align} \mathbb{P}_{x,y}\left(\mathcal{E}_N\leqslant\varepsilon\right)\geqslant 1 -\frac{\mathbb{E}_{x,y}\mathcal{E}_N}{\varepsilon}. \end{align}\] Thus \(\Omega_{\varepsilon,N}\cap\left\{\mathcal{E}_N\leqslant\varepsilon\right\}\neq \emptyset\) whenever \(\frac{\mathbb{E}_{x,y}\mathcal{E}_N}{\varepsilon}<\frac{\varepsilon}{3c}\) and \(\varepsilon\in(0,c)\). Taking \(\varepsilon=2(c\mathbb{E}_{x,y}\mathcal{E}_N)^{1/2}\) and using 134 , these two conditions hold for all sufficiently large \(N\). Hence, for such \(N\), we can choose \((x_N,y_N)\) so that \[\begin{align} \mathcal{E}_N(x_N,y_N)\leqslant 2(c\mathbb{E}_{x,y}\mathcal{E}_N)^{1/2}\qquad\text{and}\qquad A_N(x_N,y_N)\leqslant\mathbb{E}_{x,y} A_N(x,y) + 2(c\mathbb{E}_{x,y}\mathcal{E}_N)^{1/2}. \end{align}\] Together with 134 , this proves 130 , 131 , and 132 . The second sequence is obtained by applying the same argument to \(-A_N\) in place of \(A_N\). ◻

Proof. Using the Lipschitz continuity of \(\overline{F}_N(t,\cdot)\) from Proposition 7 and Remark 9, the bound on the difference between \(\overline{F}_N\) and \(\overline{F}_N^x\) in 108 , and the estimate \(|q_N(y)|_{L^\infty}\leqslant N^{-\gamma/2}\), we can find constants \(C,c'>0\) such that \[\begin{align} \label{e46F95N95compare95moving95q95N} \left|\overline{F}_N(t,q)-\overline{F}_N^x(t,q+q_N(y))\right|\leqslant CN^{-c'} \end{align}\tag{136}\] uniformly in \(N,x,y\). Set \[\begin{align} Q_j(y):=q+q_j(y),\qquad j\geqslant 1. \end{align}\] We apply Proposition 11 with the moving path \(Q_j(y)\). For \(1\leqslant j\leqslant N-1\), we have \[\begin{align} &-(j+1)\overline{F}_{j+1}^x(t,Q_{j+1}(y))+j\overline{F}_j^x(t,Q_j(y)) \\ &\qquad=-(j+1)\overline{F}_{j+1}^x(t,Q_j(y))+j\overline{F}_j^x(t,Q_j(y)) -(j+1)\left(\overline{F}_{j+1}^x(t,Q_{j+1}(y))-\overline{F}_{j+1}^x(t,Q_j(y))\right) \\ &\qquad=A_j(x,y)+O\left(j^{-\gamma}\right)+O\left(j\,|q_{j+1}(y)-q_j(y)|_{L^1}\right), \end{align}\] uniformly in \(x\) and \(y\). Here we used the uniform Lipschitz continuity of \(\overline{F}_j^x(t,\cdot)\) and the definition \(A_j(x,y)=A_j(x,Q_j(y))\). Summing over \(1\leqslant j\leqslant N-1\), the left-hand side telescopes and gives \[\begin{align} -N\overline{F}_N^x(t,Q_N(y))+\overline{F}_1^x(t,Q_1(y)) =\sum_{j=1}^{N-1}A_j(x,y)+O\left(\sum_{j=1}^{N-1}j^{-\gamma}\right)+O\left(\sum_{j=1}^{N-1}j\,|q_{j+1}(y)-q_j(y)|_{L^1}\right). \end{align}\] By the definition of \(q_j(y)\) in 121 , we can write \(q_j(y)=j^{-\gamma/2}B(y)\) with \(|B(y)|_{L^1}\) uniformly bounded in \(y\). Hence \[\begin{align} |q_{j+1}(y)-q_j(y)|_{L^1}\leqslant C\left(j^{-\gamma/2}-(j+1)^{-\gamma/2}\right)\leqslant Cj^{-1-\gamma/2}. \end{align}\] It follows that \[\begin{align} \frac{1}{N}\sum_{j=1}^{N-1}j^{-\gamma}=O(N^{-\gamma})=o(1) \end{align}\] and \[\begin{align} \frac{1}{N}\sum_{j=1}^{N-1}j\,|q_{j+1}(y)-q_j(y)|_{L^1}\leqslant\frac{C}{N}\sum_{j=1}^{N-1}j^{-\gamma/2}=O(N^{-\gamma/2})=o(1). \end{align}\] Since \(\overline{F}_1^x(t,Q_1(y))\) is uniformly bounded, we obtain \[\begin{align} \overline{F}_N^x(t,q+q_N(y))=-\frac{1}{N}\sum_{j=1}^{N-1}A_j(x,y)+o(1), \end{align}\] uniformly in \(x\) and \(y\). Combining this with 136 , and then averaging over \((x,y)\), gives 135 . ◻

Proof of Theorem 3. Fix \(t>0\). We first prove the upper bound in 117 for \(q\in\mathcal{Q}_{\uparrow,c}^{D}\cap \mathcal{Q}_\infty^{D}\subseteq\mathcal{Q}_{\infty,\uparrow}^{D}\) (see 54 ) with some \(c>0\), and then obtain the general case by continuity.

Let \((x_N,y_N)_{N\in\mathbb{N}}\) be the sequence in \([0,3]^{\mathbb{N}^4}\times [0,3]^\mathbb{N}\) given by Lemma 17, so that 130 , 131 , and 132 hold. Let \((N_k)_{k\in\mathbb{N}}\) be a subsequence along which \(A_N(x_N,y_N)\) attains the liminf in 130 . Then \[\begin{align} \limsup_{N\to\infty}\overline{F}_N(t,q) &\stackrel{\eqref{e46F95N61-147NsumA95j}}{\leqslant} \limsup_{N\to\infty}-\frac{1}{N}\sum_{j=1}^{N-1}\mathbb{E}_{x,y}A_j(x,y) \notag\\ &\leqslant\limsup_{N\to\infty}-\mathbb{E}_{x,y}A_N(x,y) \stackrel{\eqref{e46liminfA60liminfEA2}}{\leqslant} \limsup_{N\to\infty}-A_N(x_N,y_N) = \lim_{k\to\infty}-A_{N_k}(x_{N_k},y_{N_k}). \label{e46limsupF60-limA95k2} \end{align}\tag{137}\] For the rest of this part, set \[\begin{align} q_k := q+q_{N_k}(y_{N_k}),\qquad k\in\mathbb{N}. \end{align}\] For each continuous path \(\kappa\in\mathcal{Q}_\infty^{D}\), the derivative formulas 101 and 102 give \[\begin{align} \left\langle\kappa,\,\partial_q\widetilde{F}_{N_k}^{x_{N_k}}\left(t,q_k\right)\right\rangle_{L^2}&= \mathbb{E}\left\langle\kappa(\alpha\wedge\alpha')\cdot \mathsf{diag}\left(\tfrac{1}{N_k}\sigma\sigma'^\intercal\right)\right\rangle_{N_k,x_{N_k},q_k}^\circ,\\ \left\langle\kappa,\,\partial_q\overline{F}_{N_k+1}^{x_{N_k}}(t,q_k)\right\rangle_{L^2}&= \mathbb{E}\left\langle\kappa(\alpha\wedge\alpha')\cdot \mathsf{diag}(\tau\tau'^\intercal)\right\rangle_{N_k+1,x_{N_k},q_k}, \end{align}\] where \(\tau\) in the second line is the last vector spin at size \(N_k+1\), and the Gibbs measures are defined in 99 and 97 , respectively.

By 122 , the sequence \(q_k\) converges pointwise to \(q\). Together with 131 , this allows us to apply Proposition 13. Passing to a further subsequence, still denoted by \((N_k)_{k\in\mathbb{N}}\), we can find \(p\in \mathcal{Q}_{\infty,\leqslant 1}^{D}\) such that, for every continuous path \(\kappa\in\mathcal{Q}_\infty^{D}\), \[\begin{align} -\lim_{k\to\infty}A_{N_k}(x_{N_k},y_{N_k}) & = \mathscr{P}_{t,q}(p), \tag{138}\\ \lim_{k\to\infty}\left\langle\kappa,\,\partial_q\widetilde{F}_{N_k}^{x_{N_k}}(t,q_k)\right\rangle_{L^2}&= \mathbb{E}\left\langle\kappa(\alpha\wedge\alpha')\cdot p(\alpha\wedge\alpha')\right\rangle_{\mathfrak R}=\left\langle\kappa, p \right\rangle_{L^2}, \tag{139}\\ \lim_{k\to\infty}\left\langle\kappa,\,\partial_q\overline{F}_{N_k+1}^{x_{N_k}}(t,q_k)\right\rangle_{L^2}&= \mathbb{E}\left\langle\kappa(\alpha\wedge\alpha')\cdot \mathsf{diag}(\tau\tau'^\intercal)\right\rangle_{\mathfrak R,\, q +t \nabla\xi(p)}\stackrel{\eqref{e46dqpsi61}}{=}\left\langle\kappa, \partial_q\psi(q+t\nabla\xi(p))\right\rangle_{L^2}. \tag{140} \end{align}\] The desired upper bound in 117 , with \(p^+=p\), follows from 137 and 138 .

