Let $\Omega$ be a bounded domain in $\mathbb{R}^n$.

Let $u \in L^2(0,T;V)$ be the weak solution of the heat equation $$u_t - \Delta u = 0$$ $$u(0) = u_0$$ where $u_0$ is bounded initial data. Here either $V=H^1_0(\Omega)$ or $v=H^1(\Omega)$ and we take zero Neumann data.

How may I prove an $L^\infty-L^1$ smoothing effect of this form:

$$\lVert u(t) \rVert_{L^\infty} \leq Ct^{-\gamma}\lVert u_0 \rVert_{L^1}^\alpha$$ for all $t > 0$?

I have seen a proof when we have the porous medium equation, but it is complicated (even when picking the linear case) so I wondered if there is an easier method. Maybe something using Sobolev inequalities.

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    $\begingroup$ Take a look at Chapter 2 in the book by Davies, Heat kernels and spectral theory $\endgroup$ – Piero D'Ancona Jan 8 '16 at 23:48
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    $\begingroup$ Where did the $\alpha$ exponent come from? It seems to me that just by scaling, you can't expect anything except $\alpha = 1$. $\endgroup$ – Nate Eldredge Jan 9 '16 at 1:03
  • $\begingroup$ @NateEldredge I agree it should be one but could you tell me what you mean with your scaling argument? $\endgroup$ – Deer Jan 9 '16 at 22:22
  • $\begingroup$ @Deer: Suppose the inequality holds as you stated it. Let $u_0$ be any nonzero function and $u(t)$ the corresponding solution, which does not vanish identically at any time $t$. For any constant $b$, since the heat equation is linear, the scaled function $bu$ is a solution with initial condition $bu_0$. Then the claimed inequality says $\|b u(t)\|_{L^\infty} \le C t^{-\gamma} \|b u_0\|_{L^1}^\alpha$. Rearranging, $\|u(t)\|_{L^\infty} \le C b^{\alpha - 1} t^{-\gamma} \|u_0\|^\alpha_{L^1}$. ... $\endgroup$ – Nate Eldredge Jan 10 '16 at 1:45
  • $\begingroup$ If $\alpha > 1$ or $\alpha < 1$, we can send $b \to 0$ or $b \to +\infty$ to conclude $\|u(t)\|_{L^\infty} = 0$ which is absurd. $\endgroup$ – Nate Eldredge Jan 10 '16 at 1:45

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