Proof. Recall from 2.1 the finite collection of triples \((x_r,U_r,U'_r)_{r=1,\dotsc,L}\) such that \(U'_r\) is contained in a compact subset of \(U_r\) and the smooth mappings \(x_r,x_r^{-1}\) have uniformly bounded derivatives (as mappings between subsets of \(\mathcal{M}\) and \(\mathbb{R}^n\)). We begin by choosing \(\ell_0\) sufficiently small so that \(\ell(\gamma)\leq\ell_0\) implies that \(\gamma\) lies within a single such subset \(U'_r\) for some \(r\).

Now let \(\gamma\) be an arbitrary such closed curve of length \(\ell(\gamma)\leq\ell_0\) and suppose that \(\gamma\) lies within \(U'_r\). The composition \[\begin{align} \tilde{\gamma} = x_r \circ \gamma \end{align}\] defines a piecewise smooth closed curve in \(\mathbb{R}^n\), and it is readily verified that the associated 1-currents are related by \([[\tilde{\gamma}]]=(x_r)_\# [[\gamma]]\), \([[\gamma]]=(x_r^{-1})_\# [[\tilde{\gamma}]]\). The closedness of the curves implies that \[\begin{align} \partial_1 [[\tilde{\gamma}]]=0 . \end{align}\] Using the notation of Federer?*4.2.10 and pp. 406–408, we have \([[\tilde{\gamma}]]\in I_1(\mathbb{R}^n)\), and there exist \(\tilde{C}<\infty\) and \(\tilde{S} \in I_2(\mathbb{R}^n)\subset \mathbf{M}_2(\mathbb{R}^n)\) (a generalized spanning surface) such that \[\begin{align} \partial_2 \tilde{S}&= [[\tilde{\gamma}]] , \\ \mathbb{M}([[\tilde{S}]])&\leq \tilde{C}\ell( \tilde{\gamma})^2 . \end{align}\] Moreover that result guarantees that the support of \(\tilde{S}\) lies within distance \(C \ell(\gamma)\) of the image of \(\tilde{\gamma}\). In particular, by shrinking the value \(\ell_0\) further if necessary, we may make this distance sufficiently small to guarantee \(\operatorname*{supp}\tilde{S} \subset x_r(U_r)\).

Finally, the push-forward of \(\tilde{S}\) defines a \(2\)-current \[\begin{align} S_\gamma = (x_r^{-1})_\# \tilde{S} \in\mathbf{M}_2 \end{align}\] satisfying the desired claims. ◻

Proof. For all \((n-k-1)\)-forms \(\zeta\in C^1(\Lambda^{n-k-1})\), Stokes’ theorem and \(\mathrm{d}_{n-1}(\zeta\wedge\omega)=\mathrm{d}_{n-k-1}\zeta\wedge\omega + (-1)^{n-k-1}\zeta\wedge \mathrm{d}_k\omega\) give \[\partial_{n-k} T_\omega(\zeta) = \int_\mathcal{M}\mathrm{d}_{n-k-1}\zeta\wedge\omega = (-1)^{n-k} \int_\mathcal{M}\zeta\wedge\mathrm{d}_k\omega = (-1)^{n-k} T_{\mathrm{d}_k\omega}(\zeta) . \qedhere\] ◻

Proof. The first bound 44 is immediate. The other bounds all follow by estimating the integrals with a dyadic expansion over annuli such that \(d_\mathcal{M}(\cdot,q) \approx \sqrt{t} 2^j\), using the fact that \(V(\left\{p\colon d_\mathcal{M}(\cdot,q)\leq r\right\})\lesssim r^n\) and merging higher powers of \(t\) into lower ones since \(0<t\leq 1\). ◻

Proofs. We recall the identification \(e^{t\Delta_{1,c}} T=T_{\omega_t}\) established in 3. In particular, recalling 41 and integrating 42 over the manifold, we have \[\begin{align} \left\Vert e^{t\Delta_{1,c}} T \right\Vert_{L^1} &= \left\Vert\omega_t\right\Vert_{L^1} = \int_{\mathcal{M}} \left\Vert\omega_t(p)\right\Vert_p \, \mathrm{d}V(p) \\ &\leq C\int_{\mathcal{M}} \int_\mathcal{M}K_t(p,q)\,\mathrm{d}\mu_T(q) \, \mathrm{d}V(p)\\ &\leq C \sup_{q \in \mathcal{M}} \left\Vert K_t(\cdot,q)\right\Vert_{L^1}\mathbb{M}(T)\\ &\lesssim \mathbb{M}(T) , \end{align}\] where the last step used 45 . Similarly, taking the supremum instead as in 44 , \[\begin{align} \left\Vert e^{t\Delta_{1,c}} T \right\Vert_{L^{\infty}} &= \left\Vert\omega_t\right\Vert_{L^\infty} = \sup_{p \in \mathcal{M}} \left\Vert\omega_t(p)\right\Vert_p \\ &\leq C \sup_{p \in \mathcal{M}}\int_\mathcal{M}K_t(p,q)\,\mathrm{d}\mu_T(q)\\ &\leq C \sup_{p,q \in \mathcal{M}} K_t(p,q) \, \mu_T(\mathcal{M})\\ &\lesssim t^{-n/2}\mathbb{M}(T) . \qedhere \end{align}\] ◻

Proof. Let \(h(r)=-g'(r)\) and note that the assumptions imply \[g(\rho)=\int_\rho^\infty h(r)\,\mathrm{d}r=\int_0^\infty h(r)\mathbb{1}_{\left\{r>\rho\right\}}\,\mathrm{d}r\] for all \(\rho\in\left[0,\infty\right)\). Interchange the order of integration to find \[\begin{align} \int_\mathcal{M}\left\vert f(q)\right\vert \mathrm{d}\mu(q) &\leq \int_\mathcal{M}g(d_\mathcal{M}(p,q)) \, \mathrm{d}\mu(q) \\& = \int_0^\infty \int_\mathcal{M}h(r)\mathbb{1}_{\left\{r> d_\mathcal{M}(p,q)\right\}} \, \mathrm{d}\mu(q) \, \mathrm{d}r \\& = \int_0^\infty h(r) \mu(\left\{q\in\mathcal{M}\colon d_\mathcal{M}(p,q)< r\right\}) \, \mathrm{d}r \\& \leq \int_0^\infty h(r) (r\left\Vert\mu\right\Vert_{\mathrm{Morrey}}) \, \mathrm{d}r . \qedhere \end{align}\] ◻

Proof. By 3, \[\begin{align} \left\Vert e^{t\Delta_{1,c}} T \right\Vert_{L^{\infty}} &= \sup_{p\in\mathcal{M}} \left\Vert\omega_t\right\Vert_p \\& \leq C\int_\mathcal{M}(t^{-n/2}\exp\left( -d_\mathcal{M}(p,q)^2/(4t) \right)+1) \,\mathrm{d}\mu_T(q) . \end{align}\] The constant term in the integrand gives a contribution \(C\mu_T(\mathcal{M})\), which we can bound in terms of the Morrey norm by compactness: \[\mu_T(\mathcal{M})\leq \operatorname{diam}(\mathcal{M})\left\Vert\mu_T\right\Vert_{\mathrm{Morrey}} .\] For the rest of the integral, we can apply the special case of 7 in which \(g(r)=t^{-n/2}e^{-r^2/(4t)}\). Then \(h(r)=\frac{1}{2}rt^{-(n+2)/2}e^{-r^2/(4t)}\) and we find \(\int_0^\infty rh(r)\,\mathrm{d}r=t^{-(n-1)/2}\sqrt{\pi}\). Thus \[\begin{align} \left\Vert e^{t\Delta_{1,c}} T \right\Vert_{L^{\infty}} &\leq C \left\Vert\mu_T\right\Vert_{\mathrm{Morrey}} t^{-(n-1)/2}\sqrt{\pi} + C \operatorname{diam}(\mathcal{M})\left\Vert\mu_T\right\Vert_{\mathrm{Morrey}} . \end{align}\] By definition \(\left\Vert\mu_T\right\Vert_{\mathrm{Morrey}}=\left\Vert T\right\Vert_{\mathrm{Morrey}}\), and the second term can be merged into the first because \(t\leq 1\). ◻

Proofs. By the Interpolation [1](#Interpolation){reference-type=“ref+Label” reference=“Interpolation”}, \[\begin{align} \left\Vert e^{t\Delta_{1,c}} T\right\Vert_{L^{n/(n-\alpha),1}} \leq C \left\Vert e^{t\Delta_{1,c}} T\right\Vert_{L^1}^{1-\alpha/n} \left\Vert e^{t\Delta_{1,c}} T\right\Vert_{L^\infty}^{\alpha/n} . \end{align}\] The upper bound from 2 then follows by using the \(L^1\) bound from 5 and the \(L^\infty\) bound from 8. Integrating as in 3 then leads to a factor \(t^{\alpha/(2n)-1}\), which ensures that the integral is finite. ◻

Proof. By 11, we have \[\begin{align} \left\Vert e^{t\Delta_{1,c}} [[\gamma]] \right\Vert_{L^1} &= \sup_{\xi\in C(\Lambda^1), \|\xi\|_{L^\infty}\leq 1} e^{t\Delta_{1,c}} [[\gamma]](\xi)\\ &= \sup_{\xi\in C(\Lambda^1), \|\xi\|_{L^\infty}\leq 1} [[\gamma]](e^{t\Delta_{1}} \xi) \end{align}\] By the smallness assumption on the curve, we may apply 2 to obtain \[\begin{align} \left\vert[[\gamma]](e^{t\Delta_{1}} \xi)\right\vert = \left\vert S_\gamma(\mathrm{d}_1 e^{t\Delta_{1}} \xi)\right\vert \leq C\ell(\gamma)^2\left\Vert\mathrm{d}_1 e^{t\Delta_{1}} \xi\right\Vert_{L^\infty}, \end{align}\] and then 9, the assumption \(\left\Vert\xi\right\Vert_{L^\infty}\leq 1\), and the bound 47 give \[\begin{align} \left\Vert e^{t\Delta_{1,c}} [[\gamma]] \right\Vert_{L^1} &\lesssim \ell(\gamma)^2 \sup_{p \in \mathcal{M}} \int_\mathcal{M}\tilde{K}_t(p,q) \,\mathrm{d}V(q)\\ &\lesssim t^{-1/2}\ell(\gamma)^2 . \qedhere \end{align}\] ◻

Proof. As in the proof of 2, the Interpolation [1](#Interpolation){reference-type=“ref+Label” reference=“Interpolation”}gives \[\begin{align} \left\Vert e^{t\Delta_{1,c}} [[\gamma]] \right\Vert_{L^{n/(n-\alpha),1}} \leq C \left\Vert e^{t\Delta_{1,c}} [[\gamma]] \right\Vert_{L^1}^{1-\alpha/n} \left\Vert e^{t\Delta_{1,c}} [[\gamma]] \right\Vert_{L^\infty}^{\alpha/n} . \end{align}\] Bound the \(L^1\) norm using 10 and bound the \(L^\infty\) norm using 6 with \(T=[[\gamma]]\), recalling that \(\mathbb{M}([[\gamma]])\leq\ell(\gamma)\). ◻

Proof. 5 indicates a change in scaling behavior at time scales of order \(\ell(\gamma)^2\). With this in mind, make the substitutions \(s=t/\ell(\gamma)^2\) and later \(u=s \left\Vert\gamma\right\Vert_{\mathrm{Morrey}}^{2\alpha/n}\), replacing the upper limit of integration by \(\infty\), to find \[\begin{align} &\int_0^1 t^{\alpha/2 - 1} \left\Vert e^{t\Delta_{n-1}}[[\gamma]]\right\Vert_{L^{n/(n-\alpha),1}} \mathrm{d}t \\&\quad \leq C \ell(\gamma) \int_0^1 \min\left( \left(\frac{t \left\Vert\gamma\right\Vert_{\mathrm{Morrey}}^2}{\ell(\gamma)^2}\right)^{\alpha/(2n)} , \left(\frac{t}{\ell(\gamma)^2}\right)^{-(n-\alpha)/(2n)} \right) \frac{\mathrm{d}t}{t} \\&\quad \leq C \ell(\gamma) \int_0^\infty \min\left( \left(s\left\Vert\gamma\right\Vert_{\mathrm{Morrey}}^2\right)^{\alpha/(2n)} , s^{-(n-\alpha)/(2n)} \right) \frac{\mathrm{d}s}{s} \\&\quad = C \ell(\gamma) \int_0^\infty \min\left( \left(u\left\Vert\gamma\right\Vert_{\mathrm{Morrey}}^{2(n-\alpha)/n}\right)^{\alpha/(2n)} , u^{-(n-\alpha)/(2n)}\left\Vert\gamma\right\Vert_{\mathrm{Morrey}}^{\alpha(n-\alpha)/n^2} \right) \frac{\mathrm{d}u}{u} \\&\quad = C \ell(\gamma) \left\Vert\gamma\right\Vert_{\mathrm{Morrey}}^{\alpha(n-\alpha)/n^2} \left( \int_0^1 u^{\alpha/(2n)-1}\,\mathrm{d}u + \int_1^\infty u^{-(1+(n-\alpha)/(2n))}\,\mathrm{d}u \right) \\&\quad = C \ell(\gamma) \left\Vert\gamma\right\Vert_{\mathrm{Morrey}}^{\alpha(n-\alpha)/n^2} \left( \frac{2n}{\alpha} + \frac{2n}{n-\alpha} \right) . \qedhere \end{align}\] ◻

Proof. Let \(\xi\in C(\Lambda^j)\) with \(\left\Vert\xi\right\Vert_{L^{n/\alpha,1}}\leq 1\). Then using 11, \[\left\vert S(\xi)\right\vert = \lim_{m\to\infty}\left\vert S_m(\xi)\right\vert \leq \liminf_{m\to\infty} \left\Vert S_m\right\Vert_{L^{n/(n-\alpha),1}}\] and taking the supremum over such \(\xi\) shows 50 . For 51 , let \(\xi\in C(\Lambda^j)\) be arbitrary. Then also \(e^{t\Delta_j}\xi \in C(\Lambda^j)\), so \[e^{t\Delta_{j,c}} S_m(\xi) = S_m(e^{t\Delta_j}\xi) \to S(e^{t\Delta_j}\xi) = e^{t\Delta_{j,c}} S(\xi)\] verifies the weak-star convergence, and the last statement follows by 50 . ◻

