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Let $M$ be a compact and closed smooth Riemannian manifold, and consider weak solution $u$ of the equation $$u_t - \Delta u = f$$ given $f \in L^2(Q)$ and $u(0)=u_0 \in L^\infty(M)$.

I'm looking for a reference or sketch proof using Nash-Moser-De-Giorgi iterations (and not heat kernels or semigroup stuff, since I wish to adapt it to a different setting) of an $L^\infty-L^2$ smoothing effect for $u$, i.e., $$\lVert u \rVert_{L^\infty(Q)} \leq C(\lVert f \rVert_{L^2(Q)}, \lVert u_0 \rVert_{L^\infty(M)})$$ where $Q:=[0,T]\times M$.

Note that $f$ is only in $L^2$.

I tried books by Davies etc. but for the proofs involving bounded domains it seems the proofs really depend on the Sobolev inequalities, which varies on manifolds.

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  • $\begingroup$ It's not clear to me that an $L^2$ bound on $f$ suffices. If you assume an $L^\infty$ bound, then you can use the maximum principle, which does not require the Sobolev inequality. The NMD iteration gives a time-dependent $L^\infty$ bound on $u$ in terms of the $L^q$ bound of $u_0$. Is that what you want? $\endgroup$ – Deane Yang Mar 17 '16 at 17:55
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    $\begingroup$ Suppose $u_0$ is constant (and therefore smooth on $M$) and $f$ is an $L^2$ function on $M$ and independent of $t$. Then the unique solution to the heat equation is the solution to the ODE $\partial_t u = f$, which is $u = u_0 + tf$, which is only in $L^2$ and not necessarily in $L^\infty$. $\endgroup$ – Deane Yang Mar 17 '16 at 20:26
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    $\begingroup$ @DeaneYang: something's slightly wacky. To get the ODE reduction you want $f$ independent of $x\in M$, right? And to get the $L^\infty$ blow-up in finite time you want $f = f(t)$ be such that $\int_0^T f(t) \mathrm{d}t = \infty$. But then $f \not\in L^2([0,T])$. $\endgroup$ – Willie Wong Mar 17 '16 at 20:46
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    $\begingroup$ Just out of curiosity: what is a setting in which a heat equation cannot be treated by "semigroup stuff" methods? $\endgroup$ – Delio Mugnolo Mar 18 '16 at 5:18
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    $\begingroup$ @DelioMugnolo I have certain time-dependencies in my equation, which AFAIK are not particularly easily amenable to such methods. I'm not saying they cannot be treated though. $\endgroup$ – EDA Mar 18 '16 at 17:32
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To do De Giorgi-Nash-Moser it is important that $f$ is in $L^q$ for $q$ larger than half the dimension. A heuristic is scaling: if $v$ solves $-\Delta v = f$, then the right side for the rescaling $v(\epsilon x)$ is $\epsilon^2 f(\epsilon x)$, whose $L^q$ norm is like $\epsilon^{2-n/q}$, so zooming in helps when $q > n/2$.

When in dimension $4$ or higher, one can take e.g. $v = \sum h_k\varphi(2^k x)$ for some smooth $\varphi$ that is $1$ in $B_1$ and vanishes outside $B_2$. Taking $h_k = 1/k$ makes $v$ unbounded ($\log\log$ growth near the origin) while $f := -\Delta v$ is $L^2$ (the integral of $f^2$ is like $\sum h_k^2 2^{(4-n)k}$).

It seems to me that for such a choice of $f$, the solution $u$ will immediately become unbounded since $u - v$ is unbounded initially and solves the heat equation.

Note also that indeed, in Deane's computation, the (reciprocal) exponent on $f$ looks like $n/2$ for large $p$. It is important to use that $f \in L^q$ for $q > n/2$, for instance as follows (in the elliptic case for simplicity): Let $A_p = \left(\int u^{p\chi}\right)^{\frac{1}{p \chi}}$, with $\chi = n/(n-2)$. Applying Holder to the last term in the first line of Deane's computation gives an inequality like $$A_p \leq (pC(f,S))^{1/p} A_{\gamma p},$$ where $\gamma < 1$ depends on $q/(q-1) < \chi$, and $C(f,S)$ depends on $f$ and the Sobolev constant. Iteration gives $$A_{2\gamma^{-k}} \leq C(f,S)^{\sum j\gamma^j} \|u\|_{L^2(Q)}.$$ Taking $k \rightarrow \infty$ gives an $L^{\infty}$ bound in terms of the desired quantities and the Sobolev constant. I'm not sure how to remove dependence on the Sobolev constant.