It remains to verify that \(p\) satisfies the critical relation in 116 . Since each \(\mathcal{D}_{N,i}(x_N,y_N)\) is nonnegative, 124 and 132 imply that \(\lim_{N\to\infty}\mathcal{D}_{N,i}(x_N,y_N)=0\) for every \(i\in\mathbb{N}\). Hence, \[\begin{align} \lim_{N\to\infty} \left\langle q_i,\;\partial_q \widetilde{F}_N^{x_N}(t, q+q_N(y_N)) - \partial_q \overline{F}_{N+1}^{x_N}(t, q+q_N(y_N))\right\rangle_{L^2} =0,\qquad\forall i \in \mathbb{N}. \end{align}\] Combining this with 139 and 140 , with \(\kappa=q_i\), gives \[\begin{align} \left\langle q_i, p\right\rangle_{L^2} = \left\langle q_i, \partial_q\psi(q+t\nabla\xi(p))\right\rangle_{L^2},\qquad\forall i \in\mathbb{N}. \end{align}\] By 119 , we obtain \[\begin{align} p=\partial_q \psi(q+t\nabla\xi(p)), \end{align}\] which is the critical relation in 116 with \(p^+=p\).

We have so far assumed that \(q\in\mathcal{Q}_{\infty,\uparrow}^{D}\) for some \(c>0\). For a general \(q\in\mathcal{Q}_1^{D}\), choose a sequence \((q_n)_{n\in\mathbb{N}}\) such that \(q_n\in \mathcal{Q}_{\infty,\uparrow}^{D}\) with \(q_n\to q\) in \(L^1\). By the previous argument, there is an associated sequence \((p_n)_{n\in\mathbb{N}}\) such that \((q_n,p_n)\) satisfies 116 and gives the upper bound on \(\limsup_{N\to\infty}\overline{F}_N(t,q_n)\) in 117 . By the compactness of paths in Lemma 7, after passing to a subsequence, we may assume that \(p_n\) converges in \(L^1\) to some \(p\). Using the Lipschitz continuity of \(\overline{F}_N\) from Proposition 7 and the continuity estimates in Lemma 10, we can send \(n\to\infty\) and obtain both the upper bound in 117 and the critical relation in 116 for this \(p\) at the original path \(q\). This completes the proof of the upper bound.

The lower bound in 117 is proved in the same way. The only change is to use the sequence \((x'_N,y'_N)\) from Lemma 17, which gives the lower bound in 133 , and then reverse the corresponding inequalities in 137 . ◻

Proof. Fix a regularization and let \(\mathsf{H}\) be the associated Lipschitz Hamiltonian. Denote by \(L_\mathsf{H}\) its Lipschitz constant, and let \(\mathbf{1}\in L^2([0,1),\mathbb{R}^\mathscr{S})\) be the constant path with all coordinates equal to one. Set \(C_\mathsf{H}:=L_\mathsf{H}|\mathbf{1}|_{L^2}\). We show that \[v^+(t,q):=v(t,q)+\rho\ell(q)+\rho C_\mathsf{H}t\] is a viscosity supersolution. Indeed, if a smooth test function \(\phi\) touches \(v^+\) from below at \((t,q)\), then \(\phi-\rho\ell-\rho C_\mathsf{H}t\) touches \(v\) from below at \((t,q)\). Hence \[\partial_t\phi(t,q)-\rho C_\mathsf{H}-\mathsf{H}\left(\partial_q\phi(t,q)-\rho\mathbf{1}\right)\geqslant 0.\] Using the Lipschitz continuity of \(\mathsf{H}\), we get \[\begin{align} \partial_t\phi(t,q)-\mathsf{H}(\partial_q\phi(t,q)) &\geqslant\rho C_\mathsf{H}+\mathsf{H}\left(\partial_q\phi(t,q)-\rho\mathbf{1}\right)-\mathsf{H}(\partial_q\phi(t,q)) \\ &\geqslant\rho C_\mathsf{H}-\rho L_\mathsf{H}|\mathbf{1}|_{L^2}=0. \end{align}\] Thus \(v^+\) is a supersolution. Similarly, \[v^-(t,q):=v(t,q)-\rho\ell(q)-\rho C_\mathsf{H}t\] is a subsolution. By 147 , we have \(v^-(0,\cdot)\leqslant u(0,\cdot)\leqslant v^+(0,\cdot)\). The comparison principle therefore gives \(v^-\leqslant u\leqslant v^+\), which is exactly 148 . ◻

Proof of Proposition 18. We set \(\mathcal{C}_{D}:=\mathcal{Q}_2^{D}\), \(\mathcal{C}_\mathscr{S}:=\mathcal{Q}_2^\mathscr{S}\), \(\mathcal{H}_{D}:=L^2([0,1),\mathbb{R}^{D})\), and \(\mathcal{H}_\mathscr{S}:=L^2([0,1),\mathbb{R}^\mathscr{S})\). Define bounded linear maps \(A,\Pi:\mathcal{H}_{D}\to\mathcal{H}_\mathscr{S}\) by \[\begin{align} (Aq)_s=\frac{1}{|\mathsf{D}_s|}\sum_{d\in\mathsf{D}_s}q_d, \qquad (\Pi q)_s=\frac{1}{{D}}\sum_{d\in\mathsf{D}_s}q_d . \end{align}\] Thus \(Aq=q^{\mathrm{avg}}\). The adjoint of \(A\) is given by \((A^*\eta)_d=\eta_s/|\mathsf{D}_s|\) for \(d\in\mathsf{D}_s\), and \(\Pi({D}A^*\eta)=\eta\).

Fix a regularization \(\overline{\xi}\) of \(\xi\), and define \(\overline{\xi}^{\mathrm{vec}}:\mathbb{R}^{D}_+\to\mathbb{R}\) by \[\begin{align} \overline{\xi}^{\mathrm{vec}}(a):={D}\overline{\xi}(\Pi a). \end{align}\] Then \(\overline{\xi}^{\mathrm{vec}}\) is a regularization of \(\xi^{\mathrm{vec}}\). Let \(\mathsf{H}^\mathscr{S}\) and \(\mathsf{H}^{\mathrm{vec}}\) be the Hamiltonians associated with \(\overline{\xi}\) and \(\overline{\xi}^{\mathrm{vec}}\), respectively.

Step 1. Identification of the averaged Hamiltonian. We claim that \[\begin{align} \label{e46avg95H95identity} \mathsf{H}^{\mathrm{vec}}({D}A^*\eta)={D}\mathsf{H}^\mathscr{S}(\eta), \qquad \eta\in\mathcal{H}_\mathscr{S}. \end{align}\tag{149}\] Indeed, if \(r\in \mathcal{C}_{D}\cap({D}A^*\eta+\mathcal{C}_{D}^*)\), then \(\Pi r\in \mathcal{C}_\mathscr{S}\). Moreover, for every \(h\in \mathcal{C}_\mathscr{S}\), \[\begin{align} \left\langle\Pi r-\eta,h\right\rangle_{\mathcal{H}_\mathscr{S}} = \left\langle r-{D}A^*\eta, \left(\frac{h_s}{{D}}\right)_{d\in\mathsf{D}_s} \right\rangle_{\mathcal{H}_{D}} \geqslant 0 . \end{align}\] Hence \(\Pi r\in \mathcal{C}_\mathscr{S}\cap(\eta+\mathcal{C}_\mathscr{S}^*)\), and so \[\begin{align} \int_0^1\overline{\xi}^{\mathrm{vec}}(r(s))\,\mathrm{d}s = {D}\int_0^1\overline{\xi}(\Pi r(s))\,\mathrm{d}s \stackrel{\eqref{e46def95H95spin95glass}}{\geqslant} {D}\mathsf{H}^\mathscr{S}(\eta). \end{align}\] Taking the infimum over \(r\) gives \(\mathsf{H}^{\mathrm{vec}}({D}A^*\eta)\geqslant{D}\mathsf{H}^\mathscr{S}(\eta)\). Conversely, if \(p\in \mathcal{C}_\mathscr{S}\cap(\eta+\mathcal{C}_\mathscr{S}^*)\), then \(r:={D}A^*p\) belongs to \(\mathcal{C}_{D}\), satisfies \(\Pi r=p\), and for every \(k\in \mathcal{C}_{D}\), \[\begin{align} \left\langle r-{D}A^*\eta,k\right\rangle_{\mathcal{H}_{D}} = {D}\left\langle p-\eta,Ak\right\rangle_{\mathcal{H}_\mathscr{S}} \geqslant 0 . \end{align}\] Thus \(r\in \mathcal{C}_{D}\cap({D}A^*\eta+\mathcal{C}_{D}^*)\). Taking the infimum over \(p\) gives the reverse inequality, and proves 149 .