Proof of 13. Let \(T\) be a 1-current with \(\partial_1 T=0\), and let \(\gamma_{i,m}\), \(\lambda_{i,m}\), and \(M\) be the curves, scalars, and universal constant from 5, so that the 1-currents \[S_m = \sum_{i=1}^m \lambda_{i,m} \frac{[[\gamma_{i,m}]]}{\ell(\gamma_{i,m})}\] satisfy \(S_m\to T\) weakly-star as \(m\to\infty\), by 24 . By the triangle inequality, \[\begin{gather} \int_0^1 t^{\alpha/2-1} \left\Vert e^{t\Delta_{1,c}} S_m\right\Vert_{L^{n/(n-\alpha),1}} \mathrm{d}t \\ \leq \sum_{i=1}^m \left\vert\lambda_{i,m}\right\vert \ell(\gamma_{i,m})^{-1} \int_0^1 t^{\alpha/2-1} \left\Vert e^{t\Delta_{1,c}}[[\gamma_{i,m}]]\right\Vert_{L^{n/(n-\alpha),1}} \mathrm{d}t . \end{gather}\] To control the effect of the factor \(\ell(\gamma_{i,m})^{-1}\), split the sum. For short curves, 6 gives a uniform bound that is linear in \(\ell(\gamma_{i,m})\), whereas for long curves, the sublinear bound from 3 is sufficient: \[\begin{align} &\int_0^1 t^{\alpha/2-1} \left\Vert e^{t\Delta_{1,c}} S_m\right\Vert_{L^{n/(n-\alpha),1}} \mathrm{d}t \notag\\&\quad \leq \sum_{\substack{i\in\left\{1,\dotsc,m\right\}\colon \\ \ell(\gamma_{i,m})\leq \ell_0}} \left\vert\lambda_{i,m}\right\vert \ell(\gamma_{i,m})^{-1} [C \ell(\gamma_{i,m}) M^{\alpha(n-\alpha)/n^2}] \notag\\&\qquad + \sum_{\substack{i\in\left\{1,\dotsc,m\right\}\colon \\ \ell(\gamma_{i,m}) > \ell_0}} \left\vert\lambda_{i,m}\right\vert \ell(\gamma_{i,m})^{-1} [C\ell(\gamma_{i,m})^{1-\alpha/n} M^{\alpha/n}] &\text{by \eqref{morrey95bound}} \notag\\&\quad \leq C \sum_{i=1}^m \left\vert\lambda_{i,m}\right\vert \left[ M^{\alpha(n-\alpha)/n^2} + (M/\ell_0)^{\alpha/n} \right] \notag\\&\quad \leq 2C \left[ M^{\alpha(n-\alpha)/n^2} + (M/\ell_0)^{\alpha/n} \right] \mathbb{M}(T) &\text{by \eqref{mass}} . \end{align}\] Since this bound is uniform in \(m\), the result follows from 11. ◻

Proof of 12 for orientable manifolds. Let \(F\in C^1(\Lambda^{n-1})\) satisfy \(\mathrm{d}_{n-1} F=0\). Since the manifold is orientable, we may form the corresponding \(1\)-current \(T_F\), which satisfies \(\partial_1 T_F=0\) by 8. Thus 13 applies, and along with [TomegaPropagatorsCompatible,NormDualityCurrentsForms] gives \[\begin{align} &\int_0^1 t^{\alpha/2-1} \left\Vert e^{t\Delta_{n-1}} F\right\Vert_{L^{n/(n-\alpha),1}(\Lambda^{n-1})} \mathrm{d}t \\&\quad = \int_0^1 t^{\alpha/2-1} \left\Vert e^{t\Delta_{1,c}} T_F\right\Vert_{L^{n/(n-\alpha),1}(\Lambda^{n-1})} \mathrm{d}t \\&\quad \leq C \mathbb{M}(T_F) = C\left\Vert F\right\Vert_{L^1(\Lambda^{n-1})} . \end{align}\] Since this inequality holds for all \(F\in C^1(\Lambda^{n-1})\) satisfying \(\mathrm{d}_{n-1}F = 0\), we may argue as in ISS? to extend the inequality to \(F\in L^1(\Lambda^{n-1})\) satisfying \(\mathrm{d}_{n-1}F = 0\). ◻

Proof. Let \((x_r,U_r)_{r=1}^L\) be the finite atlas constructed in 2.1 and \((\chi_r)_{r=1}^L\) be the partition of unity subordinate to \((U'_r)_{r=1}^L\). Fix points \(y_r\in U_r\), \(r\in\left\{1,\dotsc,L\right\}\). By connectedness, for all \(r,r'\in\left\{1,\dotsc,L\right\}\), we may fix geodesics \(\tilde{\gamma}^{(r,r')}\) in \(\mathcal{M}\) from \(y_r\) to \(y_{r'}\), possibly of length 0 if \(y_r=y_{r'}\). For \(p\in U_r, p'\in U_{r'}\), define \(\gamma^{(r,r')}_{p,p'}\) to be the concatenation of the geodesics \(\gamma_{p,y_r}^{(r)}\), \(\tilde{\gamma}^{(r,r')}\), and \(\gamma_{y_{r'},p'}^{(r')}\). Thus \(\gamma^{(r,r')}_{p,p'}\) is a piecewise smooth curve from \(p\) to \(p'\). In particular, \[\partial_1[[\gamma^{(r,r')}_{p,p'}]](f) = f(p')-f(p)\] for all \(f\in C(\mathcal{M})\). Furthermore, since each curve \(\gamma^{(r,r')}_{p,p'}\) is a concatenation of three geodesics from a compact metric space, \[\ell(\gamma^{(r,r')}_{p,p'}) \leq 3\operatorname{diam}(\mathcal{M}) , \quad \left\Vert\big.\smash{\gamma^{(r,r')}_{p,p'}}\right\Vert_{\mathrm{Morrey}} \leq 6 ;\] see ChenGoodmanHernandezSpectorPreprint?*Corollary 2.3 for the second assertion. It is readily verified that \([[\gamma_{p,y_r}^{(r)}]]\) and \([[\gamma_{y_{r'},p'}^{(r')}]]\), and hence \([[\gamma^{(r,r')}_{p,p'}]]\), depend continuously on \((p,p')\in (U_r,U_{r'})\). Finally we define \[g_{p,q} = \sum_{r,r'=1}^L\chi^{(r)}(p) \chi^{(r')}(q) [[\gamma^{(r,r')}_{p,q}]] ,\] where in the sum we can restrict to \(r,r'\) for which \(p\in U_r\), \(q\in U_{r'}\), without loss of generality. The assertions then follow from linearity and the triangle inequality. ◻

Proof. First suppose \(\mathcal{M}\) is connected. Let \(\nu\in\Lambda^n\) denote the volume form of \(\mathcal{M}\), and let \(F\in C^\infty(\Lambda^{n-1})\) be a given smooth \((n-1)\)-form. We may write the \(n\)-form \(\mathrm{d}_{n-1} F\) as \(\mathrm{d}_{n-1} F = \varphi\, \nu\), where \(\varphi=\star_n\mathrm{d}_{n-1} F\in C^\infty(\mathcal{M})\) is a smooth function. By Stokes’ theorem, \[\int_\mathcal{M}\mathrm{d}_{n-1} F = 0 , \quad\text{i.e.,}\quad \int_\mathcal{M}\varphi(p)\,\mathrm{d}V(p) = 0 ,\] and consequently \[\int_\mathcal{M}\varphi^+(p)\,\mathrm{d}V(p) = \int_\mathcal{M}\varphi^-(p)\,\mathrm{d}V(p) = \tfrac{1}{2} \left\Vert\varphi\right\Vert_{L^1} = \tfrac{1}{2}\left\Vert\mathrm{d}_{n-1} F\right\Vert_{L^1} ,\] where \(\varphi^+\) and \(\varphi^-\) denote the positive and negative parts of \(\varphi\), respectively.

If \(\mathrm{d}_{n-1} F=0\) then \(\partial_1 T_F=0\) already and we may set \(R_F=0\). Otherwise, define \[K(p,q) = \frac{\varphi^-(p)\varphi^+(q)}{\frac{1}{2} \left\Vert\varphi\right\Vert_{L^1}} .\] Then \(K\) is a transport kernel on \(\mathcal{M}\times\mathcal{M}\) transporting \(\varphi^-\) to \(\varphi^+\) in the sense that \(K(p,q)\geq 0\) for all \(p,q\in\mathcal{M}\) and \[\begin{align} \int_\mathcal{M}K(p,q)\,\mathrm{d}V(q) &= \varphi^-(p) &\text{for all }p\in\mathcal{M} , \\ \int_\mathcal{M}K(p,q)\,\mathrm{d}V(p) &= \varphi^+(q) &\text{for all }q\in\mathcal{M} . \end{align}\] Define \[R_F = \iint_{\mathcal{M}\times\mathcal{M}} g_{p,q} K(p,q) \,\mathrm{d}V(p)\,\mathrm{d}V(q) .\] Then for \(f\in C(\mathcal{M})\), \[\begin{align} \partial_1 R_F(f) &= \iint_{\mathcal{M}\times\mathcal{M}} \partial_1 g_{p,q}(f) K(p,q) \,\mathrm{d}V(p)\,\mathrm{d}V(q) \\& = \iint_{\mathcal{M}\times\mathcal{M}} [f(q)-f(p)] \frac{\varphi^-(p)\varphi^+(q)}{\frac{1}{2} \left\Vert\varphi\right\Vert_{L^1}} \,\mathrm{d}V(p)\,\mathrm{d}(q) \\& = \int_\mathcal{M}f(q)\varphi^+(q) \frac{\left\Vert\varphi^-\right\Vert_{L^1}}{\frac{1}{2}\left\Vert\varphi\right\Vert_{L^1}} \,\mathrm{d}V(q) - \int_\mathcal{M}f(p)\varphi^-(p) \frac{\left\Vert\varphi^+\right\Vert_{L^1}}{\frac{1}{2}\left\Vert\varphi\right\Vert_{L^1}} \,\mathrm{d}V(p) \\& = \int_\mathcal{M}f \, [\varphi^- - \varphi^+]\,\mathrm{d}V \\& = -\int_\mathcal{M}f \varphi\, \nu = -\int_\mathcal{M}f\,\mathrm{d}_{n-1} F = -T_{\mathrm{d}_{n-1} F}(f) = \partial_1 T_F(f) , \end{align}\] where the last step used 8. The bounds on \(\mathbb{M}(R_F)\) and \(\left\Vert R_F\right\Vert_{\mathrm{Morrey}}\) follow from the triangle inequality, the uniform bounds on \(\mathbb{M}(g_{p,q})\) and \(\left\Vert g_{p,q}\right\Vert_{\mathrm{Morrey}}\), and the observation that \[\begin{align} \iint_{\mathcal{M}\times\mathcal{M}} C K(p,q)\,\mathrm{d}V(p)\,\mathrm{d}V(q) &= C\frac{\left\Vert\varphi^-\right\Vert_{L^1} \left\Vert\varphi^+\right\Vert_{L^1}}{\frac{1}{2}\left\Vert\varphi\right\Vert_{L^1}} \\& = \tfrac{1}{2}C\left\Vert\varphi\right\Vert_{L_1} = \tfrac{1}{2}C\left\Vert\mathrm{d}_1 F\right\Vert_{L^1} . \end{align}\]

Finally if \(\mathcal{M}\) is compact but not connected, then it consists of a finite number of connected components; apply the preceding construction to each component separately and take the sum. ◻

Proof of 14 for orientable manifolds. By 13, the 1-current \(T_F-R_F\) satisfies \(\partial_1 (T_F-R_F)=0\) and \[\begin{align} \mathbb{M}(T_F-R_F) &\leq \mathbb{M}(T_F) + \mathbb{M}(R_F) \\& \lesssim \left\Vert F\right\Vert_{L^1(\Lambda^{n-1})} + \left\Vert\mathrm{d}_{n-1} F\right\Vert_{L^1(\Lambda^n)} . \end{align}\] By 13 we conclude that \[\int_0^1 t^{\alpha/2-1}\| e^{t\Delta_{1,c}} (T_F-R_F)\|_{L^{n/(n-\alpha),1}(\Lambda^{n-1})} \, \mathrm{d}t \lesssim \|F\|_{L^1(\Lambda^{n-1})}+\|\mathrm{d}F\|_{L^1(\Lambda^n)} .\] On the other hand, 3 implies that \[\begin{align} \int_0^1 t^{\alpha/2-1}\| e^{t\Delta_{1,c}} R_F\|_{L^{n/(n-\alpha),1}(\Lambda^{n-1})} \, \mathrm{d}t &\lesssim \mathbb{M}(R_F)^{1-\alpha/n} \left\Vert R_F\right\Vert_{\mathrm{Morrey}}^{\alpha/n} \\& \lesssim \left\Vert\mathrm{d}F\right\Vert_{L^1(\Lambda^{n})}^{1-\alpha/n+\alpha/n} \end{align}\] by the bounds in 13. Combining these two upper bounds, \[\begin{align} \int_0^1 t^{\alpha/2-1}\| e^{t\Delta_{1,c}} T_F \|_{L^{n/(n-\alpha),1}(\Lambda^{n-1})} \, \mathrm{d}t \lesssim \|F\|_{L^1(\Lambda^{n-1})}+\|\mathrm{d}F\|_{L^1(\Lambda^n)} \end{align}\] also. Then the identities \[\begin{align} \left\Vert e^{t\Delta_{1,c}} T_F\right\Vert_{L^{n/(n-\alpha),1}(\Lambda^{n-1})} &= \left\Vert T_{e^{t\Delta_{n-1}}F} \right\Vert_{L^{n/(n-\alpha),1}(\Lambda^{n-1})} \\& =\left\Vert e^{t\Delta_{n-1}}F \right\Vert_{L^{n/(n-\alpha),1}(\Lambda^{n-1})} \end{align}\] from [TomegaPropagatorsCompatible,NormDualityCurrentsForms] complete the proof. ◻

Proof. Note that if \(\omega\) is written in coordinates on \(U_r\) as \(\omega=\sum_{\left\vert K\right\vert} \omega_K \, \mathrm{d}x_r^K\), then \(\omega_K=\omega\bigl( (\frac{\partial}{\partial x_r^i} \bigr)_{i\in K})\) recovers the coordinate functions. Thus, abbreviating \(\eta_J=\omega\wedge \mathrm{d}y_r^{J^c}\) for \(\left\vert J\right\vert=j\) and \(\tilde{\omega}=\psi_{k,j,r}(\varphi_{k,j,r}(\omega))=\psi_{k,j,r}(\vec{\eta})\), it suffices to show that \(\tilde{\omega}_K=\omega_K\) on \(U'_r\), for all \(\left\vert K\right\vert=k\). We have \[\begin{align} \tilde{\omega}_K = \tilde{\omega}((Y_r)_K) &= \frac{1}{\binom{n-k}{j-k}} \sum_{\left\vert J\right\vert=j} \eta_J((Y_r)_K, (Y_r)_{J^c}) \\& = \frac{1}{\binom{n-k}{j-k}} \sum_{\left\vert J\right\vert=j, K\subset J} \eta_J((Y_r)_K, (Y_r)_{J^c}) \end{align}\] because, in any term for which \(K\) is not a subset of \(J\), we will have \(K\cap J^c\neq\emptyset\) and hence the arguments to \(\eta_J\) will contain at least one repeated vector field \(Y_{r,i}\). By 53 and the fact that \(\eta_J=\omega\wedge \mathrm{d}y_r^{J^c}\), it follows that \[\begin{align} \eta_J((Y_r)_K, (Y_r)_{J^c}) = \omega((Y_r)_K) \, \mathrm{d}y_r^{J^c}((Y_r)_{J^c}) = \omega((Y_r)_K) = \omega_K \end{align}\] on \(U'_r\). Hence \[\begin{align} \tilde{\omega}_K = \frac{1}{\binom{n-k}{j-k}} \sum_{\left\vert J\right\vert=j, K\subset J} \omega_K = \omega_{K} \end{align}\] on \(U'_r\), since the choice of \(J\) corresponds to choosing the \(j-k\) elements of \(J\setminus K\) out of the \(n-k\) choices in \([n]\setminus K\). ◻