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  • $\begingroup$ Thanks for the answer. If $n$ is small enough, I can't seem to prove the result. Using Deane's answer, in the end I get that on the RHS $C^{1-\frac 1p}p^{1-\frac 1p}\left(\int_0^T \left(\int_\Omega |f|^{(np)/(n+2p-2)}\right)^{(n+2p-2)/(np)}\right)^{\frac 1p}$ which I don't see how to take the limit as $p \to \infty$, due to the factor outside the integral. Do you have any ideas? $\endgroup$ – EDA Mar 19 '16 at 19:50
  • $\begingroup$ This looks right. Scaling argument gives the same. You need an $L^{n/2}$ bound. $\endgroup$ – Deane Yang Mar 20 '16 at 10:40
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    $\begingroup$ A sketch of the iteration is now included, giving an $L^{\infty}$ estimate that depends on the Sobolev constant- I'm not sure how to remove this dependence! $\endgroup$ – Connor Mooney Mar 20 '16 at 22:32
  • $\begingroup$ The question is whether the the $L^\infty$ bound of $u$ on $Q$ in terms of the $L^\infty$ bound of $u_0$ on $M$. You have an $L^2$ bound of $u$ on the right side of your inequality. I'm pretty sure if the Moser iteration is done carefully, where you bound the $L^\infty$ norm of $u$ in terms of the $L^p$ norm of $u_0$ and $f$ and then let $p \rightarrow \infty$, then the Sobolev constant disappears. But I still agree it will depend on the $L^{n/2}$ norm of $f$. $\endgroup$ – Deane Yang Mar 21 '16 at 4:44
  • $\begingroup$ I agree- by your computation $\|u\|_{L^p(Q)}$ are controlled by $\|u_0\|_{L^{\infty}}$ and $\|f\|_{L^q(Q)}$, with constant degenerating with $p$. Thus, the iteration outlined above, "starting at $p$," gives an estimate depending on the desired quantities, $p$, and $S$. Unfortunately, the dependency on the Sobolev constant and $p$ oppose each other. I'm not sure how to set up the iteration in a way that fixes this. $\endgroup$ – Connor Mooney Mar 21 '16 at 23:47
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Here's a rough calculation that might give you what you want (but the constant depends on the Sobolev constant!): The basic iterative step in Moser iteration for this particular equation is: \begin{align*} \frac{1}{p}\frac{d}{dt}\int |u|^p &\le \int |u|^{p-2}u\Delta u + |u|^{p-1}f\\ &\le -\int |\nabla |u|^{p/2}|^2 + |u|^{p-1}f\\ &\le -C^{-1}\left(\int |u|^{pn/(n-2)}\right)^{(n-2)/n}\\ & + \left(\int |u|^{pn/n-2)}\right)^{(n-2)(p-1)/(pn)}\left(\int |f|^{(n+2p-2)/(np)}\right)^{np/(n+2p-2)}\\ &\le (2C)^{p-1}\left(\int |f|^{(n+2p-2)/(np)}\right)^{n/(n+2p-2)}. \end{align*} This implies an inequality for $\|u\|_p$, which you can take the limit $p\rightarrow 0$. It appears to me that you don't need to iterate.

Actually, there's a good chance the Sobolev constant factor converges to $1$ in the limit $p\rightarrow\infty$, so the final $L^\infty$ inequality does not depend on the Sobolev constant. In fact, I used that observation once.

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  • $\begingroup$ You probably will need to fix the exponents. $\endgroup$ – Deane Yang Mar 17 '16 at 21:42
  • $\begingroup$ Thanks! I'll go through it. When you used the Sobolev inequality to get the third inequality to the second didn't you miss a lower order term involving some power of $u$ (I think the $p$th power)? $\endgroup$ – EDA Mar 17 '16 at 23:17
  • $\begingroup$ Ah, yes. Due to that, the volume will appear in the inequality, too. $\endgroup$ – Deane Yang Mar 17 '16 at 23:39
  • $\begingroup$ Hmm, I'm not sure that it works unfortunately. Your exponents for the integral over $f$ should be reciprocals of what you've written, and the outer exponent blows up as $p \to \infty$. $\endgroup$ – EDA Mar 18 '16 at 17:31
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    $\begingroup$ I agree the exponents for f and the integral should be replaced by their reciprocals. Unfortunately, I don't have time right now to work out the details carefully, but I don't see why the exponents should blow up. But I might be using the wrong choice of exponents for the Holder inequality. $\endgroup$ – Deane Yang Mar 18 '16 at 21:45

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