# $L^{\infty}$ estimate for heat equation with $L^2$ initial data

Is there a way to show in a relatively simple manner that for a Lipschitz bounded, connected, open domain $$\Omega\subset\mathbb{R}^N$$ for any $$f\in L^2(\Omega)$$ the solution of the problem:

$$\begin{cases}\dfrac{\partial u}{\partial t}=\Delta u, & (t,x)\in (0,T)\times \Omega\\ \dfrac{\partial u}{\partial\nu}=0, & (t,x)\in (0,T)\times\partial\Omega \\ u(0,x)=f(x), & x\in\Omega\end{cases}$$

is bounded and $$\Vert u(t,\cdot)\Vert_{L^{\infty}(\Omega)}\leq c\cdot t^{-N/4}\cdot \Vert f\Vert_{L^2(\Omega)}$$ for any $$t\in (0,T)$$, for some $$c$$ depending only on $$\Omega$$?

• Use the integral representation in terms of the fundamental solution? Commented Jan 12 at 20:35
• In Theorem 6.3 and 6.4 this inequality in Ouhabaz (page 158) this inequality is equivalent with proving that $\Vert u\Vert^2_{\frac{2d}{d-2}}\leq c\cdot a(u,u),\ \forall u\in D(a)$. But this is not true in our case since $a(u,u)=\int_{\Omega}|\nabla u |^2\ dx$ and $D(a)=H^1(\Omega)$. For example $u=$constant do not satisfy such an inequality. Commented Jan 13 at 4:40
• One has to be a bit careful here, there a two (non-equivalent) forms of ultracontractivity. The one holds for the heat semigroup on full space and requires the $L^1$-$L^\infty$ bound from the question for all $t\geq 0$. As noted in Denis' answer, this cannot possible hold for the Neumann Laplacian on bounded domains. The second form of ultracontractivity requires the $L^1$-$L^\infty$ bound from the question only on bounded intervals with a constant $c$ depending on $T$. This form of ultracontractivity is true for the Neumann Laplacian on bounded domains with Lipschitz boundary. Commented Jan 13 at 9:53
• The equivalent characterization in terms of a Sobolev inequality takes the form $\lVert u\rVert_{2d/(d-1)}^2\leq c(a(u,u)+\lVert u \rVert_2^2)$, which is true for bounded domains with Lipschitz boundary. Commented Jan 13 at 9:54
• As explained by @MaoWao one has to apply Ouhabaz results to the form $a+1$ which gives global ultracontractivity to for the scaled semigroup and local for the original one. Commented Jan 13 at 12:05

The inequality is obviously false, because if $$f\equiv c$$ is a constant, then $$u\equiv c$$ remains constant. Since $$\Omega$$ is bounded, this $$f$$ is $$L^2(\Omega)$$, and yet $$\|u(t)\|_\infty=|c|$$ does not decay as $$t\to+\infty$$.
I suppose that you make a confusion with the heat equation in the hole space ($$\Omega={\mathbb R}^N$$, of course the boundary condition disappears). Then $$u(t)=H_t\star f$$ where the heat kernel is given by $$H_t(x)=\frac1{(2\pi t)^{N/2}}K\left(\frac x{\sqrt t}\right),\qquad K(y):=\exp\frac{-|y|^2}4.$$ From Young inequality (or just Hölder) $$\|u(t)\|_\infty\le\|H_t\|_2\|f\|_2=\frac c{T^{N/4}}\|f\|_2.$$
Turning back to the case of a bounded domain, the general rule is an exponential decay as $$t\to+\infty$$. For instance, the spectrum of $$\Delta$$ under the Dirichlet boundary condition $$u|_{\partial\Omega}=0$$ is real, bounded above by $$-\mu_D<0$$, and we get $$\|u(t)\|_p=O(e^{-\mu_D t})$$ for every $$p\ge2$$. In the case of the Neumann boundary condition, we must subtract the mean of the initial data: $$\|u(t)-\frac1{|\Omega|}\int_\Omega f(x)dx\|_p=O(e^{-\mu_N t}),\qquad\forall p\ge2.$$ Here $$\mu_N$$ is the least non-zero eigenvalue of $$-\Delta$$.