Proof. Both \(\varphi_{k,r}\) and \(\psi_{k,r}\) are linear over scalar-valued functions, so \[\Psi_k(\Phi_k(\omega)) = \sum_{r=1}^L\psi_{k,r}( \chi_r \varphi_{k,r}(\omega) ) = \sum_{r=1}^L\chi_r \psi_{k,r}(\varphi_{k,r}(\omega)) = \sum_{r=1}^L\chi_r \omega = \omega\] where we used 14 and the fact that \(\chi_r\) is supported in \(U'_r\). ◻

Proof. The first statement follows immediately by considering vector entries separately: \[\mathrm{d}_{n-j+k} (\omega\wedge \mathrm{d}y_r^{J^c}) = \mathrm{d}_k\omega \wedge \mathrm{d}y_r^{J^c}\] for each \(r,J\). When \(j\geq k+1\), the second statement follows from the first because of 15. ◻

Proof. The first assertion follows because the \(\mathrm{d}y_r^i\) and \(Y_{r,i}\) form a finite collection of smooth 1-forms and vector fields on a compact manifold, and therefore have uniformly bounded norms. The remaining assertions follow because the \(L^1\) and \(L^{n/(n-\alpha),1}\) norms depend monotonically on the pointwise norms \(\left\Vert\cdot\right\Vert_p\). ◻

Proof of 16. The argument works by comparing the \(k\)-form heat propagator \(e^{t\Delta_k}\omega\) to the quantity \(\Psi_k(e^{t\vec{\Delta}_{n-1}}\Phi_k(\omega))\) involving an \((n-1)\)-form propagator: \[\begin{align} \int_0^1& t^{\alpha/2-1} \big\Vert e^{t\Delta_k}\omega \big\Vert_{L^{n/(n-\alpha),1}} \,\mathrm{d}t \notag\\& \leq \int_0^1 t^{\alpha/2-1} \big\Vert\Psi_k(e^{t\vec{\Delta}_{n-1}}\Phi_k(\omega)) \big\Vert_{L^{d/(d-\alpha),1}} \,\mathrm{d}t \notag\\&\quad + \int_0^1 t^{\alpha/2-1} \big\Vert e^{t\Delta_k}\omega - \Psi_k(e^{t\vec{\Delta}_{n-1}}\Phi_k(\omega)) \big\Vert_{L^{d/(d-\alpha),1}} \,\mathrm{d}t \notag\\& =: I+\mathit{II} . \label{IandII} \end{align}\tag{54}\] In particular, the estimate for \(I\) follows directly from \((n-1)\)-form bounds: \[\begin{align} \label{PromotedPropagatedBound} I&= \int_0^1 t^{\alpha/2-1} \big\Vert \Psi_{k}(e^{t\vec{\Delta}_{n-1}}\Phi_{k}(\omega)) \big\Vert_{L^{n/(n-\alpha),1}} \,\mathrm{d}t \\& \lesssim \int_0^1 t^{\alpha/2-1} \big\Vert e^{t\vec{\Delta}_{n-1}}\Phi_{k}(\omega) \big\Vert_{L^{n/(n-\alpha),1}} \,\mathrm{d}t &&\text{by \Cref{PhiPsiBounded}} \notag\\& \lesssim \left\Vert\Phi_{k}(\omega)\right\Vert_{L^1} + \big\Vert\vec{\mathrm{d}}_{n-1}\Phi_{k}(\omega)\big\Vert_{L^1} &&\text{by \Cref{int01Bound95General}} \notag\\& \lesssim \left\Vert\Phi_{k}(\omega)\right\Vert_{L^1} + \big\Vert\Phi_{k+1}(\mathrm{d}_k \omega)\big\Vert_{L^1} &&\text{by \Cref{dPhiRelation}} \notag\\& \lesssim \left\Vert\omega\right\Vert_{L^1} + \left\Vert\mathrm{d}_k\omega\right\Vert_{L^1} &&\text{by \Cref{PhiPsiBounded}} . \notag \end{align}\tag{55}\]

To estimate \(\mathit{II}\), we recall several results concerning the heat propagator on forms, see for instance BerGetVer1992?*Theorem 2.30. It can be shown that the heat propagator is an integral operator, and we write \(\mathrm{p}_{t}^k(p,q)(\cdot) \in\mathrm{Hom}(\wedge^k(T^*_q\mathcal{M}), \wedge^k(T^*_p\mathcal{M}))\) for the kernel corresponding to \(e^{t\Delta_k}\), i.e., \[\label{k95heat95kernel} e^{t\Delta_k}\omega(p) = \int_\mathcal{M}\mathrm{p}_{t}^k(p,q)(\omega(q)) \,\mathrm{d}V(q) .\tag{56}\] Since \(\Psi_k\) and \(\Phi_k\) operate pointwise, we have similarly \[\begin{align} \Psi_k(e^{t\vec{\Delta}_{n-1}}\Phi_k(\omega)) &= \int_\mathcal{M}\widetilde{\mathrm{p}}_{t}^{k}(p,q) (\omega(q))\,\mathrm{d}V(q) , \end{align}\] where \[\label{Psi95Phi95kernel} \widetilde{\mathrm{p}}_{t}^{k}(p,q) = \Psi_k\vert_p \circ \mathbf{p}_{t}^{n-1}(p,q) \circ \Phi_k\vert_q .\tag{57}\] In other words, the kernel for \(\Psi_k \circ e^{t\vec{\Delta}_{n-1}} \circ \Phi_k\) is the composition of the linear maps \(\Psi_k\vert_p\), \(\mathbf{p}_{t}^{n-1}(p,q)\), and \(\Phi_k\vert_q\), where \(\mathbf{p}_{t}^{n-1}(p,q)\) means the linear map \(\mathrm{p}_{t}^{n-1}(p,q)\) applied elementwise to a vector with entries in \(\wedge^{n-1}(T^*_q\mathcal{M})\).

The kernels \(\mathrm{p}_t^k,\mathrm{p}_t^{n-1}\) are smooth for \(t>0\), and for \(t\to 0\) behave like scalar heat kernels times an asymptotic series in powers of \(t\). Namely, there are functions \(V_{k,i}(p,q)\) with values in \(\mathrm{Hom}(\wedge^{k}(T^*_q\mathcal{M}), \wedge^{k}(T^*_p\mathcal{M}))\) and depending smoothly on \(p,q\) such that, defining \[\begin{align} \mathrm{p}_t^{k,N}(p,q) &= \frac{\exp(-d_\mathcal{M}(p,q)^2/(4t))}{(4\pi t)^{n/2}} \sum_{i=0}^N t^i V_{k,i}(p,q) , \tag{58} \\ \mathrm{p}_t^{n-1,N}(p,q) &= \frac{\exp(-d_\mathcal{M}(p,q)^2/(4t))}{(4\pi t)^{n/2}} \sum_{i=0}^N t^i V_{n-1,i}(p,q) , \tag{59} \end{align}\] and by fixing \(N\) large enough (specifically, \(N\geq n/2\) suffices) we can ensure that \[\begin{align} \|\mathrm{p}_t^k(p,q) - \mathrm{p}_t^{k,N}(p,q)\|_{p,q} &\lesssim 1 , \tag{60} \\ \|\mathrm{p}_t^{n-1}(p,q) - \mathrm{p}_t^{n-1,N}(p,q)\|_{p,q} &\lesssim 1 , \tag{61} \end{align}\] uniformly over \(0<t\leq 1\) and \(p,q\in\mathcal{M}\), where \(\left\Vert\cdot\right\Vert_{p,q}\) means the operator norm on \(\mathrm{Hom}(\wedge^{k}(T^*_q\mathcal{M}), \wedge^{k}(T^*_p\mathcal{M}))\). Moreover \[\label{base95term95diagonal} V_{k,0}(p,p)=\mathrm{identity}_{\wedge^{k}(T^*_p\mathcal{M})} ,\quad V_{n-1,0}(p,p)=\mathrm{identity}_{\wedge^{n-1}(T^*_p\mathcal{M})} .\tag{62}\]

By linearity, the kernel \(\widetilde{\mathrm{p}}_t^k\) from 57 has a similar approximation \[\begin{align} \widetilde{\mathrm{p}}_t^{k,N}(p,q) &= \frac{\exp(-d_\mathcal{M}(p,q)^2/(4t))}{(4\pi t)^{n/2}} \sum_{i=0}^N t^i \widetilde{V}_{k,i}(p,q) \label{n-195kernel95approx95promoted} \end{align}\tag{63}\] where \(\widetilde{V}_{k,i}(p,q)=\Psi_k\vert_p \circ \mathbf{V}_{n-1,i}(p,q) \circ \Phi_k\vert_q\) for all \(i\), and where similarly \(\mathbf{V}_{n-1,i}(p,q)\) means \(V_{n-1,i}(p,q)\) applied entrywise to a vector with entries in \(\wedge^{n-1}(T_q^*\mathcal{M})\). The linear mappings \(\Psi_k\vert_p, \Phi_k\vert_q\) vary smoothly as functions of \(p,q\in\mathcal{M}\), and since \(\mathcal{M}\) is compact it follows that they have uniformly bounded operator norms. From the bound 61 it therefore follows that \[\label{kernel95error95promoted} \|\widetilde{\mathrm{p}}_t^k(p,q) - \widetilde{\mathrm{p}}_t^{k,N}(p,q)\|_{p,q} \lesssim 1 ,\tag{64}\] again uniformly over \(0<t\leq 1\) and \(p,q\in\mathcal{M}\).

By 15 and 62 , \[\begin{align} \widetilde{V}_{k,0}(p,p) &= \Psi_k\vert_p \circ \mathrm{identity}_{(\wedge^{n-1}(T^*_p\mathcal{M}))^{L\binom{n}{k+1}}} \circ \Phi_k\vert_p \notag\\& = \mathrm{identity}_{\wedge^k(T^*_p\mathcal{M})} , \end{align}\] and in particular \(\widetilde{V}_{k,0}(p,p)=V_{k,0}(p,p)\) for all \(p\in\mathcal{M}\). Since both \(V_{k,0}\) and \(\widetilde{V}_{k,0}\) are smooth, it follows that \[\big\Vert V_{k,0}(p,q) - \widetilde{V}_{k,0}(p,q)\big\Vert_{p,q} = O(d_\mathcal{M}(p,q)) ,\] uniformly over \(p,q\in\mathcal{M}\). Meanwhile, the functions \(V_{k,i},\widetilde{V}_{k,i}\) are continuous with compact support and therefore have uniformly bounded norms. Combining, \[\begin{align} &\left\Vert \mathrm{p}_t^{k,N}(p,q) - \widetilde{\mathrm{p}}_t^{k,N}(p,q) \right\Vert_{p,q} \notag\\&\quad \leq \frac{\exp(-d_\mathcal{M}(p,q)^2/(4t))}{(4\pi t)^{n/2}} \left[ \big\Vert V_{k,0}(p,q) - \widetilde{V}_{k,0}(p,q)\big\Vert_{p,q} \Bigg.\right. \notag\\&\qquad\qquad\qquad\qquad\qquad\qquad \left. + \sum_{i=1}^N t^i \bigl( \big\Vert V_{k,i}(p,q)\big\Vert_{p,q} + \big\Vert\widetilde{V}_{k,i}(p,q)\big\Vert_{p,q} \bigr) \right] \notag\\&\quad \lesssim \frac{\exp(-d_\mathcal{M}(p,q)^2/(4t))}{(4\pi t)^{n/2}} [ d_\mathcal{M}(p,q) + t ] , \label{promotion95error95N} \end{align}\tag{65}\] uniformly over \(0<t\leq 1\) and \(p,q\in\mathcal{M}\), where the last bound has used \(t^i\leq t\) for all \(i\geq 1\). Further combining 60 , 64 , and 65 , \[\label{promotion95error95total} \left\Vert\mathrm{p}_t^k(p,q) - \widetilde{\mathrm{p}}_t^k(p,q)\right\Vert_{p,q} \lesssim \frac{\exp(-d_\mathcal{M}(p,q)^2/(4t))}{(4\pi t)^{n/2}} [ d_\mathcal{M}(p,q) + t ] + 1 .\tag{66}\]

Considered as a function of \(p\) with \(q\) fixed, the upper bound in 66 has Lorentz norm bounded by \[\label{UpperBoundLorentz} \left\|\frac{\exp(-d_\mathcal{M}(\cdot,q)^2/(4t))}{(4\pi t)^{n/2}} \left(d_\mathcal{M}(\cdot,q) + t \right)+1\right\|_{L^{n/(n-\alpha),1}} \lesssim t^{-\alpha/2}t^{1/2} + 1 .\tag{67}\] We can verify 67 using the Interpolation [1](#Interpolation){reference-type=“ref+Label” reference=“Interpolation”}as follows. Recall from 46 that the non-constant part of 67 has \(L^1\) norm of order \(t^{1/2}\), while it is easy to check that its \(L^\infty\) norm is of order \(t^{1/2-n/2}\). Thus \[\left\Vert \frac{\exp(-d_\mathcal{M}(\cdot,q)^2/(4t))}{(4\pi t)^{n/2}} \left(d_\mathcal{M}(\cdot,q) + t \right) \right\Vert_{L^{n/(n-\alpha),1}} \lesssim (t^{1/2})^{1-\alpha/n} (t^{1/2-n/2})^{\alpha/n} ,\] giving the term \(t^{-\alpha/2+1/2}\) as required. Meanwhile the constant term gives a trivial contribution by the compactness of the manifold.