Step 2. The subsolution inequality. Let \(\phi\) be a smooth test function such that \(g-\phi\) has a local maximum at \((t_0,q_0)\in(0,\infty)\times \mathcal{C}_{D}\). Replacing \(\phi\) by \(\phi+|t-t_0|^2+|q-q_0|_{\mathcal{H}_{D}}^2\), we may assume that the maximum is strict. Thus, for some \(r\in(0,t_0)\), \[\begin{align} \label{e46avg95strict95max} g(t,q)-\phi(t,q) \leqslant g(t_0,q_0)-\phi(t_0,q_0) - |(t,q)-(t_0,q_0)|_{\mathbb{R}\times\mathcal{H}_{D}}^2 \end{align}\tag{150}\] on \(K_r^{D}:=B_r(t_0,q_0)\cap(\mathbb{R}_+\times \mathcal{C}_{D})\). Put \(p_0:=Aq_0\), and choose \(R>0\) such that \(Aq\in K_R^\mathscr{S}:=B_R(p_0)\cap \mathcal{C}_\mathscr{S}\) whenever \((t,q)\in K_r^{D}\). Here, \(B_r(t_0,q_0)\) and \(B_R(p_0)\) are metric balls defined in the obvious way.

For \(\varepsilon>0\), define on \(K_r^{D}\times K_R^\mathscr{S}\) \[\begin{align} \label{e46avg95Phi95eps} \Phi_\varepsilon(t,q,p) := {D}u(t,p)-\phi(t,q)-\frac{{D}}{2\varepsilon}|p-Aq|_{\mathcal{H}_\mathscr{S}}^2 . \end{align}\tag{151}\] By Stegall’s variational principle (Theorem 6), there is \(\zeta_\varepsilon=(\tau_\varepsilon,\iota_\varepsilon,\varrho_\varepsilon)\in\mathbb{R}\times\mathcal{H}_{D}\times\mathcal{H}_\mathscr{S}\) with \(|\zeta_\varepsilon|\leqslant\varepsilon^2\) such that \[\begin{align} \Phi_\varepsilon(t,q,p)+\tau_\varepsilon t+\left\langle\iota_\varepsilon,q\right\rangle_{\mathcal{H}_{D}} +\left\langle\varrho_\varepsilon,p\right\rangle_{\mathcal{H}_\mathscr{S}} \end{align}\] attains its maximum at some \((t_\varepsilon,q_\varepsilon,p_\varepsilon)\in K_r^{D}\times K_R^\mathscr{S}\).

Set \(d_\varepsilon:=|p_\varepsilon-Aq_\varepsilon|_{\mathcal{H}_\mathscr{S}}\) and \(\Delta_\varepsilon:=|(t_\varepsilon,q_\varepsilon)-(t_0,q_0)|_{\mathbb{R}\times\mathcal{H}_{D}}\). Since \((t_\varepsilon,q_\varepsilon,p_\varepsilon)\) maximizes the perturbed functional and \(p_0=Aq_0\), comparison with \((t_0,q_0,p_0)\) gives \[\begin{align} 0 &\leqslant\Phi_\varepsilon(t_\varepsilon,q_\varepsilon,p_\varepsilon)-\Phi_\varepsilon(t_0,q_0,p_0) +\tau_\varepsilon(t_\varepsilon-t_0)+\left\langle\iota_\varepsilon,q_\varepsilon-q_0\right\rangle_{\mathcal{H}_{D}} +\left\langle\varrho_\varepsilon,p_\varepsilon-p_0\right\rangle_{\mathcal{H}_\mathscr{S}} \\ &= {D}u(t_\varepsilon,p_\varepsilon)-\phi(t_\varepsilon,q_\varepsilon) -\frac{{D}}{2\varepsilon}d_\varepsilon^2-{D}u(t_0,p_0)+\phi(t_0,q_0) \\ &\qquad +\tau_\varepsilon(t_\varepsilon-t_0)+\left\langle\iota_\varepsilon,q_\varepsilon-q_0\right\rangle_{\mathcal{H}_{D}} +\left\langle\varrho_\varepsilon,p_\varepsilon-p_0\right\rangle_{\mathcal{H}_\mathscr{S}}. \end{align}\] Adding and subtracting \({D}u(t_\varepsilon,Aq_\varepsilon)\), and using \(g(t,q)={D}u(t,Aq)\) and \(p_0=Aq_0\), we obtain \[\begin{align} 0 &\leqslant\left(g(t_\varepsilon,q_\varepsilon)-\phi(t_\varepsilon,q_\varepsilon)\right) -\left(g(t_0,q_0)-\phi(t_0,q_0)\right) +{D}\left(u(t_\varepsilon,p_\varepsilon)-u(t_\varepsilon,Aq_\varepsilon)\right) \\ &\qquad -\frac{{D}}{2\varepsilon}d_\varepsilon^2 +\tau_\varepsilon(t_\varepsilon-t_0)+\left\langle\iota_\varepsilon,q_\varepsilon-q_0\right\rangle_{\mathcal{H}_{D}} +\left\langle\varrho_\varepsilon,p_\varepsilon-p_0\right\rangle_{\mathcal{H}_\mathscr{S}}. \end{align}\] By the strict maximum condition 150 , \[\begin{align} \left(g(t_\varepsilon,q_\varepsilon)-\phi(t_\varepsilon,q_\varepsilon)\right) -\left(g(t_0,q_0)-\phi(t_0,q_0)\right) \leqslant-\Delta_\varepsilon^2 . \end{align}\] Moreover, \(p_\varepsilon,Aq_\varepsilon\in K_R^\mathscr{S}\), so the Lipschitz continuity of \(u\) on bounded sets gives \[\begin{align} {D}\left(u(t_\varepsilon,p_\varepsilon)-u(t_\varepsilon,Aq_\varepsilon)\right) \leqslant C|p_\varepsilon-Aq_\varepsilon|_{\mathcal{H}_\mathscr{S}}=Cd_\varepsilon. \end{align}\] Since \(|\zeta_\varepsilon|\leqslant\varepsilon^2\) and all the points considered stay in the fixed bounded set \(K_r^{D}\times K_R^\mathscr{S}\), the linear perturbation is bounded by \[\begin{align} \left| \tau_\varepsilon(t_\varepsilon-t_0)+\left\langle\iota_\varepsilon,q_\varepsilon-q_0\right\rangle_{\mathcal{H}_{D}} +\left\langle\varrho_\varepsilon,p_\varepsilon-p_0\right\rangle_{\mathcal{H}_\mathscr{S}} \right| \leqslant C\varepsilon^2 . \end{align}\] Combining the previous estimates yields \[\begin{align} 0\leqslant-\Delta_\varepsilon^2+Cd_\varepsilon-\frac{{D}}{2\varepsilon}d_\varepsilon^2+C\varepsilon^2 . \end{align}\] Equivalently, \[\begin{align} \label{e46avg95convergence95est} |(t_\varepsilon,q_\varepsilon)-(t_0,q_0)|_{\mathbb{R}\times\mathcal{H}_{D}}^2 +\frac{{D}}{2\varepsilon}|p_\varepsilon-Aq_\varepsilon|_{\mathcal{H}_\mathscr{S}}^2 \leqslant C|p_\varepsilon-Aq_\varepsilon|_{\mathcal{H}_\mathscr{S}}+C\varepsilon^2 . \end{align}\tag{152}\] We now deduce convergence. By Young’s inequality, \[\begin{align} Cd_\varepsilon\leqslant\frac{{D}}{4\varepsilon}d_\varepsilon^2+C\varepsilon. \end{align}\] Using this in 152 and absorbing the term \(\frac{{D}}{4\varepsilon}d_\varepsilon^2\) into the left-hand side gives \[\begin{align} \Delta_\varepsilon^2+\frac{{D}}{4\varepsilon}d_\varepsilon^2\leqslant C\varepsilon+C\varepsilon^2 . \end{align}\] Hence \(\Delta_\varepsilon\to0\) and \(d_\varepsilon\to0\), that is, \[\begin{align} \label{e46avg95convergence} \lim_{\varepsilon\to0}(t_\varepsilon,q_\varepsilon)=(t_0,q_0), \qquad \lim_{\varepsilon\to0} p_\varepsilon-Aq_\varepsilon=0\quad\text{in }\mathcal{H}_\mathscr{S}. \end{align}\tag{153}\] Since \(A\) is bounded, this also implies \(Aq_\varepsilon\to Aq_0=p_0\) and therefore \(p_\varepsilon\to p_0\).

