Proof (Proof of theorem 2). By the elliptic theory, there exists a discrete, countably infinity set of eigenvalues \(\{\mu_1\leq\mu_2\leq\dots\}\) with associated eigenfunctions \(\{\phi_i\}\) for the Laplacian, that is, for each \(i\), we have

\[ \Delta\phi_i = -\mu_i\phi_i \]

such that the \(\phi_i\)’s form an orthonormal basis with respect to the \(L^2\) norm. See Section 8.12 of (Gilbarg and Trudinger 2001). Let \(f_0 \in L^2(M)\). Since the \(\phi_i\)’s are an orthonormal basis for \(L^2(M\)), we can write

\[ f_0(x) = \sum\limits_{i=1}^\infty a_i\phi_i(x), \text{ with } a_i = \int_M f_0\phi_i dx. \]

This is because

\[ a_i = \int_M a_i \phi_i^2 dx = \int_M \left(\sum\limits_{j=1}^\infty a_j\phi_j\right)\phi_i dx = \int_M f_0\phi_idx, \]

since

\[ \int_M \phi_i\phi_j dx = \begin{cases} &0, \text{ if } i \neq j, \\ &1, \text{ if } i = j. \end{cases} \]

Then we can see that defining \(H(x,y,t) = \sum\limits_{i=1}^\infty e^{-\mu_i t}\phi_i(x)\phi_i(y)\) and in turn defining \(f(x,t)\) by

\[\begin{aligned} f(x,t) &= \int_M H(x,y,t)f_0(y)dy = \int_M \sum\limits_{i=1}^\infty e^{-\mu_i t}\phi_i(x)\phi_i(y)f_0(y)dy \\ &= \sum\limits_{i=1}^\infty e^{-\mu_i t}\left(\int_M f_0(y)\phi_i(y)dy\right)\phi_i(x) = \sum\limits_{i=1}^\infty e^{-\mu_i t}a_i\phi_i(x) \end{aligned}\]

formally satisfies (2.1) - (2.3). We check this below.

For each \(x \in \partial M\), \(\phi_i(x) = 0\), so \(f\) formally satisifies the Dirichlet boundary condition. Also note that

\[ \lim\limits_{t\to0}f(x,t) = \lim\limits_{t\to0}\sum\limits_{i=1}^\infty e^{-\mu_i t}a_i\phi_i(x) = \sum\limits_{i=1}^\infty\lim\limits_{t\to0}e^{-\mu_i t}a_i\phi_i(x) = \sum\limits_{i=1}^\infty a_i\phi_i(x) = f_0(x), \]

so \(f\) formally satisfies the initial data condition. Finally, we compute

\[ \frac{\partial}{\partial_t}\left(\sum\limits_{i=1}^\infty e^{-\mu_i t}a_i\phi_i\right) = \sum\limits_{i=1}^\infty \frac{\partial}{\partial_t}\left(e^{-\mu_i t}a_i\phi_i\right) = \sum\limits_{i=1}^\infty -e^{-\mu_i t}a_i\mu_i\phi_i = \sum\limits_{i=1}^\infty e^{-\mu_i t}a_i\Delta\phi_i, \]

so we can see that \(f\) formally satisfies the heat equation.

It remains to justify these computations by showing that \(H\) as defined above is well-defined, i.e. that the sum converges uniformly, and also to show uniqueness, positivity and the final condition in the statement of the theorem.

First we show uniqueness using the maximum principle. Suppose \(f, \bar{f} \in L^2(M)\) both solve the heat equation arising from heat kernels \(H, \bar{H}\) respectively, such that \(\lim\limits_{t\to0}f(x,t) = f_0(x) = \lim\limits_{t\to0}\bar{f}(x,t)\) and that both satisfy the Dirichlet boundary condition. Then consider \(f(x,t) - \bar{f}(x,t)\) on \(\Gamma = M\times\{0\} \times \partial M \times (0,\infty))\). On \(M\times \{0\}\), we have that

\[ f(x,0) - \bar{f}(x,0) = f_0(x) - f_0(x) = 0, \]

and on \(\partial M \times (0,\infty)\), since \(f, \bar{f}\) satisfy the Dirichlet boundary condition, we have

\[ f(x,t) - \bar{f}(x,t) = 0 - 0 = 0. \]

So \(f - \bar{f} \equiv 0\) on \(\Gamma\), which by the maximum pinciple implies \(f - \bar{f} \equiv 0\) on \(M \times (0,\infty)\). Then this implies

\[ 0 = \int_M \left(H - \bar{H}\right)f_0dy \Leftrightarrow H = \bar{H}. \]