Thus applying Minkowski’s inequality for integrals to the pointwise upper bound \[\left\Vert e^{t\Delta_k}\omega(p) - \Psi_k(e^{t\vec{\Delta}_{n-1}}\Phi_k(\omega)) (p)\right\Vert_{p} \leq \int_\mathcal{M}\left\Vert\mathrm{p}_t^k(p,q) - \widetilde{\mathrm{p}}_t^k(p,q)\right\Vert_{p,q} \left\Vert\omega(q)\right\Vert_q \, \mathrm{d}V(q)\] yields \[\begin{align} &\big\Vert e^{t\Delta_k}\omega - \Psi_k(e^{t\vec{\Delta}_{n-1}}\Phi_k(\omega)) \big\Vert_{L^{n/(n-\alpha),1}} \\&\quad \lesssim \int_\mathcal{M}\left\|\frac{\exp(-d_\mathcal{M}(\cdot,q)^2/(4t))}{(4\pi t)^{n/2}} \left(d_\mathcal{M}(\cdot,q) + t^{1/2} \right)+1\right\|_{L^{n/(n-\alpha),1}}\left\Vert\omega(q)\right\Vert_q \,\mathrm{d}V(q) \\&\quad \lesssim \left(t^{-\alpha/2}t^{1/2} +1\right)\left\Vert\omega\right\Vert_{L^1} . \end{align}\] Finally, returning to 54 , we have \[\begin{align} \mathit{II} &= \int_0^1 t^{\alpha/2} \big\Vert e^{t\Delta_k}\omega - \Psi_k(e^{t\vec{\Delta}_{n-1}}\Phi_k(\omega)) \big\Vert_{L^{n/(n-\alpha),1}} \frac{\mathrm{d}t}{t} \\& \lesssim \int_0^1 \left( t^{1/2} + t^{\alpha/2} \right)\left\Vert\omega\right\Vert_{L^1} \frac{\mathrm{d}t}{t} \\& \lesssim \left\Vert\omega\right\Vert_{L^1} \end{align}\] which along with 55 completes the proof. ◻

Proof of 15. Now suppose \(\mathrm{d}_k\omega=0\) is given. Then 16 implies that \(\vec{\mathrm{d}}_{n-1}\Phi_k(\omega)=0\), i.e., every entry of \(\Phi_k(\omega)\) is a closed \((n-1)\)-form to which 12 applies. Then the bound 55 can be replaced by \[\begin{align} \label{PromotedPropagatedBoundClosed} I&= \int_0^1 t^{\alpha/2-1} \big\Vert \Psi_{k}(e^{t\vec{\Delta}_{n-1}}\Phi_{k}(\omega)) \big\Vert_{L^{n/(n-\alpha),1}} \,\mathrm{d}t \\& \lesssim \int_0^1 t^{\alpha/2-1} \big\Vert e^{t\vec{\Delta}_{n-1}}\Phi_{k}(\omega) \big\Vert_{L^{n/(n-\alpha),1}} \,\mathrm{d}t &&\text{by \Cref{PhiPsiBounded}} \notag\\& \lesssim \left\Vert\Phi_{k}(\omega)\right\Vert_{L^1} &&\text{by \Cref{int01Bound95dF610}} \notag\\& \lesssim \left\Vert\omega\right\Vert_{L^1} &&\text{by \Cref{PhiPsiBounded}} , \notag \end{align}\tag{68}\] using 12 instead of 14. The rest of the proof is identical, and we find that \(\mathit{II}\lesssim \left\Vert\omega\right\Vert_{L^1}\) and thus \(I+\mathit{II}\lesssim \left\Vert\omega\right\Vert_{L^1}\), as required. ◻

Proof. Define \(\epsilon=t_1/2\), so that for any \(t\geq t_1\) we may express \[\begin{align} t= \epsilon +(t-t_1)+ \epsilon \end{align}\] and use the semigroup property of the heat propagator to obtain \[\begin{align} e^{t\Delta_k} \xi = e^{\epsilon\Delta_k} e^{(t-t_1)\Delta_k} e^{\epsilon\Delta_k} \xi . \end{align}\] By 3 and Hölder’s inequality we deduce \[\begin{align} \sup_{p\in \mathcal{M}} \left\Vert e^{t\Delta_k} \xi(p)\right\Vert_p &\leq C \sup_{p\in \mathcal{M}} \int_\mathcal{M}K_\epsilon(p,q) \left\Vert e^{(t-t_1)\Delta_k} e^{\epsilon\Delta_k} \xi(q)\right\Vert_q \,\mathrm{d}V(q) \notag\\ &\lesssim \|e^{(t-t_1)\Delta_k} e^{\epsilon\Delta_k}\xi\|_{L^2(\Lambda^k)} \label{LargetLinftyFromL2} . \end{align}\tag{69}\] A similar argument shows that \[\label{LargetL2FromL1} \|e^{\epsilon\Delta_k}\xi\|_{L^2(\Lambda^k)} \lesssim \|\xi\|_{L^1(\Lambda^k)}\tag{70}\] and therefore it remains to bound \(\|e^{(t-t_1)\Delta_k} e^{\epsilon\Delta_k}\xi\|_{L^2(\Lambda^k)}\) in terms of \(\|e^{\epsilon\Delta_k}\xi\|_{L^2(\Lambda^k)}\). The proof is similar to BerGetVer1992?*p. 88. Let \(\{\varphi_i\}_{i=1}^\infty\) be an orthonormal basis for \(L^2(\Lambda^k)\) consisting of eigenforms of \(\Delta_k\) with corresponding eigenvalues \(\{\lambda_i\}_{i=1}^\infty\). The Laplacian \(\Delta_k\) is positive semi-definite with a purely discrete spectrum, so finitely many eigenvalues \(\lambda_i\) may be zero but the remaining eigenvalues satisfy \(\min\left\{\lambda_i\colon \lambda_i\neq 0\right\}=\delta>0\). Using the relation \[\begin{align} e^{(t-t_1)\Delta_k}\varphi_i = \exp(-\lambda_i (t-t_1)) \varphi_i \end{align}\] along with Parseval’s identity, \[\begin{align} \|e^{(t-t_1)\Delta_k} e^{\epsilon\Delta_k}\xi\|_{L^2(\Lambda^k)} &=\left(\sum_{i=1}^\infty \savebox{\@brx}{\(\m@th{\langle}\)} \mathopen{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}e^{(t-t_1)\Delta_k} e^{\epsilon\Delta_k}\xi,\varphi_i\savebox{\@brx}{\(\m@th{\rangle}\)} \mathclose{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}^2 \right)^{1/2}\\ &= \left(\sum_{i=1}^\infty \exp(-2\lambda_i (t-t_1)) \savebox{\@brx}{\(\m@th{\langle}\)} \mathopen{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}e^{\epsilon\Delta_k}\xi,\varphi_i\savebox{\@brx}{\(\m@th{\rangle}\)} \mathclose{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}^2 \right)^{1/2}. \end{align}\] By assumption, \(\xi\) is orthogonal to the space of harmonic forms, so \(e^{\epsilon\Delta_k}\xi\) is as well and therefore all of the terms with \(\lambda_i=0\) vanish. This implies \[\begin{align} \|e^{(t-t_1)\Delta_k} e^{\epsilon\Delta_k}\xi\|_{L^2(\Lambda^k)} \leq e^{-\delta(t-t_1)} \|e^{\epsilon\Delta_k}\xi\|_{L^2(\Lambda^k)} , \end{align}\] and together with 69 and 70 this completes the proof. ◻

Proof. Because of the compactness of the manifold, \(L^\infty\) bounds immediately give \(L^1\) bounds, whereupon the Interpolation [1](#Interpolation){reference-type=“ref+Label” reference=“Interpolation”}gives the Lorentz bound \[\| e^{t\Delta_k} F\|_{L^{n/(n-\alpha),1}(\Lambda^k)}\lesssim V(\mathcal{M})^{1-\alpha/n} \| e^{t\Delta_k} F \|_{L^{\infty}(\Lambda^k)} .\] Thus 18 with \(t_1=1\) gives \[\begin{align} \int_1^\infty t^{\alpha/2-1} \|e^{t\Delta_k} F \|_{L^{n/(n-\alpha),1}(\Lambda^k)}\,\mathrm{d}t &\lesssim \int_1^\infty t^{\alpha/2-1} \|e^{t\Delta_k} F \|_{L^{\infty}(\Lambda^k)}\,\mathrm{d}t \\ &\lesssim \int_1^\infty t^{\alpha/2-1} \exp(-c_3t) \|F\|_{L^1(\Lambda^k)}\,\mathrm{d}t \\ &\lesssim \|F\|_{L^1(\Lambda^k)} . \qedhere \end{align}\] ◻

Proof of 6 for orientable manifolds. Throughout, let the \(k\)-form \(F\in L^1\cap\mathcal{H}^\perp(\Lambda^k)\) be orthogonal to the space of harmonic \(k\)-forms. If \(\mathrm{d}_k F=0\), the result follows immediately from [int01Bound_Closed_kform,int1inftyBound].

To handle \(\mathrm{d}^*_k F\) when \(k\in\left\{1,\dotsc,n\right\}\), apply Hodge duality. Define the \((n-k)\)-form \[\tilde{F} = \star_k F .\] The Hodge star is an isometry; preserves the Hodge Laplacian and hence also the heat propagator, Riesz transform, and orthogonality to harmonic forms (apart from the degree of the forms concerned); and interchanges \(\mathrm{d}\) with \(\mathrm{d}^*\) modulo signs, see 19 and 85 . Thus \[\label{starInterchanges} \begin{align} \big\Vert\tilde{F}(p)\big\Vert_p &= \big\Vert F(p)\big\Vert_p \quad\text{for all }p\in\mathcal{M} , \\ \big\Vert\tilde{F}\big\Vert_{L^1(\Lambda^{n-k})} &= \big\Vert F\big\Vert_{L^1(\Lambda^k)} \\ e^{t\Delta_{n-k}}\tilde{F} &= \star_k e^{t\Delta_k}F , \\ \big\Vert e^{t\Delta_{n-k}}\tilde{F}\big\Vert_{L^{n/(n-\alpha),1}(\Lambda^{n-k})} &= \big\Vert e^{t\Delta_k}F\big\Vert_{L^{n/(n-\alpha),1}(\Lambda^k)} , \\ \mathrm{d}_{n-k}\tilde{F} &= (-1)^k \star_{k-1} \mathrm{d}^*_k F , \\ \big\Vert\mathrm{d}_{n-k}\tilde{F}\big\Vert_{L^1(\Lambda^{n-k+1})} &= \big\Vert\mathrm{d}^*_k F\big\Vert_{L^1(\Lambda^{k-1})} . \end{align}\tag{71}\] In particular, the assumption \(\mathrm{d}^*_k F=0\) gives \(\mathrm{d}_{n-k}\tilde{F}=0\), so we may apply our preceding result to the closed \(\tilde{k}\)-form \(\tilde{F}\), where \(\tilde{k}=n-k\in\left\{1,\dotsc,n-1\right\}\), to find \[\begin{align} \int_0^\infty t^{\alpha/2-1} \big\Vert e^{t\Delta_k}F\big\Vert_{L^{n/(n-\alpha),1}(\Lambda^k)} \, \mathrm{d}t &= \int_0^\infty t^{\alpha/2-1} \big\Vert e^{t\Delta_{n-k}}\tilde{F}\big\Vert_{L^{n/(n-\alpha),1}(\Lambda^{n-k})} \, \mathrm{d}t \\& \lesssim \big\Vert\tilde{F}\big\Vert_{L^1(\Lambda^{n-k})} = \big\Vert F\big\Vert_{L^1(\Lambda^k)} . \qedhere \end{align}\] ◻

Proof of 1 for orientable manifolds. The bound 9 follows immediately from 6 by Minkowski’s inequality for integrals. ◻

Proof of 4 for orientable manifolds. The bound 17 follows immediately from 16 and 7. The bound 18 follows from 17 by the same Hodge duality argument as in the proof of 6, noting from 71 that the term \(\big\Vert\mathrm{d}_{n-k}\tilde{F}\big\Vert_{L^1(\Lambda^{n-k+1})}\) in the resulting upper bound reduces to the term \(\left\Vert\mathrm{d}^*_k F\right\Vert_{L^1(\Lambda^{k-1})}\) in 18 . ◻

Proofs for non-orientable manifolds. Now let \(\mathcal{M}\) be a general compact manifold, not necessarily orientable, and let \(\widehat{\mathcal{M}}\) be the orientable double cover of \(\mathcal{M}\). That is, \(\widehat{\mathcal{M}}\) is an orientable Riemannian manifold and there is a natural local isometry \(\pi\colon\widehat{\mathcal{M}}\to\mathcal{M}\) such that \(\widehat{\pi}\) is two-to-one, i.e., \(\widehat{\pi}^{-1}(\left\{p\right\})\) contains exactly two elements of \(\widehat{\mathcal{M}}\) for each \(p\in\mathcal{M}\); see LeeSmoothManifolds?*Chapter 15. Since \(\mathcal{M}\) is compact, so is \(\widehat{\mathcal{M}}\). The conclusions of [RieszClosedCoclosed_k,divinLresult,BesovClosedCoclosed,int01Bound_dF=0,int01Bound_General,int01Bound_Closed_kform,int01Bound_General_kform] therefore apply to \(\widehat{\mathcal{M}}\).

Let \(F\in C^\infty(\Lambda^k\mathcal{M})\) be a smooth \(k\)-form on \(\mathcal{M}\), and write \(\widehat{F}=\widehat{\pi}^\# F\in C^\infty(\Lambda^k\widehat{\mathcal{M}})\) for its pull-back to \(\widehat{\mathcal{M}}\). Here we write \(\Lambda^k\mathcal{M},\Lambda^k\widehat{\mathcal{M}}\) to distinguish forms on \(\mathcal{M}\) or \(\widehat{\mathcal{M}}\). By general properties of pull-backs and exterior derivatives, \(\mathrm{d}_k \widehat{F} = \widehat{\pi}^\# \mathrm{d}_k F\), where on the left-hand side the exterior derivative is taken in \(\widehat{\mathcal{M}}\). Moreover, since \(\widehat{\pi}\) is a local isometry, it follows that \(\widehat{\Delta}_k \widehat{F} = \widehat{\pi}^\# \Delta_k F\), where \(\widehat{\Delta}_k\) means the Hodge Laplacian on \(\widehat{\mathcal{M}}\), and consequently \(e^{t\widehat{\Delta}_k}\widehat{F} = \widehat{\pi}^\# e^{t\Delta_k} F\).

Note that in any compact Riemannian manifold, a \(k\)-form \(\omega\) is orthogonal to the space of harmonic \(k\)-forms for that manifold if and only if its heat propagator tends to 0 as \(t\to\infty\). Since \(e^{t\widehat{\Delta}_k}\widehat{F} = \widehat{\pi}^\# e^{t\Delta_k} F\), it follows that if \(F\) is orthogonal to the space of harmonic \(k\)-forms for \(\mathcal{M}\), then \(\widehat{F}\) is orthogonal to the space of harmonic \(k\)-forms for \(\widehat{\mathcal{M}}\).

