# Reference request: harnack inequality for distributional solutions of the heat equation

Dear Math Overflowers,

I'm looking for references on the parabolic Harnack inequality for distributional solutions of the heat equation on the whole space $$\partial_t u=\Delta u\quad\text{and}\quad u(0)=u_0\quad\Leftrightarrow\quad u(t,x)=\int\limits_{R^d}\Gamma_t(x-y)u_0(y)dy.$$ Here $\Gamma_t(z)=\frac{e^{-\frac{|z|^2}{4t}}}{(4\pi t)^{d/2}}$ is the usual heat kernel and my initial data $u_0\geq 0$ is only in $L^1\cap L^p(R^d)$ for some $p>1$.

More precisely, I'm interested in the case when the initial datum $u_0$ is supported on a positive finite measure set $E=supp(u_0)$ and $u_0|_E\geq M>0$ for some constant $M$. I would like to control $u(t,x)\geq (\ldots)$ for points $x\in supp(u_0)$ and small times $t>0$, which really looks like Harnack inequality to me. The thing is that I have no information at all on the initial support (except that it has finite measure so it may not even be bounded) and I only have mere $L^1\cap L^p$ regularity for $u_0$. I strongly doubt that I can get pointwise estimates, but maybe $L^p$ estimates for $|u(t)|_{L^p(E)}\geq (\ldots)$?

Thank you in advance!

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## 1 Answer

Without knowing how the initial data is "spread," we can't get either pointwise or $L^p$ estimates independent of the measure of the initial support. Take for example the building block solution $v$ with initial data $\chi_{B_{\epsilon}}$. It follows from the representation formula that at time $t = \epsilon$, $v \leq C\epsilon^{n/2}$ on $B_{\epsilon}$. If we add together $\epsilon^{-n}$ translations of $v$ sufficiently "spread apart" (depending on $\epsilon$) to get $u$, it is easy to see (by exponential spatial decay) that the same bound will hold at all the points where $u$ was initially supported, which has measure like $1$.

No pointwise estimate follows immediately. To see no $L^p$, if we compute at $t = \epsilon$ we get $$\int_{E} u^p \leq C\epsilon^{-n} \int_{B_{\epsilon}} \epsilon^{np/2} \leq C\epsilon^{np/2}.$$

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great answer @Connor Mooney, thanks a lot! But darn, I whish you were wrong ;-) – leo monsaingeon Oct 24 '13 at 16:22