To see positivity on the interior, assume \(f_0 \geq 0\) on \(M\). We will see by the strong maximum principle that \(f(x,t)\) is striclty positive on \((M\setminus\partial M)\times(0,\infty)\). First, if \(f_0 \equiv 0\) on \(M\), then since \(f(x,t) = 0\) on \(\partial M\) (the Dirichlet condition), that means \(f(x,t) \equiv 0\) on \((\partial M \times (0,\infty)) \cup (M\times \{0\})\). So by the maximum principle, that implies \(f(x,t) \equiv 0\) everywhere. So that case is trivial. Now suppose there is some \(x_0 \in M\) such that \(f_0(x_0) > 0\). Then by smoothness of \(f\) (justified later when we talk about well-definedness of \(H\)), \(f(x,t) > 0\) near the point \((x_0, 0)\). Now there are two cases:

1. If \(f \equiv \text{const}\), then that implies \(f > 0\) everywhere.
2. If \(f\) is not constant, then applying the strong maximum principle to \(-f\), we have that \(-f\) does not achieve its maximum anywhere in the interior, i.e. in \((M \setminus \partial M) \times (0,\infty)\), which implies that \(f\) does not achieve its minimum anywhere in \((M\setminus\partial M) \times (0,\infty)\). Since \(f \geq 0\) on the boundary (\(f_0 \geq 0\) on \(M\) and \(f(x,t) = 0\) on \(\partial M \times (0,\infty)\)), that means \(f\) is strictly positive in the interior.

Again we rely on the maximum principle to show that

\[ \int_M H(x,y,t)dy \leq 1. \]

Taking \(f_0 \equiv 1\), we have that the integral \(\int_M H(x,y,t) dy = f(x,t)\) is a solution to the heat equation with boundary values \(f_0 \equiv 1\) on \(M\times\{0\}\) and \(f(x,t) = 0\) on \(\partial M \times (0,\infty)\) due to the Dirichlet condition. So by the maximum principle, since the maximum of \(f\) is attained on the boundary, this implies that \(\int_M H(x,y,t)dy \leq 1\).

Finally, it remains to justify the above formal calculations (i.e. show that \(H\) is well-defined) and subsequent regularity of \(H\), and thus that of \(f\). To do this, we want to show that the sum

\[ H(x,y,t) = \sum\limits_{i=1}^\infty e^{-\mu_i t}\phi_i(x)\phi_i(y) \]

converges absolutely. To do this, we will use the Sobolev inequality to obtain a bound on the supremum norm of the eigenfunctions \(\phi_i\) and lemma 3 to obtain a bound on the eigenvalues \(\mu_i\). Recall in dimension \(m \geq 3\), the Sobolev inequality states that there is a constant \(C_{\mathcal{SD}} > 0\) depending only on \(M\), such that for each \(f \in H^{1,2}_c(M)\), we have

\[ \int_M |\nabla f|^2 \geq C_{\mathcal{SD}}\left(\int_M |f|^{\frac{2m}{m-2}}\right)^{\frac{m-2}{m}}. \]

From now on, we will only cover the \(m \geq 3\) case. The \(m=2\) case is analogous using the corresponding Sobolev inequality for that dimension. Let \(u \in H^{1,2}_c(M)\) be a nonnegative function, \(k \geq 2\) be some constant. Then integrating by parts gives

\[\begin{aligned} \int_M u^{k-1}\Delta u &= -\int_M \langle \nabla\left(u^{k-1}\right),\nabla u\rangle \\ &= -(k-1) \int_M u^{k-2}|\nabla u|^2 \\ &= \frac{-4(k-1)}{k^2}\int_M \left|\nabla\left(u^{k/2}\right)\right|^2, \end{aligned}\]

since

\[ \left|\nabla\left(u^{k/2}\right)\right|^2 = \left(\frac{k}{2}\right)^2\left(u^{k/2-1}\right)^2|\nabla u|^2 = \frac{k^2}{4}u^{k-2}|\nabla u|^2. \]

Then applying the Sobolev inequality, we have that

\[\begin{aligned} \int_M u^{k-1}\Delta u &= \frac{-4(k-1)}{k^2}\int_M \left|\nabla\left(u^{k/2}\right)\right|^2 \\ &\leq \frac{-4(k-1)C_{\mathcal{SD}}}{k^2}\left(\int_M |u|^{\frac{k}{2}\frac{2m}{m-2}}\right)^{\frac{m-2}{m}} \\ &\leq \frac{-2C_{\mathcal{SD}}}{k^2}\left(\int_M |u|^{\frac{km}{m-2}}\right)^{\frac{m-2}{m}}. \end{aligned}\](4.1)