The local isometry property also implies that \(\mathrm{d}^*_k \widehat{F} = \widehat{\pi}^\# \mathrm{d}^*_k F\). For the same reason, pointwise norms coincide: for all \(\widehat{p}\in\widehat{\mathcal{M}}\), \[\begin{align} \big\Vert\widehat{F}(\widehat{p})\big\Vert_{\widehat{p}} &= \big\Vert F(\widehat{\pi}(\widehat{p}))\big\Vert_{\widehat{\pi}(\widehat{p})} , \\ \big\Vert\mathrm{d}_k\widehat{F}(\widehat{p})\big\Vert_{\widehat{p}} &= \big\Vert\mathrm{d}_k F(\widehat{\pi}(\widehat{p}))\big\Vert_{\widehat{\pi}(\widehat{p})} , \\ \big\Vert\mathrm{d}^*_k\widehat{F}(\widehat{p})\big\Vert_{\widehat{p}} &= \big\Vert\mathrm{d}^*_k F(\widehat{\pi}(\widehat{p}))\big\Vert_{\widehat{\pi}(\widehat{p})} , \\ \big\Vert e^{t\widehat{\Delta}_k}\widehat{F}(\widehat{p})\big\Vert_{\widehat{p}} &= \left\Vert e^{t\Delta_k}F(\widehat{\pi}(\widehat{p}))\right\Vert_{\widehat{\pi}(\widehat{p})} . \end{align}\] Integrate over \(\widehat{p}\), noting that because of the double cover, the push-forward under \(\widehat{\pi}\) of volume measure on \(\widehat{\mathcal{M}}\) is twice the volume measure on \(\mathcal{M}\). Hence \[\begin{align} \big\Vert\widehat{F}\big\Vert_{L^1(\Lambda^k\widehat{\mathcal{M}})} &= 2\big\Vert F\big\Vert_{L^1(\Lambda^k\mathcal{M})} , \\ \big\Vert\mathrm{d}_k\widehat{F}\big\Vert_{L^1(\Lambda^{k+1}\widehat{\mathcal{M}})} &= 2\big\Vert\mathrm{d}_k F\big\Vert_{L^1(\Lambda^{k+1}\mathcal{M})} , \\ \big\Vert\mathrm{d}^*_k\widehat{F}\big\Vert_{L^1(\Lambda^{k-1}\widehat{\mathcal{M}})} &= 2\big\Vert\mathrm{d}^*_k F\big\Vert_{L^1(\Lambda^{k-1}\mathcal{M})} . \end{align}\] Similar reasoning shows that Lorentz norms in \(\mathcal{M}\) and \(\widehat{\mathcal{M}}\) are also related by a constant factor, \[\begin{align} \big\Vert e^{t\widehat{\Delta}_k}\widehat{F}\big\Vert_{L^{n/(n-\alpha),1}(\Lambda^{n-1}\widehat{\mathcal{M}})} &= C\left\Vert e^{t\Delta_{n-1}}F\right\Vert_{L^{n/(n-\alpha),1}(\Lambda^{n-1}\mathcal{M})} , \end{align}\] where careful calculation shows that \(C=2^{1-\alpha/n}\).

Consequently, the bounds in [RieszClosedCoclosed_k,divinLresult,BesovClosedCoclosed,int01Bound_dF=0,int01Bound_General,int01Bound_Closed_kform,int01Bound_General_kform] for \(\mathcal{M}\) all follow from the corresponding bounds for \(\widehat{\mathcal{M}}\). ◻

Proof of 2. If \(\omega \in C^\infty(\Lambda^k)\) then, by 5 and 33 , \[\begin{align} \mathcal{I}_{2,k} (-\Delta_k\omega) &= \int_0^\infty e^{t\Delta_k}(-\Delta_k\omega)\,\mathrm{d}t \label{left95inverse}\\ &= \lim_{\epsilon \to 0}\int_{\epsilon} ^{1/\epsilon} -\Delta_ke^{t\Delta_k}\omega\,\mathrm{d}t \notag\\ &= \lim_{\epsilon \to 0}\int_{\epsilon} ^{1/\epsilon} -\frac{\partial}{\partial t} e^{t\Delta_k}\omega\,\mathrm{d}t \notag\\ &= \lim_{\epsilon \to 0}\left[e^{\epsilon\Delta_k}\omega -e^{\epsilon^{-1}\Delta_k}\omega\right] \notag\\ &=P_{\mathcal{H}^\perp}\omega , \notag \end{align}\tag{72}\] where \(P_{\mathcal{H}^\perp}\omega=\omega-P_{\mathcal{H}}\omega\) denotes orthogonal projection onto \(\mathcal{H}^\perp(\Lambda^k)\).

Now consider \(F\in L^1\cap\mathcal{H}^\perp(\Lambda^k)\). Then \[\begin{align} Z=-\mathcal{I}_{2,k} F \end{align}\] is well-defined, and we claim that it is a weak solution of 10 in the sense that \[\begin{align} \savebox{\@brx}{\(\m@th{\langle}\)} \mathopen{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}Z, \Delta_k\omega\savebox{\@brx}{\(\m@th{\rangle}\)} \mathclose{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}} = \savebox{\@brx}{\(\m@th{\langle}\)} \mathopen{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}F ,\omega\savebox{\@brx}{\(\m@th{\rangle}\)} \mathclose{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}} \end{align}\] for all \(\omega \in C^\infty(\Lambda^k)\). This follows from 72 and the self-adjoint property 31 of the Riesz potentials: \[\begin{align} \savebox{\@brx}{\(\m@th{\langle}\)} \mathopen{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}Z, \Delta_k \omega\savebox{\@brx}{\(\m@th{\rangle}\)} \mathclose{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}} &= \savebox{\@brx}{\(\m@th{\langle}\)} \mathopen{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}-\mathcal{I}_{2,k} F ,\Delta_k\omega\savebox{\@brx}{\(\m@th{\rangle}\)} \mathclose{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}} \\ &=\savebox{\@brx}{\(\m@th{\langle}\)} \mathopen{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}} F ,-\mathcal{I}_{2,k}\Delta_k\omega\savebox{\@brx}{\(\m@th{\rangle}\)} \mathclose{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}} \\ &=\savebox{\@brx}{\(\m@th{\langle}\)} \mathopen{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}} F ,P_{\mathcal{H}^\perp}\omega\savebox{\@brx}{\(\m@th{\rangle}\)} \mathclose{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}\\ &=\savebox{\@brx}{\(\m@th{\langle}\)} \mathopen{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}} P_{\mathcal{H}^\perp}F ,\omega\savebox{\@brx}{\(\m@th{\rangle}\)} \mathclose{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}\\ &=\savebox{\@brx}{\(\m@th{\langle}\)} \mathopen{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}F ,\omega\savebox{\@brx}{\(\m@th{\rangle}\)} \mathclose{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}. \end{align}\] Thus \(Z\) is the required solution, readily verified to be unique since any two solutions of 10 differ by a harmonic \(k\)-form. Either of the assumptions \(\mathrm{d}_k F=0\) or \(\mathrm{d}^*_k F=0\) allows us to apply 1 and deduce the desired bound. ◻

Proof of 3. For \(F,G\) smooth and orthogonal to the spaces of harmonic forms, we may define \[\label{hodge95solution} Z=\mathrm{d}_{k-1} \mathcal{I}_{2,k-1} F + \mathrm{d}^*_{k+1} \mathcal{I}_{2,k+1} G .\tag{73}\] Then, recalling 32 and 72 and using \(\mathrm{d}^*_{k-1} F=0\), we have \[\begin{align} \mathrm{d}^*_k Z &= \mathrm{d}^*_k \mathrm{d}_{k-1} \mathcal{I}_{2,k-1} F + \mathrm{d}^*_k \mathrm{d}^*_{k+1} \mathcal{I}_{2,k+1} G \\ &= \mathrm{d}^*_k \mathrm{d}_{k-1} \mathcal{I}_{2,k-1} F \\ &= \mathcal{I}_{2,k-1} \mathrm{d}^*_k \mathrm{d}_{k-1} F \\ &= \mathcal{I}_{2,k-1} \left(\mathrm{d}^*_k \mathrm{d}_{k-1} + \mathrm{d}_{k-2} \mathrm{d}^*_{k-1}\right) F \\ &= \mathcal{I}_{2,k-1}(- \Delta_{k-1} F) = F , \end{align}\] and similarly \(\mathrm{d}_k Z=G\). When \(F,G\) are not smooth, it can be shown that 73 still defines a weak solution to the Hodge system, via a duality argument similar to the proof of 2.

For the purpose of estimation, it is useful to introduce the Riesz transforms, which can be defined by \[\label{Rdef} \begin{align} R_j \omega &:= \frac{1}{\Gamma(1/2)}\int_0^\infty t^{1/2-1}\mathrm{d}_j e^{t\Delta_j} \omega\, \mathrm{d}t , \\ R^*_j \omega &:= \frac{1}{\Gamma(1/2)}\int_0^\infty t^{1/2-1} \mathrm{d}^*_j e^{t\Delta_j} \omega \,\mathrm{d}t \end{align}\tag{74}\] whenever \(\omega \in L^r(\Lambda^k)\) or \(\omega\in L^{r,1}(\Lambda^k)\) for some \(r>1\). If in addition \(\omega \in \mathcal{H}^\perp(\Lambda^j)\), this definition is equivalent to \[\label{R} \begin{align} R_j \omega &\equiv \mathrm{d}_j \mathcal{I}_{1,j} \omega,\\ R^*_j \omega &\equiv \mathrm{d}^*_j \mathcal{I}_{1,j} \omega, \end{align}\tag{75}\] so recalling from 23 that \(\mathcal{I}_{1,j} \mathcal{I}_{1,j} = \mathcal{I}_{2,j}\), we can express the solution \(Z\) as \[\label{Z61RIF43RIG} Z = R_{k-1} \mathcal{I}_{1,k-1} F + R^*_{k+1} \mathcal{I}_{1,k+1} G .\tag{76}\] As noted by Strichartz Strichartz?, the Riesz transforms are bounded as linear operators \(R_{k-1} \colon L^r(\Lambda^{k-1}) \to L^r(\Lambda^{k})\) and \(R^*_{k+1} \colon L^r(\Lambda^{k+1}) \to L^r(\Lambda^{k})\) for all \(1<r<\infty\). By interpolation, it follows that they are also bounded as linear operators \(R_{k-1} \colon L^{r,s}(\Lambda^{k-1}) \to L^{r,s}(\Lambda^{k})\) and \(R^*_{k+1} \colon L^{r,s}(\Lambda^{k+1}) \to L^{r,s}(\Lambda^{k})\) on Lorentz spaces, for all \(1<r<\infty\) and \(1\leq s \leq \infty\).

Combining the observations above, 1 gives \[\begin{align} \|Z\|_{L^{n/(n-1),1}(\Lambda^k)} &\leq \| R_{k-1} \mathcal{I}_{1,k-1} F \|_{L^{n/(n-1),1}(\Lambda^k)} + \| R^*_{k+1} \mathcal{I}_{1,k+1} G \|_{L^{n/(n-1),1}(\Lambda^k)} \\& \lesssim \|\mathcal{I}_{1,k-1} F \|_{L^{n/(n-1),1}(\Lambda^{k-1})} + \| \mathcal{I}_{1,k+1} G \|_{L^{n/(n-1),1}(\Lambda^{k+1})} \\& \lesssim \|F\|_{L^1(\Lambda^{k-1})}+\|G\|_{L^1(\Lambda^{k+1})} . \qedhere \end{align}\] ◻

Proof of 1. Let \(\mathcal{M}= \mathbb{T}^3\). In this setting, we can identify both 2-forms and 1-forms with vector-valued functions, and both 3-forms and 0-forms with scalar functions, \[\begin{align} C^\infty(\Lambda^2\mathbb{T}^3) \cong C^\infty(\Lambda^1\mathbb{T}^3) &\cong C^\infty(\mathbb{T}^3;\mathbb{R}^3) , \\ C^\infty(\Lambda^3\mathbb{T}^3) \cong C^\infty(\Lambda^0\mathbb{T}^3) &\cong C^\infty(\mathbb{T}^3;\mathbb{R}) . \end{align}\] Under these identifications, the co-exterior derivative and exterior derivative on 2-forms correspond to the curl and divergence, respectively: \[\begin{align} \mathop{\mathrm{curl}}&\cong \mathrm{d}_2^* \colon C^\infty(\Lambda^2\mathbb{T}^3) \to C^\infty(\Lambda^1\mathbb{T}^3) , \\ \mathop{\mathrm{div}}&\cong \mathrm{d}_2 \colon C^\infty(\Lambda^2\mathbb{T}^3) \to C^\infty(\Lambda^3\mathbb{T}^3) . \end{align}\] Thus, identifying \(Z\) as a 2-form and \(F\) as a 1-form, the div-curl system 14 reads \[\begin{align} \mathrm{d}_2^* Z&= F , \\ \mathrm{d}_2 Z &= 0 . \end{align}\] For 1-forms, the identification of curl and divergence with exterior derivative and co-exterior derivative is reversed, \[\begin{align} \mathop{\mathrm{curl}}&\cong \mathrm{d}_1 \colon C^\infty(\Lambda^1\mathbb{T}^3) \to C^\infty(\Lambda^2\mathbb{T}^3) , \\ \mathop{\mathrm{div}}&\cong \mathrm{d}^*_1 \colon C^\infty(\Lambda^1\mathbb{T}^3) \to C^\infty(\Lambda^0\mathbb{T}^3) , \end{align}\] so the compatibility condition \(\mathop{\mathrm{div}}F=0\) can be expressed as \(\mathrm{d}^*_1 F=0\). The harmonic 1-forms on \(\mathbb{T}^3\) correspond to the constant vector fields, so the orthogonality condition from 3 reduces to the vector integral equality \(\int_{\mathbb{T}^3} F\,\mathrm{d}x = 0\). We conclude that 3 applies with \(k=2\), and yields \[\begin{align} \|Z\|_{L^{3/2,1}(\Lambda^2)} \lesssim \|F\|_{L^1(\Lambda^1)} . \end{align}\] The identifications between forms and vector-valued functions are isometries, and therefore this inequality can be expressed in terms of \(Z,F\) as vectors: \[\begin{align} \|Z\|_{L^{3/2,1}(\mathbb{T}^3;\mathbb{R}^3)} \lesssim \|F\|_{L^1(\mathbb{T}^3;\mathbb{R}^3)} . \end{align}\] Finally, Hölder’s inequality on the Lorentz scale (see 22) gives, for \(a \in \mathbb{T}^3\), \[\begin{align} \int_{\mathbb{T}^3} \frac{|Z(x)|}{|x-a|}\,dx &\lesssim \|Z\|_{L^{3/2,1}(\mathbb{T}^3;\mathbb{R}^3)} \left\|\frac{1}{|\cdot-a|} \right\|_{L^{3,\infty}(\mathbb{T}^3;\mathbb{R}^3)} \\ &\lesssim \|F\|_{L^1(\mathbb{T}^3;\mathbb{R}^3)} \end{align}\] where the last inequality follows by checking that the weak-\(L^3\) norm (or quasi-norm) of the scalar function \(f(x)=1/\left\vert x-a\right\vert\) is finite and independent of \(a\). ◻