For small \(\varepsilon\), the maximum is therefore local relative to the cones. Keeping \(q=q_\varepsilon\) fixed, \(u\) is touched from above at \((t_\varepsilon,p_\varepsilon)\) by \[\begin{align} \chi_\varepsilon(t,p) := \frac{1}{{D}}\phi(t,q_\varepsilon) + \frac{1}{2\varepsilon}|p-Aq_\varepsilon|_{\mathcal{H}_\mathscr{S}}^2 - \frac{\tau_\varepsilon}{{D}}t - \frac{1}{{D}}\left\langle\varrho_\varepsilon,p\right\rangle_{\mathcal{H}_\mathscr{S}}. \end{align}\] At \((t_\varepsilon,p_\varepsilon)\), \[\begin{align} \label{e46avg95chi95derivatives} \partial_t\chi_\varepsilon = \frac{1}{{D}}\left(\phi_t(t_\varepsilon,q_\varepsilon)-\tau_\varepsilon\right), \qquad \partial_p\chi_\varepsilon = \frac{p_\varepsilon-Aq_\varepsilon}{\varepsilon} - \frac{\varrho_\varepsilon}{{D}} =:\eta_\varepsilon. \end{align}\tag{154}\] Since \(u\) is a viscosity subsolution, \[\begin{align} \label{e46avg95u95sub} \phi_t(t_\varepsilon,q_\varepsilon)-\tau_\varepsilon-{D}\mathsf{H}^\mathscr{S}(\eta_\varepsilon)\leqslant 0 . \end{align}\tag{155}\]

Keeping instead \(t=t_\varepsilon\) and \(p=p_\varepsilon\) fixed, \(q_\varepsilon\) minimizes over a metric ball centered at \(q_\varepsilon\) in \(\mathcal{C}_{D}\) the function \[\begin{align} q\mapsto \phi(t_\varepsilon,q) + \frac{{D}}{2\varepsilon}|p_\varepsilon-Aq|_{\mathcal{H}_\mathscr{S}}^2 - \left\langle\iota_\varepsilon,q\right\rangle_{\mathcal{H}_{D}}. \end{align}\] Denote this function by \(\ell(q)\). Then, this minimality implies that \(\frac{\mathrm{d}}{\mathrm{d}\delta}\left(\ell(q_\varepsilon+\delta q')-\ell(q_\varepsilon)\right)\big|_{\delta=0^+}\geqslant 0\) and thus we have \(\left\langle q', \partial_q \ell(q_\varepsilon)\right\rangle_{\mathcal{H}_{D}}\geqslant 0\) for every \(q'\) in the convex cone \(\mathcal{C}_{D}\). Notice that we used the fact that \(q_\varepsilon+\delta q'\in \mathcal{C}_{D}\) for every \(\delta>0\) since \(\mathcal{C}_{D}\) is a cone. By the definition of the dual cone in 142 , we conclude that \(\partial_q \ell(q_\varepsilon)\in \mathcal{C}_{D}^*\). Equivalently, we get \[\begin{align} \label{e46avg95normal95sub} \partial_q\phi(t_\varepsilon,q_\varepsilon) - {D}A^*\eta_\varepsilon - A^*\varrho_\varepsilon - \iota_\varepsilon \in \mathcal{C}_{D}^* . \end{align}\tag{156}\] Since \(\mathsf{H}^{\mathrm{vec}}\) is \(\mathcal{C}_{D}^*\)-increasing, 149 and 156 imply \[\begin{align} \mathsf{H}^{\mathrm{vec}}\left(\partial_q\phi(t_\varepsilon,q_\varepsilon)-A^*\varrho_\varepsilon-\iota_\varepsilon\right) \geqslant \mathsf{H}^{\mathrm{vec}}({D}A^*\eta_\varepsilon) = {D}\mathsf{H}^\mathscr{S}(\eta_\varepsilon). \end{align}\] Together with 155 , this gives \[\begin{align} \phi_t(t_\varepsilon,q_\varepsilon)-\tau_\varepsilon - \mathsf{H}^{\mathrm{vec}}\left(\partial_q\phi(t_\varepsilon,q_\varepsilon)-A^*\varrho_\varepsilon-\iota_\varepsilon\right) \leqslant 0 . \end{align}\] Letting \(\varepsilon\to0\), using 153 , the continuity of the differential of \(\phi\), the Lipschitz continuity of \(\mathsf{H}^{\mathrm{vec}}\), and \(\zeta_\varepsilon\to0\), yields \[\begin{align} \phi_t(t_0,q_0)-\mathsf{H}^{\mathrm{vec}}(\partial_q\phi(t_0,q_0))\leqslant 0 . \end{align}\] Thus \(g\) is a viscosity subsolution.

Step 3. The supersolution inequality. The supersolution argument is the same with the signs reversed, so we only record the changes. Suppose that \(g-\phi\) has a local minimum at \((t_0,q_0)\). After replacing \(\phi\) by \(\phi-|t-t_0|^2-|q-q_0|_{\mathcal{H}_{D}}^2\), we may assume that the minimum is strict. Apply Stegall’s principle (Theorem 6) to \[\begin{align} \label{e46avg95Psi95eps} \Psi_\varepsilon(t,q,p) := \phi(t,q)-{D}u(t,p)-\frac{{D}}{2\varepsilon}|p-Aq|_{\mathcal{H}_\mathscr{S}}^2 . \end{align}\tag{157}\] The same comparison as above gives 153 . At the maximizer \((t_\varepsilon,q_\varepsilon,p_\varepsilon)\), keeping \(q=q_\varepsilon\) fixed, \(u\) is touched from below by \[\begin{align} \widetilde{\chi}_\varepsilon(t,p) := \frac{1}{{D}}\phi(t,q_\varepsilon) - \frac{1}{2\varepsilon}|p-Aq_\varepsilon|_{\mathcal{H}_\mathscr{S}}^2 + \frac{\tau_\varepsilon}{{D}}t + \frac{1}{{D}}\left\langle\varrho_\varepsilon,p\right\rangle_{\mathcal{H}_\mathscr{S}}. \end{align}\] Thus, with \[\begin{align} \label{e46avg95chitilde95derivatives} \partial_t\widetilde{\chi}_\varepsilon = \frac{1}{{D}}\left(\phi_t(t_\varepsilon,q_\varepsilon)+\tau_\varepsilon\right), \qquad \partial_p\widetilde{\chi}_\varepsilon = -\frac{p_\varepsilon-Aq_\varepsilon}{\varepsilon} + \frac{\varrho_\varepsilon}{{D}} =:\eta_\varepsilon, \end{align}\tag{158}\] the viscosity supersolution property of \(u\) gives \[\begin{align} \label{e46avg95u95super} \phi_t(t_\varepsilon,q_\varepsilon)+\tau_\varepsilon-{D}\mathsf{H}^\mathscr{S}(\eta_\varepsilon)\geqslant 0 . \end{align}\tag{159}\] The first-order condition in the \(q\) variable is now \[\begin{align} \label{e46avg95normal95super} {D}A^*\eta_\varepsilon - \partial_q\phi(t_\varepsilon,q_\varepsilon) - A^*\varrho_\varepsilon - \iota_\varepsilon \in \mathcal{C}_{D}^* . \end{align}\tag{160}\] Using again the monotonicity of \(\mathsf{H}^{\mathrm{vec}}\) and 149 , we obtain \[\begin{align} {D}\mathsf{H}^\mathscr{S}(\eta_\varepsilon) = \mathsf{H}^{\mathrm{vec}}({D}A^*\eta_\varepsilon) \geqslant \mathsf{H}^{\mathrm{vec}}\left(\partial_q\phi(t_\varepsilon,q_\varepsilon)+A^*\varrho_\varepsilon+\iota_\varepsilon\right). \end{align}\] Together with 159 , this gives \[\begin{align} \phi_t(t_\varepsilon,q_\varepsilon)+\tau_\varepsilon - \mathsf{H}^{\mathrm{vec}}\left(\partial_q\phi(t_\varepsilon,q_\varepsilon)+A^*\varrho_\varepsilon+\iota_\varepsilon\right) \geqslant 0 . \end{align}\] Letting \(\varepsilon\to0\) yields \[\begin{align} \phi_t(t_0,q_0)-\mathsf{H}^{\mathrm{vec}}(\partial_q\phi(t_0,q_0))\geqslant 0 . \end{align}\] Thus \(g\) is a viscosity supersolution.

Step 4. Initial condition and uniqueness. We have proved that \(g\) is a viscosity solution of the Hamilton–Jacobi equation associated with the regularization \(\overline{\xi}^{\mathrm{vec}}\). Since the notion of viscosity solution is independent of the choice of regularization, \(g\) solves \[\begin{align} \partial_t g-\int_0^1\xi^{\mathrm{vec}}(\partial_q g)=0 \qquad\text{on }\mathbb{R}_+\times \mathcal{C}_{D}. \end{align}\] The initial condition is immediate: \[\begin{align} g(0,q)={D}u(0,Aq)={D}\psi(Aq)={D}\psi(q^{\mathrm{avg}}), \qquad q\in \mathcal{C}_{D}. \end{align}\] Finally, \(q\mapsto {D}\psi(Aq)\) is Lipschitz because \(A\) is bounded. It is also \(\mathcal{C}_{D}^*\)-increasing: if \(q'-q\in \mathcal{C}_{D}^*\), then for every \(h\in \mathcal{C}_\mathscr{S}\), \[\begin{align} \left\langle A(q'-q),h\right\rangle_{\mathcal{H}_\mathscr{S}} = \left\langle q'-q,A^*h\right\rangle_{\mathcal{H}_{D}} \geqslant 0 , \end{align}\] since \(A^*h\in \mathcal{C}_{D}\). Hence \(A(q'-q)\in \mathcal{C}_\mathscr{S}^*\), and the monotonicity of \(\psi\) gives the claim. By Theorem 4, with the normalization remark if necessary, \(g\) is the unique Lipschitz viscosity solution with initial condition \(g(0,q)={D}\psi(q^{\mathrm{avg}})\). ◻

Proof. We first prove the result when \(\lambda_\infty\) is rational.