Choosing \(u = |\phi_i|\), then \(u\) satisfies \(\Delta u = -\mu_i u\) and (4.1) becomes

\[ \frac{-2C_\mathcal{SD}}{k}\left(\int_M|\phi_i|^{\frac{km}{m-2}}\right)^{\frac{m-2}{m}} \geq \int_M |\phi_i|^{k-1}\Delta|\phi_i| = -\int_M \mu_i |\phi_i|^k, \]

which is equivalent to

\[ \int_M |\phi_i|^k \geq \frac{2C_\mathcal{SD}}{k\mu_i}\left(\int_M |\phi_i|^{\frac{km}{m-2}}\right)^{\frac{m-2}{m}}. \](4.2)

Setting \(\beta = \frac{m}{m-2}\), the right hand side above becomes

\[ \frac{2C_\mathcal{SD}}{k\mu_i}\left(\int_M |\phi_i|^{k\beta}\right)^{\frac{1}{\beta}} = \left(\left(\frac{2C_\mathcal{SD}}{k\mu_i}\right)^{\frac{1}{k}}\right)^k\Vert \phi_i \Vert_{k\beta}^k, \]

and the left hand side is \(\Vert\phi_i\Vert_k^k\), so (4.2) can be rewritten as

\[ \left(\frac{2C_\mathcal{SD}}{k\mu_i}\right)^\frac{1}{k}\Vert\phi_i\Vert_{k\beta} \leq \Vert\phi_i\Vert_k. \](4.3)

Setting \(k = 2\beta^j\) for \(j = 0,1,2,\dots\), we obtain

\[ \Vert\phi_i\Vert_{2\beta^{j+1}} \leq \left(\frac{\beta^j\mu_i}{2C_\mathcal{SD}}\right)^{\frac{1}{2\beta^j}}\Vert\phi_i\Vert_{2\beta^j}. \]

Remembering that \(\Vert\phi_i\Vert_2 = 1\) we have that

\[\begin{aligned} \Vert\phi_i\Vert_{2\beta} &\leq \left(\frac{\beta\mu_i}{2C_\mathcal{SD}}\right)^{\frac{1}{2\beta}}, \\ \Vert\phi_i\Vert_{2\beta^2} &\leq \left(\frac{\beta^2\mu_i}{2C_\mathcal{SD}}\right)^{\frac{1}{2\beta^2}}\left(\frac{\beta\mu_i}{2C_\mathcal{SD}}\right)^{\frac{1}{2\beta}}, \\ &\vdots \\ \Vert\phi_i\Vert_{2\beta^j} &\leq \prod\limits_{\ell = 0}^j \left(\frac{\beta^\ell\mu_i}{2C_\mathcal{SD}}\right)^{\frac{1}{2\beta^\ell}}. \end{aligned}\]

Then letting \(j\to\infty\) and using the fact that \(\lim\limits_{p\to\infty}\nparallel\phi_i\Vert_p = \Vert\phi_i\Vert_\infty\) where \(\nparallel f \Vert_p\) denotes the average \(p\)-norm, we have

\[ \Vert\phi_i\Vert_\infty \leq C_1\mu_i^\frac{m}{4}. \](4.4)

So we have obtained a bound on the supremum norm of the eigenfunctions.

Next we have to bound the eigenvalues \(\mu_k\) in terms of \(k\). Let \(E\) be the \(k\)-dimensional vector space spanned by the first \(k\) eigenfunctions \(\{\phi_1,\phi_2,\dots,\phi_k\}\). We will show that for some constant \(C_2 > 0\) depending only on \(m\) and \(C_\mathcal{SD}\), for any \(u \in E\), the estimate

\[ \Vert u \Vert_\infty \leq C_2 \mu_k^{\frac{m-2}{2}}V(M)^\frac{m-1}{m}\nparallel u\Vert_2. \](4.5)

This estimate, combined with lemma 3 implies the following estimate

\[ \mu_k \geq C_3 k^\frac{1}{m-1}V(M)^\frac{-2}{m} \](4.6)

for some constant \(C_3 > 0\) depending only on \(m\) and \(C_\mathcal{SD}\). To establish (4.5), let \(E\) denote the \(k\)-dimensional vector space spanned by the first \(k\) eigenfunctions \(\{\phi_1,\phi_2,\dots,\phi_k\}\). Let \(u \in C^\infty\) be a function. Note that for \(u\), we have \(|\nabla u|^2 = |\nabla |u||^2\) and so

\[\begin{aligned} \Delta(u^2) &= 2u\Delta u + 2|\nabla u|^2,\\ \Delta|u|^2 &= 2|u|\Delta|u| + 2|\nabla|u||^2, \end{aligned}\]

which together imply that \(u\Delta u = |u|\Delta|u|\). So, applying (4.1) to \(|u|\), we get

\[ \int_M |u|^{k-2}u\Delta u \leq \frac{-2C_\mathcal{SD}}{k}\left(\int_M |u|^\frac{km}{m-2}\right)^\frac{m-2}{m}. \](4.7)