Proof. The codifferential can be expressed in terms of the star operator by \[\label{dAndStar} \mathrm{d}^*_j = (-1)^j \star_{j-1}^{-1} \mathrm{d}_{n-j} \star_j\tag{85}\] so that 4 and 83 give \[\begin{align} -\star_k \Delta_k \star^{-1}_k &= \star_k \mathrm{d}^*_{k+1} \mathrm{d}_k \star^{-1}_k + \star_k \mathrm{d}_{k-1} \mathrm{d}^*_k \star^{-1}_k \\& = (-1)^{k+1} \mathrm{d}_{n-k-1}\star_{k+1} \mathrm{d}_k \star^{-1}_k + (-1)^k \star_k \mathrm{d}_{k-1} \star^{-1}_{k-1} \mathrm{d}_{n-k} \\& = (-1)^{k+1} (-1)^{(k+1)(n-k-1)} (-1)^{k(n-k)} \mathrm{d}_{n-k-1} \star^{-1}_{n-k-1} \mathrm{d}_k \star_{n-k} \\&\quad + (-1)^k (-1)^{k(n-k)} (-1)^{(k-1)(n-k+1)} \star^{-1}_{n-k} \mathrm{d}_{k-1} \star_{n-k+1} \mathrm{d}_{n-k} \\& = (-1)^{(2k+1)(n-k)} (-1)^{n-k} \mathrm{d}_{n-k-1} \mathrm{d}^*_{n-k} \\&\quad + (-1)^{(2k-1)(n-k+1)} (-1)^{n-k+1} \mathrm{d}^*_{n-k+1} \mathrm{d}_{n-k} \\& = -\Delta_{n-k} . \qedhere \end{align}\] ◻

Proof of 40 and 10. For 40 , \[\begin{align} \Delta_{n-k,c}T_\omega(\eta) &= T_\omega(\Delta_{n-k}\eta) &&\text{by \eqref{LaplacianOnCurrents}} \\& = \savebox{\@brx}{\(\m@th{\langle}\)} \mathopen{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}\Delta_{n-k}\eta, \star^{-1}_{n-k}\omega\savebox{\@brx}{\(\m@th{\rangle}\)} \mathclose{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}} &&\text{by \eqref{FormAsCurrentAngleStar}} \\& = \savebox{\@brx}{\(\m@th{\langle}\)} \mathopen{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}\eta, \Delta_{n-k}\star^{-1}_{n-k}\omega\savebox{\@brx}{\(\m@th{\rangle}\)} \mathclose{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}} &&\text{by \eqref{HodgeLaplacianSelfAdjoint}} \\& = \savebox{\@brx}{\(\m@th{\langle}\)} \mathopen{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}}\eta, \star^{-1}_{n-k}\Delta_k\omega\savebox{\@brx}{\(\m@th{\rangle}\)} \mathclose{\copy\@brx\kern-0.5\wd\@brx\usebox{\@brx}} &&\text{by \Cref{DeltaAndStar}} \\& = T_{\Delta_k\omega}(\eta) &&\text{by \eqref{FormAsCurrentAngleStar}} . \end{align}\] The assertion of 40 is equivalent to ?? by the definitions of \(\Delta_{n-k,c}\) and \(T_\omega\), see 37 and 39 .

Fix \(\omega,\eta\) and let \(h(s,t)=\int_\mathcal{M}e^{s\Delta_{n-k}}\eta\wedge e^{t\Delta_k}\omega\). Then \[\begin{align} \frac{\partial h}{\partial s}(s,t) &= \int_\mathcal{M}\Delta_{n-k}e^{s\Delta_{n-k}}\eta\wedge e^{t\Delta_k}\omega \\& = \int_\mathcal{M}e^{s\Delta_{n-k}}\eta\wedge \Delta_k e^{t\Delta_k}\omega = \frac{\partial h}{\partial t}(s,t) \end{align}\] and it follows that \(h(s,t)\) depends only on \(s+t\). In particular, using 38 , \[e^{t\Delta_{n-k,c}}T_\omega(\eta) = T_\omega(e^{t\Delta_{n-k}}\eta) = h(t,0) = h(0,t) = T_{e^{t\Delta_k}\omega}(\eta) . \qedhere\] ◻

Proof of 11. The first equality of [item:FormLebesgueDuality], the duality pairing between \(L^r(\Lambda^k)\) and \(L^{r'}(\Lambda^k)\), follows from the usual one for scalar functions because, given any scalar function \(\tilde{f}\colon\mathcal{M}\to\left[0,\infty\right)\), we have \[\left\Vert\omega(p)\right\Vert_p \tilde{f}(p) = \sup_{\tilde{\omega}\in\Lambda^k\colon \left\Vert\tilde{\omega}(q)\right\Vert_q = \tilde{f}(q) \, \forall q\in\mathcal{M}} \left\langle\omega(p),\tilde{\omega}(p)\right\rangle_p\] and the maximum is attained by setting \(\tilde{\omega}(p) = \tilde{f}(p) \omega(p)/\left\Vert\omega(p)\right\Vert_p\) whenever \(\omega(p)\neq 0\). A standard approximation argument shows that the supremum is unaffected if we add the requirement that \(\tilde{\omega}\) is continuous. The second equality of [item:FormLebesgueDuality] follows by 84 and the substitution \(\xi=\star_k\tilde{\omega}\), which is a pointwise isometry.

The statement for the Lorentz norm follows for the same reason, except that the duality pairing between Lorentz norms is only up to constant factors, see 94 .

For part [item:CurrentDuality], recall the finite atlas \((x_j,U_j)_{j=1,\dotsc,L}\) and the partition of unity \((\chi_j)_{j=1,\dotsc,L}\). Define the currents \(\tilde{T}_j\) on \(\mathbb{R}^n\) given by \(\tilde{T}_j(\tilde{\xi}) = T(\chi_j \cdot (x_j)^\# \tilde{\xi})\). Then each \(\tilde{T}_j\) satisfies bounds analogous to those for \(T\), and we deduce that \(\tilde{T}_j\) is given by integration against a Radon measure that is absolutely continuous with respect to Lebesgue measure. That is, we deduce that \(\tilde{T}_j(\tilde{\xi}) = \int_{x_j(U'_j)} \left\langle\smash{\tilde{\xi}}(y), w_j(y)\right\rangle \,\mathrm{d}y\) for some \(k\)-form \(w_j\) on \(\mathbb{R}^n\) supported in \(U'_j\), and furthermore \(w_j\) is of class \(L^{r'}\) or \(L^{r',\infty}\). Finally, map back to obtain \(\omega=\sum_{j=1}^Lx_j^\# w_j\). ◻

Proof of 3. To establish that \(e^{t\Delta_{j,c}}T\) has the form \(T_{\omega_t}\), 11 shows that it is sufficient to bound \(\left\vert e^{t\Delta_{j,c}}T(\xi)\right\vert\) in terms of \(\left\Vert\xi\right\Vert_{L^1}\) for \(\xi\in C(\Lambda^j)\). To that end, recall the heat kernel \(\mathrm{p}_t^j\) and its approximation \(\mathrm{p}_t^{j,N}\) from 56 and 58 . The functions \(V_{j,i}\) are smooth on the compact manifold \(\mathcal{M}\times\mathcal{M}\) and therefore uniformly bounded, so together with the uniform error bound 60 we have \[\label{HeatKernelBoundedByK} \big\Vert\mathrm{p}_t^j(p,q)\big\Vert_{p,q} \lesssim K_t(p,q) ,\tag{86}\] uniformly over \(t\in\left(0,1\right]\) and \(p,q\in\mathcal{M}\). That is, we can take the constant \(C\) large enough so that \[\big\Vert\mathrm{p}_t^j(p,q)\big\Vert_{p,q} \leq C K_t(p,q)\] for all such \(t,p,q\). Then, applying Minkowski’s inequality to the integral 56 , \[\begin{align} \label{heatpropagator95bound} \left\Vert e^{t\Delta_j}\xi(p)\right\Vert_p &\leq \int_\mathcal{M}\big\Vert\mathrm{p}_t^j(p,q)\big\Vert_{p,q} \left\Vert\xi(q)\right\Vert_q \mathrm{d}V(q) \notag\\& \leq C\int_\mathcal{M}K_t(p,q) \left\Vert\xi(q)\right\Vert_q \mathrm{d}V(q) , \end{align}\tag{87}\] so that \[\begin{align} \left\vert e^{t\Delta_{j,c}}T(\xi)\right\vert &= \left\vert T(e^{t\Delta_j}\xi)\right\vert \notag\\ &\leq \int_\mathcal{M}\left\Vert e^{t\Delta_j}\xi(p)\right\Vert_p \mathrm{d}\mu_T(p) \notag\\ &\leq C \int_{\mathcal{M}} \int_\mathcal{M}K_t(p,q) \left\Vert\xi(q)\right\Vert_q \mathrm{d}V(q) \, \mathrm{d}\mu_T(p) \notag\\ &= C \int_{\mathcal{M}} \left\Vert\xi(q)\right\Vert_q \left(\int_\mathcal{M}K_t(p,q) \, \mathrm{d}\mu_T(p)\right) \mathrm{d}V(q) . \label{AbsetDeltaTxiBound} \end{align}\tag{88}\] In particular, \(\left\vert e^{t\Delta_{j,c}}T(\xi)\right\vert \leq \tilde{C} \left\Vert\xi\right\Vert_{L^1}\) for all \(\xi\in C(\Lambda^j)\), where \(\tilde{C}=\tilde{C}(\mathcal{M},T)<\infty\). By 11 we can identify \(e^{t\Delta_{j,c}}T=T_{\omega_t}\) for some \(\omega_t \in L^\infty(\Lambda^{n-j})\). By 10 and the semigroup property, \[\begin{align} e^{t\Delta_{j,c}}T &= e^{(t/2)\Delta_{j,c}}e^{(t/2)\Delta_{j,c}}T\\ &=e^{(t/2)\Delta_{j,c}}T_{\omega_{t/2}}\\ &= T_{e^{(t/2)\Delta_{n-j}} \omega_{t/2}} , \end{align}\] and since the heat propagator on \((n-j)\)-forms is smoothing, this shows that \(e^{t\Delta_{j,c}}T\) admits representation as a smooth \((n-j)\)-form, necessarily unique. We may therefore assume without loss of generality that \(\omega_t\) is this smooth \((n-j)\)-form, for all \(t>0\).

Finally, recalling 84 and 88 , we have \[\begin{align} \left\vert\int_\mathcal{M}\left\langle\xi(p),\star_{n-j}^{-1}\omega_t(p)\right\rangle \mathrm{d}V(p)\right\vert &= \left\vert T_{\omega_t}(\xi)\right\vert = \left\vert e^{t\Delta_{j,c}}T(\xi)\right\vert \\& \leq C \int_{\mathcal{M}} \left\Vert\xi(q)\right\Vert_q \left(\int_\mathcal{M}K_t(p,q) \, \mathrm{d}\mu_T(p)\right) \mathrm{d}V(q) \end{align}\] for all \(\xi\in C(\Lambda^j)\). By Lebesgue’s differentiation theorem we deduce \[\begin{align} \left\Vert\omega_t(p)\right\Vert_p = \left\Vert\star^{-1}_{n-j}\omega(p)\right\Vert_p \leq C\int_\mathcal{M}K_t(p,q)\,\mathrm{d}\mu_T(q) \end{align}\] for almost every \(p \in \mathcal{M}\). Since \(\omega_t\) is smooth, this inequality holds for all \(p \in \mathcal{M}\). ◻

Proof. This follows by using geodesic normal coordinates around \(q\), so that \(d_\mathcal{M}(p,q)^2\) reduces to the sum of squares of the coordinates of \(p\), for \(p\) sufficiently close to \(q\). ◻

Proof of 9. We again use the asymptotic series approximation from 58 , this time fixing \(N\geq (n+1)/2\). In the notation of BerGetVer1992?*Theorem 2.30 with \(\ell=1,k=0\), we have the bound \[\label{pt1Approx1norm} \big\Vert\mathrm{p}_t^1 - \mathrm{p}_t^{1,N}\big\Vert_1 \lesssim 1 ,\tag{89}\] where the norm \(\left\Vert\cdot\right\Vert_1\) means the supremum of values and partial derivatives up to order 1 for a function of class \(C^1\), see BerGetVer1992?*pp. 70–71.

In this case, the functions \(\mathrm{p}_t^1,\mathrm{p}_t^{1,N}\) are smooth functions on \(\mathcal{M}\times\mathcal{M}\) with values in \(\cup_{p,q\in\mathcal{M}}\mathrm{Hom}(T_q^*\mathcal{M}, T_p^*\mathcal{M})\). For an interpretation of 89 concretely in coordinates, recall the coordinate charts \((x_r,U_r)\) and partition of unity \(\chi_r\), \(r\in\left\{1,\dotsc,L\right\}\), from 2.1, and write \(\xi_t=\sum_{i=1}^n\xi_{t,r,i} \mathrm{d}x_r^i\) with \(\xi=\xi_0\). Then each mapping \(\xi\mapsto \chi_r e^{t\Delta_1}(\chi_{r'}\xi)\) can be represented in coordinates by a smooth kernel, and \[\label{xiIntegralCoordinates} \xi_{t,r,i}(p) = \sum_{r' = 1}^L\sum_{j=1}^n\int_{U_{r'}} w_{t,r,r',i,j}^1(p,q) \xi_{0,r',j}(q) \, \mathrm{d}V(q) \quad\text{for }p\in U_r ,\tag{90}\] where \(w_{t,r,r',i,j}^1\) are smooth scalar functions. If \(w_{t,r,r',i,j}^{1,N}\) are the corresponding functions for \(\mathrm{p}_t^{1,N}\), then 89 implies that \[\left\vert w_{t,r,r',i,j}^1(p,q) - w_{t,r,r',i,j}^{1,N}(p,q)\right\vert\lesssim 1 ,\quad \left\vert\frac{\partial w_{t,r,r',i,j}^1}{\partial x_r(p)^i}(p,q) - \frac{\partial w_{t,r,r',i,j}^{1,N}}{\partial x_r(p)^i}(p,q)\right\vert\lesssim 1 ,\] where \(\frac{\partial}{\partial x_r(p)^i}\) means partial differentiation in \(\mathcal{M}\times\mathcal{M}\) under the coordinate mapping \((p,q)\mapsto (x_r(p),x_{r'}(q))\in\mathbb{R}^{2n}\) on \(U_r\times U_{r'}\). Moreover the scalar functions \(w_{t,r,r',i,j}^{1,N}\) also have representations of the form 58 as finite sums of powers of \(t\) times functions to which 20 applies. (Note that we can freely shrink the support of the functions \(V_{k,i}\) from 58 to lie close enough to the diagonal so that 20 applies, without affecting the error bounds from BerGetVer1992?*Theorem 2.30, and indeed their derivation already performs such shrinking.) For \(0<t\leq 1\), the most singular such power is \(t^{-n/2}\), and we conclude that \[\left\vert\frac{\partial w_{t,r,r',i,j}^1}{\partial x_r(p)^i}(p,q)\right\vert \lesssim 1 + t^{-n/2} \frac{d_\mathcal{M}(p,q) + t}{t} \exp\left( -d_\mathcal{M}(p,q)^2/(4t) \right) = \tilde{K}_t(p,q) .\]