Step 1. The rational case with \(q\in\mathcal{Q}_\infty^\mathscr{S}\). Assume that there are \({D}\in\mathbb{N}\) and a weak partition \((\mathsf{D}_s)_{s\in\mathscr{S}}\) of \(\{1,\ldots,{D}\}\) such that \(\lambda_{\infty,s}=|\mathsf{D}_s|/{D}\) for every \(s\in\mathscr{S}\). Let \(\xi^{\mathrm{vec}}\), \(\psi^{\mathrm{vec}}\), and \(\overline{F}_N^{\mathrm{vec}}\) be the associated vector spin model. Fix \(t>0\) and \(q\in\mathcal{Q}_\infty^\mathscr{S}\), and set \(\underline q:=q^{\mathrm{vec}}\).

Applying Theorem 3 to the vector spin model at \((t,\underline q)\), we find \(\underline p^+,\underline p^-\in\mathcal{Q}_{\infty,\leqslant 1}^{D}\) such that \[\begin{align} \label{e46crit95bdd95ms95vec95crit} \underline p^\pm=\partial_{\underline q}\psi^{\mathrm{vec}}\left(\underline q+t\nabla\xi^{\mathrm{vec}}(\underline p^\pm)\right) \end{align}\tag{161}\] and \[\begin{align} \label{e46crit95bdd95ms95vec95bd} \mathscr{P}^{\mathrm{vec}}_{t,\underline q}(\underline p^-)\leqslant\liminf_{N\to\infty}\overline{F}_N^{\mathrm{vec}}(t,\underline q)\leqslant\limsup_{N\to\infty}\overline{F}_N^{\mathrm{vec}}(t,\underline q)\leqslant\mathscr{P}^{\mathrm{vec}}_{t,\underline q}(\underline p^+). \end{align}\tag{162}\] Define \(p^\pm:=(\underline p^\pm)^{\mathrm{sum}}\). Since \(\underline p^\pm\in\mathcal{Q}_{\infty,\leqslant 1}^{D}\), we have \(p^\pm\in\mathcal{Q}_{\infty,\leqslant\lambda_\infty}^\mathscr{S}\). By 92 and 161 , \[\begin{align} \label{e46crit95bdd95ms95crit95rat} p^\pm=\partial_q\psi\left(q+t\nabla\xi(p^\pm)\right). \end{align}\tag{163}\] Moreover, by Lemma 11, \[\begin{align} \label{e46crit95bdd95ms95free95energy95equiv} \lim_{N\to\infty}\left|\overline{F}_N(t,q)-{D}^{-1}\overline{F}^{\mathrm{vec}}_{\lceil N/{D}\rceil}(t,q^{\mathrm{vec}})\right|=0. \end{align}\tag{164}\] Combining 162 , 164 , and 91 , we obtain \[\begin{align} \label{e46crit95bdd95ms95bd95rat95qinf} \mathscr{P}_{t,q}(p^-)\leqslant\liminf_{N\to\infty}\overline{F}_N(t,q)\leqslant\limsup_{N\to\infty}\overline{F}_N(t,q)\leqslant\mathscr{P}_{t,q}(p^+). \end{align}\tag{165}\] This proves the rational case for \(q\in\mathcal{Q}_\infty^\mathscr{S}\).

Step 2. Extension of the rational case to \(q\in\mathcal{Q}_1^\mathscr{S}\). Let now \(q\in\mathcal{Q}_1^\mathscr{S}\), and choose \(q_k\in\mathcal{Q}_\infty^\mathscr{S}\) such that \(q_k\to q\) in \(L^1\). By Step 1, for each \(k\) there exist \(p_k^+,p_k^-\in\mathcal{Q}_{\infty,\leqslant\lambda_\infty}^\mathscr{S}\) satisfying \[\begin{align} \label{e46crit95bdd95ms95crit95k} p_k^\pm=\partial_q\psi\left(q_k+t\nabla\xi(p_k^\pm)\right) \end{align}\tag{166}\] and \[\begin{align} \label{e46crit95bdd95ms95bd95k} \mathscr{P}_{t,q_k}(p_k^-)\leqslant\liminf_{N\to\infty}\overline{F}_N(t,q_k)\leqslant\limsup_{N\to\infty}\overline{F}_N(t,q_k)\leqslant\mathscr{P}_{t,q_k}(p_k^+). \end{align}\tag{167}\] By the compactness of monotone paths (see Lemma 7), after passing to subsequences, we may assume that \(p_k^\pm\to p^\pm\) in \(L^1\) for some \(p^\pm\in\mathcal{Q}_{\infty,\leqslant\lambda_\infty}^\mathscr{S}\). Lemma 10 and 166 give \[\begin{align} \label{e46crit95bdd95ms95crit95rat95q1} p^\pm=\partial_q\psi\left(q+t\nabla\xi(p^\pm)\right). \end{align}\tag{168}\] Using the Lipschitz continuity of \(\overline{F}_N\) in Proposition 7, 167 , and Lemma 10, and then sending \(k\to\infty\), gives \[\begin{align} \label{e46crit95bdd95ms95bd95rat95q1} \mathscr{P}_{t,q}(p^-)\leqslant\liminf_{N\to\infty}\overline{F}_N(t,q)\leqslant\limsup_{N\to\infty}\overline{F}_N(t,q)\leqslant\mathscr{P}_{t,q}(p^+). \end{align}\tag{169}\] This proves the proposition when \(\lambda_\infty\) is rational.

Step 3. Approximation of the proportions. We now consider a general \(\lambda_\infty\). Let \((\lambda^{(m)})_{m\in\mathbb{N}}\) be a sequence of rational probability vectors such that \(\lambda^{(m)}\to\lambda_\infty\). Let \(\psi_m\) and \(\mathscr{P}^m\) denote the initial condition and Parisi functional corresponding to \(\lambda^{(m)}\). For each \(m\), choose a sequence of species proportions \(\lambda_N^{(m)}\) such that \(\lambda_N^{(m)}\to\lambda^{(m)}\), and write \(\overline{F}_N^m\) for the corresponding free energy.

By the rational case, for every \(m\) there exist \(p_m^+,p_m^-\in\mathcal{Q}_{\infty,\leqslant\lambda^{(m)}}^\mathscr{S}\) such that \[\begin{align} \label{e46crit95bdd95ms95crit95m} p_m^\pm=\partial_q\psi_m\left(q+t\nabla\xi(p_m^\pm)\right) \end{align}\tag{170}\] and \[\begin{align} \label{e46crit95bdd95ms95bd95m} \mathscr{P}^m_{t,q}(p_m^-)\leqslant\liminf_{N\to\infty}\overline{F}_N^m(t,q)\leqslant\limsup_{N\to\infty}\overline{F}_N^m(t,q)\leqslant\mathscr{P}^m_{t,q}(p_m^+). \end{align}\tag{171}\] By compactness, after passing to subsequences, we may assume that \(p_m^\pm\to p^\pm\) in \(L^1\). Since \(p_m^\pm\leqslant\lambda^{(m)}\) and \(\lambda^{(m)}\to\lambda_\infty\), we have \(p^\pm\in\mathcal{Q}_{\infty,\leqslant\lambda_\infty}^\mathscr{S}\). Moreover, using 63 , Lemma 9, and Lemma 10, we can pass to the limit in 170 and obtain \[\begin{align} \label{e46crit95bdd95ms95crit95general} p^\pm=\partial_q\psi\left(q+t\nabla\xi(p^\pm)\right). \end{align}\tag{172}\]

It remains to pass the bounds to the limit. By Lemma 8, \[\begin{align} \label{e46crit95bdd95ms95lambda95cont95upper} \limsup_{N\to\infty}\overline{F}_N(t,q)\leqslant\limsup_{N\to\infty}\overline{F}_N^m(t,q)+C\left(t+|q|_{L^1}+1\right)|\lambda_\infty-\lambda^{(m)}| \end{align}\tag{173}\] and \[\begin{align} \label{e46crit95bdd95ms95lambda95cont95lower} \liminf_{N\to\infty}\overline{F}_N(t,q)\geqslant\liminf_{N\to\infty}\overline{F}_N^m(t,q)-C\left(t+|q|_{L^1}+1\right)|\lambda_\infty-\lambda^{(m)}|. \end{align}\tag{174}\] Combining 171 with 173 and 174 , we get \[\begin{align} \mathscr{P}^m_{t,q}(p_m^-)-C\left(t+|q|_{L^1}+1\right)|\lambda_\infty-\lambda^{(m)}|\leqslant\liminf_{N\to\infty}\overline{F}_N(t,q) \end{align}\] and \[\begin{align} \limsup_{N\to\infty}\overline{F}_N(t,q)\leqslant\mathscr{P}^m_{t,q}(p_m^+)+C\left(t+|q|_{L^1}+1\right)|\lambda_\infty-\lambda^{(m)}|. \end{align}\] Finally, by 63 , Lemma 9, and Lemma 10, \[\begin{align} \lim_{m\to\infty}\mathscr{P}^m_{t,q}(p_m^\pm) = \mathscr{P}_{t,q}(p^\pm). \end{align}\] Letting \(m\to\infty\) yields \[\begin{align} \mathscr{P}_{t,q}(p^-)\leqslant\liminf_{N\to\infty}\overline{F}_N(t,q)\leqslant\limsup_{N\to\infty}\overline{F}_N(t,q)\leqslant\mathscr{P}_{t,q}(p^+). \end{align}\] Together with 172 , this completes the proof. ◻

Proof. We proceed in two steps. First, we prove ?? in the rational case. We then extend the result to the general case by approximation.