We now claim that for any \(u \in E\), and \(j \geq 0\), we have the following estimate

\[ \nparallel u \Vert_{2\beta^{j+1}} \leq \prod\limits_{\ell = 0}^j \left(\frac{\mu_k V(M)^{2/m}\beta^\ell}{C}\right)^\frac{1}{2\beta^\ell - 1}\nparallel u \Vert_2, \](4.8)

which we will iterate in a similar manner to the one for \(\Vert\phi_i\Vert_{2\beta^{j+1}}\) above to obtain (4.5).

To obtain (4.8), choose an \(h \in E\) such that \(h\) has the following extremal property,

\[ \frac{\Vert h \Vert_{2\beta^{j+1}}}{\Vert h \Vert_2} = \max\limits_{u \in E}\frac{\Vert u \Vert_{2\beta^{j+1}}}{\Vert u \Vert_2}. \](4.9)

Then setting \(u = h\) and \(k = 2\beta^j\) in (4.7), and then using the Holder inequality, we have that

\[\begin{aligned} \frac{C_\mathcal{SD}}{\beta^j}\left(\int_M|h|^{2\beta^{j+1}}\right)^\frac{1}{\beta} &\leq -\int_M |h|^{2\beta^j-2}h\Delta h \\ &\leq \left(\int_M |\Delta h|^{2\beta^{j+1}}\right)^\frac{1}{2\beta^j+1}\left(\int_M |h|^{2\beta^j}\right)^\frac{2\beta^j-1}{2\beta^j} V(M)^\frac{\beta-1}{2\beta^j+1}. \end{aligned}\](4.10)

To control the first term, since \(h \in E = \text{span}\{\phi_1,\phi_2,\dots,\phi_k\}\), we can write \(h = \sum\limits_{\ell = 1}^k a_\ell \phi_\ell\), and therefore \(\Delta h = -\sum\limits_{\ell=1}^k\mu_\ell a_\ell\phi_\ell\). For \(p \geq 2\) and \(1 \leq i \leq k\), observe that the function \(F(\mu_i) := \int_M |\Delta h|^p\) is convex in \(\mu_i\). To check this, we compute the second derivative and see that

\[ \frac{\partial^2 F}{\partial \mu_i^2} = \int_M \frac{\partial^2}{\partial\mu_i^2}\left|\sum\limits_{\ell = 1}^k \mu_\ell a_\ell \phi_\ell\right|^p = p(p-1)\int_M\left|\sum\limits_{\ell = 1}^k \mu_\ell a_\ell \phi_\ell\right|^{p-2}a_\ell^2\phi_\ell^2 \geq 0. \]

This means that \(F\) is bounded above by either \(F(0)\) or by \(F(\mu_k)\). Then in particular, there is a \(\{\alpha\} \subset \{1,2,\dots,k\}\) such that

\[ \int_M |\Delta h|^{2\beta^{j+1}} \leq \int_M \left|\sum\limits_\alpha \mu_k a_\alpha \phi_\alpha\right|^{2\beta^{j+1}}. \](4.11)

Then remembering that \(\sum\limits_\alpha a_\alpha \phi_\alpha \in E\) and the extremal property of \(h\) (4.9), we have that

\[\begin{aligned} \int_M \left|\sum\limits_\alpha a_\alpha \phi_\alpha\right|^{2\beta^{j+1}} &\leq \left\Vert\sum\limits_\alpha a_\alpha \phi_\alpha\right\Vert_2^{2\beta^{j+1}}\Vert h \Vert_{2\beta^{j+1}}^{2\beta^{j+1}}\Vert h \Vert_h^{-2\beta^{j+1}} \\ &\leq \left(\int_M \left|\sum\limits_\alpha a_\alpha \phi_\alpha\right|^2\right)^{\beta^{j+1}}\left(\int_M|h|^{2\beta^{j+1}}\right)\left(\int_M|h|^2\right)^{-\beta^{j+1}} \\ &\leq \left(\int_M \left|\sum\limits_\alpha a_\alpha \phi_\alpha\right|^2\right)^{\beta^{j+1}}\left(\int_M|h|^{2\beta^{j+1}}\right) \\ &\leq \left(\sum\limits_\alpha a_\alpha^2\right)^{\beta^{j+1}}\left(\int_M|h|^{2\beta^{j+1}}\right) \\ &\leq \int_M |h|^{2\beta^{j+1}}. \end{aligned}\]