Now take the partial derivatives inside the integral in 90 to find \[\frac{\partial \xi_{t,r,i}}{\partial x_r^k}(p) = \sum_{r' = 1}^L\sum_{j=1}^n\int_{U_{r'}} \frac{\partial w_{t,r,r',i,j}^1}{\partial x_r^k}(p,q) \xi_{0,r',j}(q) \, \mathrm{d}V(q) .\] The \(2\)-form \(\mathrm{d}_1\xi_t\) can be expressed in terms of a linear combination of such partial derivatives, and therefore \[\begin{align} \left\Vert\zeta(p)\right\Vert_p = \left\Vert\mathrm{d}_1 \xi_t(p)\right\Vert_p &\lesssim \sum_{r=1}^R \sum_{i=1}^n\sum_{k=1}^n\int_{U_{r'}} \left\vert\frac{\partial w_{t,r,r',i,j}^1}{\partial x_r(p)^i}(p,q)\right\vert \left\vert\xi_{0,r',j}(q)\right\vert \mathrm{d}V(q) \\& \lesssim \int_\mathcal{M}\tilde{K}_t(p,q) \left\Vert\xi(q)\right\Vert_q \mathrm{d}V(q) . \qedhere \end{align}\] ◻

Proof of 1. The inequality is trivial if \(\left\Vert f\right\Vert_{L^\infty}=0\), so assume \(\left\Vert f\right\Vert_{L^\infty}>0\). Noting that \(\int_0^\infty f^*(u)\,\mathrm{d}u = \left\Vert f\right\Vert_{L^1}\), we have \(f^{**}(t)\leq \left\Vert f\right\Vert_{L^1}/t\) for all \(t>0\). In addition, \(f^*(t)\leq \left\Vert f\right\Vert_{L^\infty}\) and consequently \(f^{**}(t)\leq\left\Vert f\right\Vert_{L^\infty}\) for all \(t>0\). Write \(t_c=\left\Vert f\right\Vert_{L^1}/\left\Vert f\right\Vert_{L^\infty}\) for the point where these upper bounds coincide. Then, for \(1<r<\infty\) and \(1\leq s<\infty\), \[\begin{align} \|f \|_{L^{r,s}(X)}^s &= \int_0^\infty [t^{1/r}f^{**}(t)]^s \frac{\mathrm{d}t}{t} \\& \leq \int_0^{t_c} t^{s/r-1} \left\Vert f\right\Vert_{L^\infty}^s \,\mathrm{d}t + \int_{t_c}^\infty t^{s/r-s-1} \left\Vert f\right\Vert_{L^1}^s \mathrm{d}t \\& = \frac{r}{s} t_c^{s/r} \left\Vert f\right\Vert_{L^\infty}^s + \frac{1}{s-s/r} t_c^{-(s-s/r)} \left\Vert f\right\Vert_{L^1}^s \\& = \left( \frac{r}{s} + \frac{1}{s-s/r} \right) \left\Vert f\right\Vert_{L^1}^{s/r} \left\Vert f\right\Vert_{L^\infty}^{s-s/r} . \end{align}\] We have therefore shown the general inequality \[\label{GeneralLorentzInterpolation} \|f \|_{L^{r,s}(X)} \leq \frac{1}{s^{1/s}} \left( \frac{1}{1/r} + \frac{1}{1-1/r} \right)^{1/s} \left\Vert f\right\Vert_{L^1}^{1/r} \left\Vert f\right\Vert_{L^\infty}^{1-1/r}\tag{96}\] and 1 is the special case \(s=1\), \(1/r = 1-\alpha/n\). ◻

Proof. Let \(f\colon X\to\mathbb{R}\) be a measurable function and define \(A_y=\left\{x\in X\colon \left\vert f(x)\right\vert>y\right\}\) for \(y\geq 0\). Then by definition \[\label{Lr1QuasinormmuAy} {\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert f \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}_{L^{r,1}} = r \int_0^\infty \mu(A_y)^{1/r} \,\mathrm{d}y\tag{99}\] and \[\label{absfxIntegral} \left\vert f(x)\right\vert = \int_0^\infty \mathbb{1}_{\left\{x\in A_y\right\}} \,\mathrm{d}y .\tag{100}\] To make use of the operator norms \(\left\Vert T\right\Vert_{L^1\to L^1}\), \(\left\Vert T\right\Vert_{L^1\to L^\infty}\), and \(\left\Vert T\right\Vert_{L^\infty\to L^\infty}\), rewrite 100 to express \(f\) as the integral with respect to the measure appearing in 99 of functions in \(L^1\cap L^\infty\): \[f = \int_0^\infty h_y \, \mu(A_y)^{1/r} \,\mathrm{d}y \quad\text{\mu-a.e., where}\quad h_y = \frac{\mathbb{1}_{A_y}}{\mu(A_y)^{1/r}} \operatorname{sgn}(f) ,\] with \(h_y\) interpreted as the zero function (modulo equality \(\mu\)-a.e.) if \(\mu(A_y)=0\). By Minkowski’s inequality for integrals, \[\label{TfLp1Integral} \left\Vert Tf\right\Vert_{L^{p,1}} \leq \int_0^\infty \left\Vert T h_y\right\Vert_{L^{p,1}} \mu(A_y)^{1/r} \,\mathrm{d}y .\tag{101}\] From the operator norms of \(T\) we obtain three bounds \[\begin{align} \left\Vert T h_y\right\Vert_{L^1} &\leq \left\Vert T\right\Vert_{L^1\to L^1} \left\Vert h\right\Vert_{L^1} = \left\Vert T\right\Vert_{L^1\to L^1} \mu(A_y)^{1-1/r} , \tag{102} \\ \left\Vert T h_y\right\Vert_{L^\infty} &\leq \left\Vert T\right\Vert_{L^1\to L^\infty} \left\Vert h\right\Vert_{L^1} = \left\Vert T\right\Vert_{L^1\to L^\infty} \mu(A_y)^{1-1/r} , \tag{103} \\ \left\Vert T h_y\right\Vert_{L^\infty} &\leq \left\Vert T\right\Vert_{L^\infty\to L^\infty} \left\Vert h\right\Vert_{L^\infty} = \left\Vert T\right\Vert_{L^1\to L^\infty} \mu(A_y)^{-1/r} . \tag{104} \end{align}\] The Interpolation [1](#Interpolation){reference-type=“ref+Label” reference=“Interpolation”}, or 96 , applied to 102 paired with each of 103 and 104 yields two complementary bounds for \(\left\Vert T h_y\right\Vert_{L^{p,1}}\): \[\label{ThyLp1TwoBounds} \left\Vert T h_y\right\Vert_{L^{p,1}} \lesssim \left\lbrace \begin{align} &\left\Vert T\right\Vert_{L^1\to L^1}^{1/p} \left\Vert T\right\Vert_{L^1\to L^\infty}^{1-1/p} \mu(A_y)^{1-1/r} , \\ &\left\Vert T\right\Vert_{L^1\to L^1}^{1/p} \left\Vert T\right\Vert_{L^\infty\to L^\infty}^{1-1/p} \mu(A_y)^{-(1/r-1/p)} . \end{align} \right.\tag{105}\] (If \(p=r=1\), 96 and 1 do not apply, but then \(\left\Vert\cdot\right\Vert_{L^{1,1}}=\left\Vert\cdot\right\Vert_{L^1}\) and 105 holds trivially.) The first upper bound is non-decreasing in \(\mu(A_y)\) for all \(r\geq 1\), and the second is non-increasing since \(p\geq r\). The cross-over occurs at \(\mu(A_y)=\left\Vert T\right\Vert_{L^\infty\to L^\infty}/\left\Vert T\right\Vert_{L^1\to L^\infty}\), and in particular there is a uniform bound \[\label{supyThyLp1} \sup_{y\geq 0} \left\Vert T h_y\right\Vert_{L^{p,1}} \leq \left\Vert T\right\Vert_{L^1\to L^1}^{1/p} \left\Vert T\right\Vert_{L^1\to L^\infty}^{1/r - 1/p} \left\Vert T\right\Vert_{L^\infty\to L^\infty}^{1-1/r} .\tag{106}\] By 101 and 99 , \[\left\Vert T f\right\Vert_{L^{p,1}} \leq \sup_y \left\Vert T h_y\right\Vert_{L^{p,1}} \int_0^\infty \mu(A_y)^{1/r}\,\mathrm{d}y \lesssim \sup_y \left\Vert T h_y\right\Vert_{L^{p,1}} {\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert f \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}_{L^{r,1}} ,\] and by 106 and 21 this proves 98 for \(r>1\). The case \(r=1\) also follows because \({\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert f \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}_{L^{1,1}}=\left\Vert f\right\Vert_{L^{1,1}}=\left\Vert f\right\Vert_{L^1}\), and 97 is the special case \(p=r\). ◻

Proof. In HRS?*Equation (2.7) on p. 1935, it is shown that \[\begin{align} \label{Stolyarov95bound} \|\mathcal{I}_\alpha f\|_{\dot{B}^{0,1}_{r,1}(\mathbb{R}^n)} &\lesssim \int_0^\infty t^{\alpha/2-1} \|e^{t\Delta} f\|_{L^{r,1}(\mathbb{R}^n)}\,\mathrm{d}t . \end{align}\tag{108}\] (Their statement is for the case \(r=n/(n-\alpha)\), but the same proof applies for all \(r\geq1\).) By 24, the Riesz potential is a homeomorphism \(I_\alpha\colon \dot{B}^{-\alpha ,1}_{r,1}(\mathbb{R}^n) \to \dot{B}^{0,1}_{r,1}(\mathbb{R}^n)\), so 108 is equivalent to the assertion that \[\label{Stolyarov95bound95equivalent} \|f\|_{\dot{B}^{-\alpha,1}_{r,1}(\mathbb{R}^n)} \lesssim \int_0^\infty t^{\alpha/2-1} \|e^{t\Delta} f\|_{L^{r,1}(\mathbb{R}^n)}\,\mathrm{d}t .\tag{109}\] It therefore suffices to establish the reverse inequality, and by density it is sufficient to prove such a bound for smooth, compactly supported functions. Let \(f \in C^\infty_c(\mathbb{R}^n)\) and set \(f_k = f\ast (\psi_{2^{k+1}} -\psi_{2^k})\). Using \(f = \sum_k f_k\) we have \[\begin{align} \int_0^\infty t^{\alpha/2-1} \| e^{t\Delta}f\|_{L^{r,1}(\mathbb{R}^n)} \,\mathrm{d}t &= \sum_{l\in \mathbb{Z}} \int_{2^{-2l}}^{2^{-2l+2}} t^{\alpha/2-1} \| e^{t\Delta}f\|_{L^{r,1}(\mathbb{R}^n)} \,\mathrm{d}t\\ &\leq \sum_{l\in \mathbb{Z}} (2^{-2l+2})^{\alpha/2} \int_{2^{-2l}}^{2^{-2l+2}} \| e^{t\Delta} f\|_{L^{r,1}(\mathbb{R}^n)} \,\frac{\mathrm{d}t}{t}\\ &\leq 2^\alpha \sum_{k \in \mathbb{Z}} \sum_{l\in \mathbb{Z}} 2^{-\alpha l} \int_{2^{-2l}}^{2^{-2l+2}} \| e^{t\Delta}f_k\|_{L^{r,1}(\mathbb{R}^n)} \,\frac{\mathrm{d}t}{t}\\ &= 2^\alpha \left(\sum_{k \in \mathbb{Z}} \sum_{l\leq k} \cdots + \sum_{k \in \mathbb{Z}} \sum_{l> k} \cdots\right)\\ &=: I + \mathit{II} . \end{align}\] For \(I\), the support condition \(\operatorname*{supp} \widehat{f_k} \subset B(0,2^{k+2})\setminus B(0,2^{k})\) allows us to use Bernstein’s inequality \[\begin{align} \| e^{t\Delta}f_k\|_{L^{r,1}(\mathbb{R}^n)} \leq 2^{-2k} \| \Delta e^{t\Delta}f_k\|_{L^{r,1}(\mathbb{R}^n)} \end{align}\] and \(\| \Delta e^{t\Delta}\|_{L^{r,1}(\mathbb{R}^n) \to L^{r,1}(\mathbb{R}^n)}\lesssim t^{-1}\) gives \[\begin{align} \int_{2^{-2l}}^{2^{-2l+2}} \| e^{t\Delta}f_k\|_{L^{r,1}(\mathbb{R}^n)} \,\frac{\mathrm{d}t}{t} &\leq \int_{2^{-2l}}^{2^{-2l+2}} 2^{-2k} \| \Delta e^{t\Delta} f_k\|_{L^{r,1}(\mathbb{R}^n)} \,\frac{\mathrm{d}t}{t} \\ &\leq \int_{2^{-2l}}^{2^{-2l+2}} 2^{-2k} \| f_k\|_{L^{r,1}(\mathbb{R}^n)} \,\frac{\mathrm{d}t}{t^2}\\ &\lesssim 2^{2l-2k} \| f_k\|_{L^{r,1}(\mathbb{R}^n)} \end{align}\] and therefore \[\begin{align} I &\leq 2^{\alpha+1} \sum_{k \in \mathbb{Z}} 2^{-2k} \| f_k\|_{L^{r,1}(\mathbb{R}^n)} \sum_{l\leq k} 2^{(2-\alpha) l} \\ &\lesssim \sum_{k \in \mathbb{Z}} 2^{-\alpha k} \| f_k\|_{L^{r,1}(\mathbb{R}^n)} . \end{align}\]

For \(\mathit{II}\), we use that the heat kernel is a contraction on \(L^{r,1}\), \(\| e^{t\Delta}f_k\|_{L^{r,1}(\mathbb{R}^n)}\leq \| f_k\|_{L^{r,1}(\mathbb{R}^n)}\), to obtain the bound \[\begin{align} \mathit{II} &\lesssim \sum_{k \in \mathbb{Z}} \| f_k\|_{L^{r,1}(\mathbb{R}^n) }\sum_{l> k} 2^{-\alpha l} \\ &\approx \sum_{k \in \mathbb{Z}} 2^{-\alpha k} \| f_k\|_{L^{r,1}(\mathbb{R}^n) } . \qedhere \end{align}\] ◻

Proof. Replacing \(f\) by \(\Delta f\) and applying 25 and 24 gives \[\begin{align} \int_0^\infty t^{\alpha/2-1} \|\Delta e^{t\Delta} f\|_{L^{r,1}(\mathbb{R}^n)}\,\mathrm{d}t \asymp \|\Delta f\|_{\dot{B}^{-\alpha,1}_{r,1}(\mathbb{R}^n)} \asymp \| f\|_{\dot{B}^{2-\alpha,1}_{r,1}(\mathbb{R}^n)} \end{align}\] for \(\alpha>0\). More generally, applying 24 \(i\) times and 25 with \(\alpha=2i-\beta>0\) and \(f\) replaced by \(\Delta^i f\) yields the result. ◻

Proof. For \(0<t\leq 1\), the operator \(T=e^{t\Delta_k}\) is given by an integral kernel bounded in terms of the function \(K_t\), see 86 and [CurrentEvolutionAsForm,KernelBounds], and it follows immediately that \[\left\Vert e^{t\Delta_k}\right\Vert_{L^1\to L^1} \lesssim \sup_q \left\Vert K_t(\cdot, q)\right\Vert_{L^1} , \quad \left\Vert e^{t\Delta_k}\right\Vert_{L^\infty\to L^\infty} \lesssim \sup_p \left\Vert K_t(p,\cdot)\right\Vert_{L^1}\] for \(0<t\leq 1\). Recalling 45 , we therefore have \[\label{PropagatorOperatorBounds} \begin{align} \left\Vert e^{t\Delta_k}\right\Vert_{L^1(\Lambda^k)\to L^1(\Lambda^k)} \lesssim 1 , \quad \left\Vert e^{t\Delta_k}\right\Vert_{L^\infty(\Lambda^k)\to L^\infty(\Lambda^k)} \lesssim 1 \end{align}\tag{112}\] for all \(0<t\leq 1\).