Step 1. The rational case. Assume first that \(\lambda_\infty\) is rational. Choose \({D}\in\mathbb{N}\) and a weak partition \((\mathsf{D}_s)_{s\in\mathscr{S}}\) of \(\{1,\ldots,{D}\}\) such that \(\lambda_{\infty,s}=|\mathsf{D}_s|/{D}\). Let \(\xi^{\mathrm{vec}}\) and \(\overline{F}_N^{\mathrm{vec}}\) be the associated vector spin model introduced in 74 –79 , and let \(f^{\mathrm{vec}}\) be the viscosity solution appearing in Theorem 5. By Proposition 18, the function \(g(t,\underline q):={D}f(t,\underline q^{\mathrm{avg}})\) is the Lipschitz viscosity solution of the Hamilton–Jacobi equation associated with \(\xi^{\mathrm{vec}}\), with initial condition \(g(0,\underline q)={D}\psi(\underline q^{\mathrm{avg}})\). Moreover, by Corollary 1, applied to the vector model at \(t=0\), we have \[\begin{align} {D}\psi(\underline q^{\mathrm{avg}})\stackrel{\eqref{e46psi61sumlambdapsi}}{=}\sum_{s\in\mathscr{S}}|\mathsf{D}_s|\psi_\circ\left((\underline q^{\mathrm{avg}})_s\right)\stackrel{\text{\eqref{e46b94avg61} \& Jensen}}{\leqslant}\sum_{s\in\mathscr{S}}\sum_{d\in\mathsf{D}_s}\psi_\circ(\underline q_d)\stackrel{\eqref{e46psi95single61}}{=}\psi^{\mathrm{vec}}(\underline q). \label{e46HJ95bdd95ms95init95comp} \end{align}\tag{175}\] Therefore \(g(0,\cdot)\leqslant f^{\mathrm{vec}}(0,\cdot)\), and the comparison principle in Proposition 15 gives \[\begin{align} {D}f(t,\underline q^{\mathrm{avg}})=g(t,\underline q)\leqslant f^{\mathrm{vec}}(t,\underline q),\qquad (t,\underline q)\in\mathbb{R}_+\times\mathcal{Q}_2^{D}. \label{e46HJ95bdd95ms95compare95vec} \end{align}\tag{176}\] Now fix \((t,q)\in\mathbb{R}_+\times\mathcal{Q}_\infty^\mathscr{S}\) and take \(\underline q=q^{\mathrm{vec}}\). Since we have \((q^{\mathrm{vec}})^{\mathrm{avg}}=q\) due to 88 , Theorem 5 and 176 yield \[\begin{align} {D}f(t,q)\leqslant f^{\mathrm{vec}}(t,q^{\mathrm{vec}})\leqslant\liminf_{N\to\infty}\overline{F}_N^{\mathrm{vec}}(t,q^{\mathrm{vec}}). \label{e46HJ95bdd95ms95vec95lower} \end{align}\tag{177}\] Using Lemma 11, and replacing \(N\) by \(\lceil N/{D}\rceil\) in the vector free energy, we obtain \[\begin{align} f(t,q)\leqslant\liminf_{N\to\infty}\overline{F}_N(t,q),\qquad (t,q)\in\mathbb{R}_+\times\mathcal{Q}_\infty^\mathscr{S}. \end{align}\] Finally, the extension from \(q\in\mathcal{Q}_\infty^\mathscr{S}\) to \(q\in\mathcal{Q}_2^\mathscr{S}\) follows by approximation. Indeed, choose \(q_k\in\mathcal{Q}_\infty^\mathscr{S}\) such that \(q_k\to q\) in \(L^2\) and in \(L^1\). The solution \(f\) is Lipschitz by Theorem 4, while \(\overline{F}_N\) is uniformly Lipschitz in \(q\) by Proposition 7. Therefore, \[\begin{align} f(t,q)\leqslant f(t,q_k)+C|q-q_k|_{L^2}\leqslant\liminf_{N\to\infty}\overline{F}_N(t,q_k)+C|q-q_k|_{L^2} \\ \leqslant\liminf_{N\to\infty}\overline{F}_N(t,q)+C|q-q_k|_{L^1}+C|q-q_k|_{L^2}. \end{align}\] Letting \(k\to\infty\) proves the rational case.

Step 2. Approximation of the proportions. We now remove the rationality assumption. Let \((\lambda^{(n)})_{n\in\mathbb{N}}\) be a sequence of rational points in \((0,1)^\mathscr{S}\) such that \(\sum_{s\in\mathscr{S}}\lambda_s^{(n)}=1\) and \(\lambda^{(n)}\to\lambda_\infty\). Let \(\psi_n(q):=\sum_{s\in\mathscr{S}}\lambda_s^{(n)}\psi_\circ(q_s)\), and let \(f_n\) be the Lipschitz viscosity solution of \[\begin{align} \partial_t f_n-\int_0^1\xi(\partial_q f_n)=0,\qquad f_n(0,\cdot)=\psi_n . \end{align}\] By the rational case, for any auxiliary sequence of species proportions \(\lambda_N^{(n)}\) satisfying \(\lambda_N^{(n)}\to\lambda^{(n)}\), we have \[\begin{align} f_n(t,q)\leqslant\liminf_{N\to\infty}\overline{F}_{N,\lambda_N^{(n)}}(t,q),\qquad (t,q)\in\mathbb{R}_+\times\mathcal{Q}_2^\mathscr{S}. \label{e46HJ95bdd95ms95rational95n} \end{align}\tag{178}\] On the other hand, Lemma 8 gives, for the original sequence \(\lambda_N\to\lambda_\infty\), \[\begin{align} \liminf_{N\to\infty}\overline{F}_{N,\lambda_N}(t,q)\geqslant\liminf_{N\to\infty}\overline{F}_{N,\lambda_N^{(n)}}(t,q)-C\left(t+|q|_{L^1}+1\right)|\lambda_\infty-\lambda^{(n)}|. \label{e46HJ95bdd95ms95lambda95cont} \end{align}\tag{179}\] Combining 178 and 179 , we get \[\begin{align} \liminf_{N\to\infty}\overline{F}_{N,\lambda_N}(t,q)\geqslant f_n(t,q)-C\left(t+|q|_{L^1}+1\right)|\lambda_\infty-\lambda^{(n)}|. \label{e46HJ95bdd95ms95before95n} \end{align}\tag{180}\] It remains to pass to the limit in \(n\). Set \[\rho_n:=\max_{s\in\mathscr{S}}|\lambda_s^{(n)}-\lambda_{\infty,s}|.\] Since \(\psi_\circ(0)=0\) and \(\psi_\circ\) is \(L^1\)-Lipschitz by Lemma 9, we have, for every \(q\in\mathcal{Q}_2^\mathscr{S}\), \[\begin{align} |\psi_n(q)-\psi(q)| &\leqslant\sum_{s\in\mathscr{S}}|\lambda_s^{(n)}-\lambda_{\infty,s}|\,|\psi_\circ(q_s)| \leqslant\rho_n\sum_{s\in\mathscr{S}}|q_s|_{L^1} =\rho_n\ell(q). \end{align}\] Applying Lemma 19 to \(f_n\) and \(f\), we get \[\label{e46fn95to95f95lambda95stability} |f_n(t,q)-f(t,q)|\leqslant\rho_n\left(\ell(q)+C_\mathsf{H}t\right),\qquad (t,q)\in\mathbb{R}_+\times\mathcal{Q}_2^\mathscr{S}.\tag{181}\] Combining 180 with 181 , we obtain \[\begin{align} \liminf_{N\to\infty}\overline{F}_{N,\lambda_N}(t,q) &\geqslant f(t,q)-\rho_n\left(\ell(q)+C_\mathsf{H}t\right)-C\left(t+|q|_{L^1}+1\right)|\lambda_\infty-\lambda^{(n)}|. \end{align}\] Letting \(n\to\infty\) gives ?? as desired. ◻

Proof of Theorem 1. By Proposition 20, we already have \[\begin{align} \label{e46t46lower} f(t,q)\leqslant\liminf_{N\to\infty}\overline{F}_N(t,q), \qquad (t,q)\in\mathbb{R}_+\times\mathcal{Q}_2^\mathscr{S}. \end{align}\tag{182}\] Theorem 4 together with Remark 17 also gives the Hopf formula for \(f\). Hence, \(f\) is equal to the right-hand side of 10 ; this yields the desired lower bound for the limit free energy. It remains to prove the matching upper bound.