So inserting this into (4.10), we have

\[\begin{aligned} \frac{C_\mathcal{SD}}{\beta^j}\left(\int_M|h|^{2\beta^{j+1}}\right)^\frac{1}{\beta} &\leq \left(\int_M |\Delta h|^{2\beta^{j+1}}\right)^\frac{1}{2\beta^j+1}\left(\int_M |h|^{2\beta^j}\right)^\frac{2\beta^j-1}{2\beta^j} V(M)^\frac{\beta-1}{2\beta^j+1} \\ &\leq V(M)^\frac{\beta-1}{2\beta^{j+1}}\Vert h \Vert_{2\beta^j}^{2\beta^j-1}\left(\int_M \left|\sum\limits_\alpha \mu_\alpha a_\alpha \phi_\alpha\right|^{2\beta^{j+1}}\right)^\frac{1}{2\beta^{j+1}} \\ &\leq V(M)^\frac{\beta-1}{2\beta^{j+1}} \Vert h \Vert_{2\beta^j}^{2\beta^j-1}\mu_k \Vert h \Vert_{2\beta^{j+1}}, \end{aligned}\]

which implies

\[ \nparallel h \Vert_{2\beta^{j+1}} \leq \left(\frac{\mu_k V(M)^{2/m}\beta^j}{C}\right)^\frac{1}{2\beta^j-1}\nparallel h \Vert_{2\beta^j}. \](4.12)

By the extremal property of \(h\), we have the same inequality for \(u \in E\) and then iterating we obtain the claimed estimate (4.8) for \(u\). Then letting \(j \to \infty\), we arrive at (4.5), which, after applying lemma 3, gives us (4.6).

Now that we have bounded \(\Vert\phi_i\Vert_\infty\) with (4.4) and \(\mu_k\) with (4.6), we have that each summand

\[ e^{-\mu_i t}\phi_i(x)\phi_i(y) \leq e^{-\mu_i t}\Vert\phi_i\Vert_\infty^2 \leq C_1 e^{-\mu_i t}\mu_i^{m/2} \leq C_1 C(t) e^{-\mu_i t/2}, \]

where the last inequality comes from the fac that \(e^{-x t}x^{m/2} \leq C(t) e^{-xt/2}\) for some constant \(C(t) > 0\) depending only on \(t\) and valid for any \(0\leq x < \infty\).

So we have that the infinite series

\[ \sum\limits_{i=1}^\infty e^{-\mu_i t}\phi_i(x)\phi_i(y) \leq C_1C(t) \sum\limits_{i=1}^\infty e^{-C_4 i^{\frac{1}{m-1}} t}, \]

which converges uniformly. So \(H\) as defined above is well-defined on \(M \times M \times [a,\infty)\) for any \(a > 0\). To see that \(H\), and therefore \(f\), is smooth, note that

\[ \int_M \langle \nabla\phi_i,\nabla\phi_j\rangle = -\int_M \phi_j \Delta\phi_i = \int_M \mu_i \phi_i\phi_j = \delta_{ij}\mu_i, \]

where \(\delta_{ij}\) is the Kronecker delta. So the integral of each partial sum

\[\begin{aligned} \int_M \left|\sum\limits_{i=1}^k e^{-\mu_i t}\phi_i(x)\nabla\phi_j(y)\right|^2dy &= \int_M \sum\limits_{i,j=1}^k e^{-(\mu_i + \mu_j)t} \phi_i(x)\phi_j(x)\langle\nabla\phi_i(y),\nabla\phi_j(y)\rangle dy \\ &= \int_M \sum\limits_{i,j=1}^k e^{-(\mu_i+\mu_j)t}\phi_i(x)\phi_j(x)\delta_{ij}\mu_i dy \\ &= \sum\limits_{i=1}^k 2^{-2\mu_i t}\mu_i \phi_i(x)^2, \end{aligned}\]

which by the argument above converges uniformly as \(k \to \infty\). So the partial sums \(\sum\limits_{i=1}^k e^{-2\mu_i t}\mu_i \phi_i(x)^2\) converge weakly in \(H^{1,2}(M)\) to \(H(x,y,t)\), and since each partial sum solves the heat equation, this means \(H(x,y,t)\) is a weak solution to the heat equation. Then regularity theory implies that \(H\) is smooth and that \(f\) therefore satisfies (2.1), (2.2) and (2.3), thereby justifying our formal calculations above.