For \(t>1\), the composition \(e^{t\Delta_k}\vert_{\mathcal{H}^\perp(\Lambda^k)}\) has exponential decay by 18, while \(e^{t\Delta_k}\vert_{\mathcal{H}(\Lambda^k)}\) acts as the identity. Thus both of these operators satisfy the bounds in 112 . Moreover, since \(\mathcal{H}(\Lambda^k)\) is a finite-dimensional vector subspace of \(L^\infty(\Lambda^k)\subset L^1(\Lambda^k)\), the projections \(P_{\mathcal{H}(\Lambda^k)}\) and \(P_{\mathcal{H}^\perp(\Lambda^k)}=\mathrm{identity}-P_{\mathcal{H}(\Lambda^k)}\) are bounded as operators on both \(L^1\) and \(L^\infty\). Thus 112 holds for all \(t>1\) also. By 21, 112 implies 110 .

A similar argument applies for \(\Delta_k e^{t\Delta_k}\). By analogy with 20 and its proof, this operator is given by an integral kernel that is bounded up to constants by \[K^{(2)}_t(p,q) = \frac{(d_\mathcal{M}(p,q)+t)^2}{t^{n/2 + 2}} \exp(-d_\mathcal{M}(p,q)^2/(4t)) + 1\] for \(0<t\leq 1\), and it can be readily verified that \[\begin{align} \sup_p \big\Vert K^{(2)}_t(p,\cdot)\big\Vert_{L^1} \lesssim 1/t , \quad \sup_q \big\Vert K^{(2)}_t(\cdot,q)\big\Vert_{L^1} \lesssim 1/t . \end{align}\] For \(t>1\), we can again use the exponential decay from 18 on \(\mathcal{H}^\perp(\Lambda^k)\), noting that \(\Delta_k e^{t\Delta_k}\) vanishes on \(\mathcal{H}(\Lambda^k)\). We conclude that \(\left\Vert\Delta_k e^{t\Delta_k}_{L^1\to L^1}\right\Vert\) and \(\left\Vert\Delta_k e^{t\Delta_k}\right\Vert_{L^\infty\to L^1\infty}\) are both of order \(1/t\), and 111 follows by 21. ◻

Proof. For \(\beta\geq 0\), the inequality is trivial and indeed the two norms are equal. To see this, note that \(i(\beta)=i(\beta-2)+1\) for all \(\beta\geq 0\), so that \(i(\beta-2)-(\beta-2)/2 = i(\beta)-\beta/2\) and \[\begin{align} \left\Vert\Delta_k F\right\Vert_{\dot{B}^{\beta-2,1}_{r,1}(\Lambda^k)} &= \int_0^\infty t^{i(\beta-2)-(\beta-2)/2-1} \left\Vert \Delta_k^{i(\beta-2)} e^{t\Delta_k} \Delta_k F\right\Vert_{L^{r,1}(\Lambda^k)} \mathrm{d}t \\ &= \int_0^\infty t^{i(\beta)-\beta/2-1} \left\Vert \Delta_k^{i(\beta-2)+1} e^{t\Delta_k} F\right\Vert_{L^{r,1}(\Lambda^k)} \mathrm{d}t \\ &= \|F\|_{\dot{B}^{\beta,1}_{r,1}(\Lambda^k)} . \end{align}\] For \(\beta<0\), use 22 and make the substitution \(t=2\tilde{t}\) to obtain \[\begin{align} \left\Vert\Delta_k F\right\Vert_{\dot{B}^{\beta-2,1}_{r,1}(\Lambda^k)} &= \int_0^\infty t^{-(\beta-2)/2-1} \left\Vert \Delta_k e^{(t/2)\Delta_k} e^{(t/2)\Delta_k} F \right\Vert_{L^{r,1}(\Lambda^k)} \mathrm{d}t \notag\\ \label{AbsorbDelta} &\lesssim \int_0^\infty (2\tilde{t})^{-\beta/2} \tilde{t}^{-1} \left\Vert e^{\tilde{t}\Delta_k} F \right\Vert_{L^{r,1}(\Lambda^k)} (2 \, \mathrm{d}\tilde{t})\\ &\lesssim\| F\|_{\dot{B}^{\beta-2,1}_{r,1}(\Lambda^k)} . \qedhere \end{align}\tag{113}\]  ◻

Proof. Apply Minkowski’s inequality to the integral representation 8 and make the substitution \(\tau=s+t,y=t/(s+t)\): \[\begin{align} &\left\Vert \mathcal{I}_{\alpha,k}F \right\Vert_{\dot{B}^{\beta+\alpha,1}_{r,1}(\Lambda^k)} \\&\quad \leq \iint_{(0,\infty)^2} s^{\alpha/2 - 1} t^{i(\beta+\alpha)-(\beta+\alpha)/2-1} \left\Vert\Delta_k^{i(\beta+\alpha)} e^{t\Delta_k} e^{s\Delta_k} F \right\Vert_{L^{r,1}(\Lambda^k)} \frac{\,\mathrm{d}s\,\mathrm{d}t}{\Gamma(\alpha/2)} \\&\quad = \int_0^\infty \int_0^1 (\tau(1-y))^{\alpha/2-1} (\tau y)^{i(\beta+\alpha)-(\beta+\alpha)/2-1} \left\Vert\Delta_k^{i(\beta+\alpha)} e^{\tau\Delta_k} F \right\Vert_{L^{r,1}(\Lambda^k)} \frac{\tau\,\mathrm{d}y\,\mathrm{d}\tau}{\Gamma(\alpha/2)} \\&\quad = \int_0^\infty \tau^{i(\beta+\alpha)-\beta/2-1} \left\Vert\Delta_k^{i(\beta+\alpha)} e^{\tau\Delta_k} F \right\Vert_{L^{r,1}(\Lambda^k)} \mathrm{d}\tau \\&\qquad\cdot \int_0^1 (1-y)^{\alpha/2-1} y^{i(\beta+\alpha)-(\beta+\alpha)/2-1} \frac{\mathrm{d}y}{\Gamma(\alpha/2)} . \end{align}\] Note that \(i(\gamma)\geq \left\lfloor\gamma/2\right\rfloor+1 > \gamma/2\) for all \(\gamma\in\mathbb{R}\), so the last \(y\)-integral converges.

Since \(\alpha>0\), we may write \(i(\beta+\alpha)=i(\beta)+m\) for some \(m\in\left\{0,1,\dotsc\right\}\). Similar to 113 , substitute \(\tau=(m+1)\tilde{\tau}\) and apply 111 \(m\) times to conclude \[\begin{align} \left\Vert \mathcal{I}_{\alpha,k}F \right\Vert_{\dot{B}^{\beta+\alpha,1}_{r,1}(\Lambda^k)} &\lesssim \int_0^\infty \tau^{i(\beta)+m-\beta/2} \left\Vert\Delta_k^{i(\beta)+m} e^{\tau\Delta_k} F \right\Vert_{L^{r,1}(\Lambda^k)} \frac{\mathrm{d}\tau}{\tau} \\& \lesssim \int_0^\infty [(m+1)\tilde{\tau}]^{i(\beta)+m-\beta/2} \tilde{\tau}^{-m}\left\Vert\Delta_k^{i(\beta)} e^{\tilde{\tau}\Delta_k} F \right\Vert_{L^{r,1}(\Lambda^k)} \frac{\mathrm{d}\tilde{\tau}}{\tilde{\tau}} \\& \lesssim \left\Vert F\right\Vert_{\dot{B}^{\beta,1}_{r,1}(\Lambda^k)} . \qedhere \end{align}\] ◻

Proof of 27. The mappings are linear so it suffices to show two-sided inequalities up to constants between the relevant norms. By density, it suffices to consider \(F \in C^\infty\cap\mathcal{H}^\perp(\Lambda^k)\).

For the Laplacian, the inequality \(\left\Vert\Delta_k F\right\Vert_{\dot{B}^{\beta-2,1}_{r,1}(\Lambda^k)} \lesssim \left\Vert F\right\Vert_{\dot{B}^{\beta,1}_{r,1}(\Lambda^k)}\) is given by 23. The reverse inequality follows from the identity \(F=\mathcal{I}_{2,k} (-\Delta_k F)\), see 72 , and 24: \[\begin{align} \| F\|_{\dot{B}^{\beta,1}_{r,1}(\Lambda^k)} &= \| \mathcal{I}_{2,k} (-\Delta_k F)\|_{\dot{B}^{\beta,1}_{r,1}(\Lambda^k)}\\ &\leq C \|\Delta_k F\|_{\dot{B}^{\beta-2,1}_{r,1}(\Lambda^k)}. \end{align}\]

Similarly for the Riesz potential, one inequality appears in 24. To show the reverse direction, we begin by showing the commutativity relation \[\label{DeltaICommuteOnHperp} -\Delta_k \mathcal{I}_{\alpha,k} \omega = \mathcal{I}_{\alpha,k}(-\Delta_k)\omega \quad\forall\omega\in C^\infty\cap\mathcal{H}^\perp(\Lambda^k), \,\forall\alpha>0 .\tag{114}\] To see this, note that 18 and 111 imply that the Laplacian can be passed through the convergent integral 8 , so that both sides of 114 reduce to \(\int_0^\infty t^{\alpha/2-1} (-\Delta_k)e^{t\Delta_k}\omega \,\mathrm{d}t\) or equivalently \(\int_0^\infty t^{\alpha/2-1} e^{t\Delta_k}(-\Delta_k)\omega \,\mathrm{d}t\).

Now set \(i=\lfloor \alpha/2 \rfloor+1\), so that \(2i>\alpha\). By 114 and 23 , \[\mathcal{I}_{2i-\alpha,k} (-\Delta_k)^i \mathcal{I}_{\alpha,k} F = \mathcal{I}_{2i-\alpha,k} \mathcal{I}_{\alpha,k} (-\Delta_k)^i F = \mathcal{I}_{2i,k}(-\Delta_k)^i F = F ,\] so that the combination of 23 and 24 imply \[\begin{align} \| F\|_{\dot{B}^{\beta,1}_{r,1}(\Lambda^k)} &= \| \mathcal{I}_{2i-\alpha,k} (-\Delta)^i \mathcal{I}_{\alpha,k}F\|_{\dot{B}^{\beta,1}_{r,1}(\Lambda^k)}\\ &\lesssim \| (-\Delta)^i \mathcal{I}_{\alpha,k}F\|_{\dot{B}^{\beta - 2i +\alpha ,1}_{r,1}(\Lambda^k)}\\ &\lesssim \|\mathcal{I}_{\alpha,k}F\|_{\dot{B}^{\beta+\alpha,1}_{r,1}(\Lambda^k)} . \qedhere \end{align}\] ◻

Proof. We begin with a Lorentz space operator norm bound for \(e^{t\Delta_k}\), similar to 22 but now between two different Lorentz spaces. Recalling 44 , \[\label{PropagatorL1LinftyOperatorBoundSmallt} \left\Vert e^{t\Delta_k}\right\Vert_{L^1(\Lambda^k)\to L^\infty(\Lambda^k)} \lesssim \sup_{p,q} K_t(p,q) \lesssim t^{-n/2} \quad\text{uniformly over 0<t\leq 1.}\tag{116}\] Because of compactness, this bound fails for \(t>1\), and indeed \(e^{t\Delta_k}\) behaves as the projection \(P_{\mathcal{H}(\Lambda^k)}\) for large \(t\). However, since \(P_{\mathcal{H}^\perp(\Lambda^k)}\) is bounded as an operator on \(L^1\) and \(L^\infty\), the composition \(e^{t\Delta_k}P_{\mathcal{H}^\perp(\Lambda^k)}\) obeys the same bound as 116 while also decaying exponentially for \(t>1\) by 18. We conclude that \[\label{ProjectedPropagatorL1Linfty} \left\Vert e^{t\Delta_k}P_{\mathcal{H}^\perp(\Lambda^k)} \right\Vert_{L^1(\Lambda^k)\to L^\infty(\Lambda^k)} \lesssim t^{-n/2}\tag{117}\] uniformly over all \(t>0\). Likewise, the bounds 112 apply to \(e^{t\Delta_k}P_{\mathcal{H}^\perp(\Lambda^k)}\), so 21 yields \[\label{ProjectedPropagatorLorentzLorentz} \left\Vert e^{t\Delta_k}P_{\mathcal{H}^\perp(\Lambda^k)} \right\Vert_{L^{r,1}\to L^{p,1}} \lesssim t^{-n(1/r-1/p)/2}\tag{118}\] for \(1<r\leq p<\infty\) fixed.

In the parametrization of 115 , the operator bound 118 gives \[\label{OperatorNormLorentzLorentz} \|e^{t\Delta_k}F\|_{L^{n/(n-\gamma),1}(\Lambda^k)} \lesssim t^{-(\gamma-\alpha)/2} \|e^{(t/2)\Delta_k}F\|_{L^{n/(n-\alpha),1}(\Lambda^k)}\tag{119}\] for \(0\leq\alpha\leq\gamma<n\) fixed and uniformly over \(F\in \dot{B}^{-\alpha,1}_{n/(n-\alpha),1}(\Lambda^k)\) and \(t>0\). The second inclusion in 115 follows by multiplying by \(t^{\gamma/2-1}\) and integrating.

For the first inclusion, we will apply 119 with \(\alpha=0\). The norm for \(\dot{B}^{0,1}_{1,1}(\Lambda^k)\) requires an extra \(\Delta_k\), and 27 provides one: \[\begin{align} \left\Vert F\right\Vert_{\dot{B}^{-\gamma,1}_{n/(n-\gamma),1}(\Lambda^k)} &\lesssim \left\Vert\Delta_k F\right\Vert_{\dot{B}^{-\gamma-2,1}_{n/(n-\gamma),1}(\Lambda^k)} \\& = \int_0^\infty t^{\gamma/2+1-1} \left\Vert\Delta_k e^{t\Delta_k} F\right\Vert_{L^{n/(n-\gamma),1}(\Lambda^k)} \,\mathrm{d}t \\& \lesssim \int_0^\infty \left\Vert\Delta_k e^{t\Delta_k} F\right\Vert_{L^{1,1}(\Lambda^k)} \,\mathrm{d}t &&\text{by \eqref{OperatorNormLorentzLorentz}} \\& = \left\Vert F\right\Vert_{\dot{B}^{0,1}_{1,1}(\Lambda^k)} . &&\qedhere \end{align}\] ◻