Fix first \(t>0\) and \(q\in\mathcal{Q}_2^\mathscr{S}\). Since \(\mathcal{Q}_2^\mathscr{S}\subseteq\mathcal{Q}_1^\mathscr{S}\), Proposition 19 gives \(p^+\in\mathcal{Q}_{\infty,\leqslant\lambda_\infty}^\mathscr{S}\) such that \[\begin{gather} p^+=\partial_q\psi\left(q+t\nabla\xi(p^+)\right),\tag{183} \\ \limsup_{N\to\infty}\overline{F}_N(t,q)\leqslant\mathscr{P}_{t,q}(p^+). \tag{184} \end{gather}\] Set \[\begin{align} \label{e46q944361} q^+:=q+t\nabla\xi(p^+). \end{align}\tag{185}\] By Corollary 1, the function \(\psi\) is convex. Hence, using 183 , for every \(q'\in\mathcal{Q}_\infty^\mathscr{S}\), \[\begin{align} \label{e46t46convex95support} \psi(q')\geqslant\psi(q^+)+\left\langle p^+,q'-q^+\right\rangle_{L^2}. \end{align}\tag{186}\] Therefore, for every \(q'\in\mathcal{Q}_\infty^\mathscr{S}\), \[\begin{align} \psi(q')+\left\langle q-q',p^+\right\rangle_{L^2}+t\int_0^1\xi(p^+(s))\,\mathrm{d}s \stackrel{\eqref{e46t46convex95support}}{\geqslant} \psi(q^+)+\left\langle q-q^+,p^+\right\rangle_{L^2}+t\int_0^1\xi(p^+(s))\,\mathrm{d}s \\ \stackrel{\eqref{e46q944361}}{=}\psi\left(q+t\nabla\xi(p^+)\right)-t\int_0^1 p^+(s)\cdot\nabla\xi(p^+(s))\,\mathrm{d}s+t\int_0^1\xi(p^+(s))\,\mathrm{d}s \stackrel{\eqref{e46sP95lambda44t44q}}{=} \mathscr{P}_{t,q}(p^+). \end{align}\] Taking the infimum over \(q'\in\mathcal{Q}_\infty^\mathscr{S}\) and then using the Hopf formula 146 for \(f\), with the admissible choice \(p=p^+\), gives \[\begin{align} \label{e46t46parisi95leq95f} \mathscr{P}_{t,q}(p^+) \leqslant \inf_{q'\in\mathcal{Q}_\infty^\mathscr{S}}\left\{\psi(q')+\left\langle q-q',p^+\right\rangle_{L^2}+t\int_0^1\xi(p^+(s))\,\mathrm{d}s\right\} \leqslant f(t,q). \end{align}\tag{187}\] Combining 184 and 187 , we obtain \[\begin{align} \label{e46t46upper} \limsup_{N\to\infty}\overline{F}_N(t,q)\leqslant f(t,q), \qquad t>0,\;q\in\mathcal{Q}_2^\mathscr{S}. \end{align}\tag{188}\] Together with 182 , this proves 10 for \(t>0\).

The case \(t=0\) follows from the Lipschitz continuity of \(\overline{F}_N\) in Proposition 7, uniformly in \(N\). Thus 10 holds for every \((t,q)\in\mathbb{R}_+\times\mathcal{Q}_2^\mathscr{S}\). Finally, the link between the right-hand side of 10 and the solution of 11 boils down to the Hopf representation 146 . This completes the proof. ◻

Proof. As observed in [9], the coefficient array may be symmetrized without changing the model. In the present notation, this means that the symmetrized coefficients in 192 satisfy \[\begin{align} \sum_{s_1,\ldots,s_p\in\mathscr{S}}\Delta^2_{s_1,\ldots,s_p}a_{s_1}\cdots a_{s_p} =\sum_{s_1,\ldots,s_p\in\mathscr{S}}\widehat\Delta^2_{s_1,\ldots,s_p}a_{s_1}\cdots a_{s_p} \end{align}\] for every \(a\in\mathbb{R}^\mathscr{S}\). The same identity with \(a=\lambda\) gives \[\begin{align} \beta_p^2=\sum_{s_1,\ldots,s_p\in\mathscr{S}}\widehat\Delta^2_{s_1,\ldots,s_p}\lambda_{s_1}\cdots\lambda_{s_p}. \end{align}\] Thus we may work with the symmetric coefficients \(\widehat\Delta^2\).

It remains to prove 190 . Write \(a_s=\lambda_s x_s\), with \(x_s\geqslant 0\). The case \(p=1\) follows directly from 193 . For \(p\geqslant 2\), the arithmetic-geometric mean inequality and the balanced condition give \[\begin{align} &\sum_{s_1,\ldots,s_p\in\mathscr{S}}\widehat\Delta^2_{s_1,\ldots,s_p}\lambda_{s_1}\cdots\lambda_{s_p}x_{s_1}\cdots x_{s_p} \\ &\qquad\leqslant\sum_{t\in\mathscr{S}}\lambda_t x_t^p\sum_{s_2,\ldots,s_p\in\mathscr{S}}\widehat\Delta^2_{t,s_2,\ldots,s_p}\lambda_{s_2}\cdots\lambda_{s_p} =\beta_p^2\sum_{t\in\mathscr{S}}\lambda_t x_t^p. \end{align}\] Summing over \(p\geqslant 1\) yields \[\begin{align} \xi(a)&\leqslant\sum_{p\geqslant 1}\beta_p^2\sum_{s\in\mathscr{S}}\lambda_s\left(\frac{a_s}{\lambda_s}\right)^p =\sum_{s\in\mathscr{S}}\lambda_s\xi_\star\left(\frac{a_s}{\lambda_s}\right). \end{align}\] The identity \(\xi(r\lambda)=\xi_\star(r)\) follows directly from the definition of \(\beta_p^2\). ◻

Proof. By Theorem 1, \[\label{e46balanced95f95hopf} f(t,q)=\sup_{p\in\mathcal{Q}_\infty^\mathscr{S}}\inf_{q'\in\mathcal{Q}_\infty^\mathscr{S}}\left\{\psi(q')+\left\langle q-q',p\right\rangle_{L^2}+t\int_0^1\xi(p(v))\,\mathrm{d}v\right\}.\tag{198}\]

We first prove the lower bound. Fix \(a\in\mathcal{Q}_\infty\) and set \(p_s=\lambda_s a\) for every \(s\in\mathscr{S}\). For every \(q'\in\mathcal{Q}_\infty^\mathscr{S}\), Jensen’s inequality and the convexity of \(\psi_\circ\) from Proposition 3 give \[\label{e46balanced95jensen95psi} \psi(q')=\sum_{s\in\mathscr{S}}\lambda_s\psi_\circ(q'_s)\geqslant\psi_\circ\left(\sum_{s\in\mathscr{S}}\lambda_s q'_s\right)=\psi_\circ((q')^\lambda).\tag{199}\] Using also \(\xi(\lambda a)=\xi_\star(a)\) due to 189 , we obtain \[\begin{align} &\psi(q')+\left\langle q-q',p\right\rangle_{L^2}+t\int_0^1\xi(p(v))\,\mathrm{d}v \\ &\qquad\geqslant\psi_\circ((q')^\lambda)+\left\langle q^\lambda-(q')^\lambda,a\right\rangle_{L^2}+t\int_0^1\xi_\star(a(v))\,\mathrm{d}v =\mathcal{J}^\star_{t,q^\lambda}((q')^\lambda,a). \end{align}\] Taking the infimum over \(q'\in\mathcal{Q}_\infty^\mathscr{S}\), and using the fact that \((q')^\lambda\) ranges over all of \(\mathcal{Q}_\infty\) by taking \(q'_s=r'\) for every \(s\), gives \[\inf_{q'\in\mathcal{Q}_\infty^\mathscr{S}}\mathcal{J}_{t,q}(q',p)\geqslant\inf_{r'\in\mathcal{Q}_\infty}\mathcal{J}^\star_{t,q^\lambda}(r',a).\] Taking the supremum over \(a\in\mathcal{Q}_\infty\) yields \(f(t,q)\geqslant u(t,q^\lambda)\).

We now prove the upper bound. Fix \(p\in\mathcal{Q}_\infty^\mathscr{S}\) and set \(a_s=p_s/\lambda_s\). By 190 , for almost every \(v\in[0,1)\), \[\xi(p(v))\leqslant\sum_{s\in\mathscr{S}}\lambda_s\xi_\star(a_s(v)).\] Therefore, for every \(q'\in\mathcal{Q}_\infty^\mathscr{S}\), \[\begin{align} &\psi(q')+\left\langle q-q',p\right\rangle_{L^2}+t\int_0^1\xi(p(v))\,\mathrm{d}v \\ &\qquad\leqslant\sum_{s\in\mathscr{S}}\lambda_s\left\{\psi_\circ(q'_s)+\left\langle q_s-q'_s,a_s\right\rangle_{L^2}+t\int_0^1\xi_\star(a_s(v))\,\mathrm{d}v\right\}. \end{align}\] Taking the infimum over \(q'\) and using that the variables \(q'_s\) are independent on the right side, we obtain \[\inf_{q'\in\mathcal{Q}_\infty^\mathscr{S}}\mathcal{J}_{t,q}(q',p)\leqslant\sum_{s\in\mathscr{S}}\lambda_s\inf_{r'\in\mathcal{Q}_\infty}\mathcal{J}^\star_{t,q_s}(r',a_s).\] Taking the supremum over \(p\in\mathcal{Q}_\infty^\mathscr{S}\), or equivalently over the collection \((a_s)_{s\in\mathscr{S}}\), gives \[f(t,q)\leqslant\sum_{s\in\mathscr{S}}\lambda_s\sup_{a_s\in\mathcal{Q}_\infty}\inf_{r'\in\mathcal{Q}_\infty}\mathcal{J}^\star_{t,q_s}(r',a_s)=\sum_{s\in\mathscr{S}}\lambda_su(t,q_s).\] This proves ?? . Taking \(q=0\) gives ?? . ◻

Proof. By Theorem 1, the left side of 200 is \(f(t,0)\). By Proposition 21, \(f(t,0)=u(t,0)\). The definition 196 of \(u\) at \(r=0\) is precisely 200 . ◻

Proof. Fix any \(0\leqslant u<v<1\). The definition of \(\mathcal{Q}_{\uparrow,c}\) gives \[\begin{align} \label{e46q40v41-q40u41} q(v) -q(u) \geqslant c(v-u) \mathrm{Id}\quad\text{and}\quad \mathsf{Ellipt}(q(v)-q(u)) \leqslant c^{-1}. \end{align}\tag{204}\] Since \(q'\in\mathcal{Q}\), we immediately have \[\begin{align} \label{e46q43q39-q-q3962} (q(v)+q'(v))-(q(u)+q'(u)) \geqslant q(v)-q(u)\geqslant c(v-u)\mathrm{Id}. \end{align}\tag{205}\] By Weyl’s inequalities, \(\lambda_{\max}\) and \(\lambda_{\min}\) are respectively sub-additive and super-additive. Therefore, given two symmetric matrices \(A,B \in \mathbf{S}^{D}\) such that \(A+B \in \mathbf{S}^{D}_{++}\), \(\lambda_{\max}(A) + \lambda_{\max}(B) \geqslant 0\), and \(\lambda_{\min}(A) + \lambda_{\min}(B) > 0\), we have \[\label{e46ellipt32bounds} \mathsf{Ellipt}(A+B) \leqslant\frac{\lambda_{\max}(A) + \lambda_{\max}(B)}{\lambda_{\min}(A) + \lambda_{\min}(B)}.\tag{206}\] The bound \(|\accentset{\large\bfseries .}{q}'|_{L^\infty}\leqslant a\) implies \[\begin{align} \label{e46kappa40v41-kappa40u4160} q'(v) - q'(u) \leqslant a(v-u)\mathrm{Id}. \end{align}\tag{207}\] Thus, \[\begin{align} \label{e46lambda40kappa40v41-kappa40u4141} \lambda_{\max}(q'(v)-q'(u)),\;\lambda_{\min}(q'(v)-q'(u)) \in [-a(v-u), a(v-u)]. \end{align}\tag{208}\] It follows that \[\begin{align} &\mathsf{Ellipt}((q(v)+q'(v))-(q(u)+q'(u))) \stackrel{\eqref{e46ellipt32bounds}\eqref{e46lambda40kappa40v41-kappa40u4141}}{\leqslant} \frac{\lambda_{\max}(q(v) - q(u)) + a(v-u)}{\lambda_{\min}(q(v) - q(u)) - a (v-u)} \\ &\stackrel{\eqref{e46q40v41-q40u41}}{\leqslant} \frac{c^{-1}(\lambda_{\min}(q(v) - q(u))-a(v-u)) + (c^{-1}+1)a(v-u)}{\lambda_{\min}(q(v) - q(u)) - a (v-u)} \leqslant c^{-1} + \frac{(c^{-1}+1)a(v-u)}{\lambda_{\min}(q(v) - q(u)) - a (v-u)} \\ &\stackrel{\eqref{e46q40v41-q40u41}}{\leqslant} c^{-1} + \frac{(c^{-1}+1)a(v-u)}{c(v-u) - a (v-u)} =c^{-1} +\frac{(c^{-1}+1)a}{c - a} = \frac{1+a}{c-a}. \end{align}\] Together with 205 , this shows that \(q+q'\in\mathcal{Q}_{\uparrow,\frac{c-a}{1+a}}\). ◻

Proof. Writing \(q' = a_i r q_i+N^{-\gamma/2}\sum_{j\neq i}a_j y_jq_j\), we obtain from 120 that \(|\accentset{\large\bfseries .}{q}'|_{L^\infty}\leqslant N^{-\gamma/2}+ r\). Hence, by Lemma 21, we can find \(c_0, N_0, r_0>0\) such that \[\begin{align} q+a_i r q_i+N^{-\gamma/2}\sum_{j\neq i}a_j y_jq_j \in \mathcal{Q}_{\uparrow,c_0},\qquad\forall i \in\mathbb{N},\;N\geqslant N_0,\;r\in[0,r_0). \end{align}\]

Fix any \(i\in\mathbb{N}\) and write \(q_* = q+N^{-\gamma/2}\sum_{j\neq i}a_j y_jq_j\) and \(\kappa = a_iq_i\). The preceding display ensures that \(q_*+r\kappa\in\mathcal{Q}_{\uparrow,c_0}\) for every \(N\geqslant N_0\) and \(r<r_0\). Set \(F(r)=\overline{F}_N(t,q_*+r\kappa)\). For \(\varepsilon>0\) small, applying the local semi-concavity result for \(\overline{F}_N\) from [3]² with \(\frac{1}{2}, q_*+r\kappa, q_*+(r+\varepsilon)\kappa, c_0\) substituted for \(r, q,q',c\) therein, we get \[\begin{align} \tfrac{1}{2}F(r)+\tfrac{1}{2}F(r+\varepsilon)- F\left(r+\tfrac{\varepsilon}{2}\right)\leqslant\tfrac{C}{4} c_0^{-2}\varepsilon^2\left|\accentset{\large\bfseries .}{\kappa}\right|_{L^2}^2 \end{align}\] for some absolute constant \(C>0\). Dividing both sides by \(\varepsilon^2\), sending \(\varepsilon\to0\), and using \(\kappa=a_iq_i\), we obtain \(\frac{\mathrm{d}^2}{\mathrm{d}r^2}F(r)\leqslant 2Cc_0^{-2}\left|a_i\accentset{\large\bfseries .}{q}_i\right|_{L^2}^2 \leqslant 2Cc_0^{-2}\), where the last inequality follows from 120 . This gives the desired result.

The same estimates hold for \(\widetilde{F}_N^x\) and \(\overline{F}^x_{N+1}\) for the same reason as in Remark 10. ◻

Sketch of proof. Define \[\begin{align} v(t,q):=u(t,q^\lambda),\qquad w(t,q):=\sum_{s\in\mathscr{S}}\lambda_s u(t,q_s). \end{align}\] We first check, in the viscosity sense, that \(v\) solves the multi-species equation 209 . Formally, if \(a=\partial_q u(t,q^\lambda)\), then \[\begin{align} \partial_{q_s}v(t,q)=\lambda_s a,\qquad s\in\mathscr{S}. \end{align}\] Thus, using the diagonal identity 189 , \[\begin{align} \xi((\partial_{q_s}v)_{s\in\mathscr{S}})=\xi(\lambda a)=\xi_\star(a), \end{align}\] and therefore the equation for \(v\) follows from the equation for \(u\).

This formal verification can be justified rigorously with the standard doubling-variable argument for viscosity solutions on infinite-dimensional cones. The only point that is not completely formal is the existence of maximizers for the penalized functional; as in the proof of Proposition 18, this is obtained by applying Stegall’s variational principle, Theorem 6. Equivalently, one repeats the proof of Proposition 18 with the bounded linear map \[\begin{align} B:\mathcal{H}_\mathscr{S}\to L^2([0,1]),\qquad Bq=\sum_{s\in\mathscr{S}}\lambda_s q_s, \end{align}\] whose adjoint is \(B^*a=(\lambda_s a)_{s\in\mathscr{S}}\). The identity \(\xi(B^*a)=\xi_\star(a)\) is exactly 189 . This proves that \(v\) is a viscosity solution, in particular a viscosity subsolution, of 209 .

Next we check that \(w\) is a viscosity supersolution of 209 . Formally, if \(a_s=\partial_q u(t,q_s)\), then \[\begin{align} \partial_{q_s}w(t,q)=\lambda_s a_s. \end{align}\] Using the one-species equation for \(u\) and the balanced comparison inequality 190 , we obtain \[\begin{align} \partial_t w(t,q)-\int_0^1\xi(\partial_q w) &=\sum_{s\in\mathscr{S}}\lambda_s\int_0^1\xi_\star(a_s(v))\,\mathrm{d}v-\int_0^1\xi((\lambda_s a_s(v))_{s\in\mathscr{S}})\,\mathrm{d}v\geqslant 0. \end{align}\] The viscosity justification is again obtained by the same Stegall perturbation argument used in Proposition 18, now applied separately to the components \(q_s\). Thus \(w\) is a viscosity supersolution.

Finally, by the transport convexity of \(\psi_\circ\) proved in Proposition 3, \[\begin{align} v(0,q)=\psi_\circ(q^\lambda)\leqslant\sum_{s\in\mathscr{S}}\lambda_s\psi_\circ(q_s)=f(0,q), \end{align}\] while \[\begin{align} w(0,q)=\sum_{s\in\mathscr{S}}\lambda_s\psi_\circ(q_s)=f(0,q). \end{align}\] The comparison principle for the Hamilton–Jacobi equation gives \(v\leqslant f\leqslant w\), which is ?? . Taking \(q=0\) gives the particular case. ◻