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The tilte says it all. Given $f\in L^1(R^d)$ (let me restrict to dimension $d\geq 3$ for convenience), what is the optimal $L^p$ regularity for solutions to $$ -\Delta u=f\hspace{3cm}(1)? $$ I'm of course aware that solutions to (1) are only defined up to harmonic functions, so I'm implicitly speaking here of the "integral" solutions defined by means of the Newton potential $$ u(x)=\int\limits_{R^d}\frac{1}{|x-y|^{d-2}}f(y)\,\mathrm{d}y\hspace{3cm}(2) $$ up to normalizing constants.

When I discussed that with one of my colleagues he claimed that the map $f\mapsto u$ is continuous from $L^1$ to $L^p$ for all $p\in [1,\frac d {d-1})$, and that the statement also holds for Radon measures $\mu$ instead of $f\in L^1$ and replacing $f(y)\,\mathrm{d}y$ by $d\mu(y)$ in (2). Unfortunately he couldn't point me to a precise reference, and a quick websearch returns quasilions of results for compactly supported $\mu$ but nothing really relevant to me for $f\in L^1$ (a priori supported in the whole space).

I apologize if the question is trivial, but I'm not familiar with potential theory. As I only need a "black box" result for a specific problem I would greatly appreciate if you could point me directly to a precise statement (e.g. theorem x.x.x page y). Please feel free to close or migrate to SE if you deem it appropriate.

Thank you in advance.

Edit: I know of course about the Hardy-Littlewood-Sobolev inequality, but the latter only gives $L^p$ information on $u$ if $f\in L^q$ with $q>1$, and weak Lebesgue $L^{p,r}$ if $q=1$ so this is off-topic here since I'm only interested in the "classical" Lebesgue $L^p$ regularity of $u$.

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    $\begingroup$ Take f to be a Dirac measure. Then u is simply the Newton potential which is not in any $L^p$ (it fails either at the origin or at infinity). I do not think it is likely that assuming f in $L^1$ changes this in an essential way. $\endgroup$
    – Michael Renardy
    Commented Jun 6, 2014 at 13:20
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    $\begingroup$ good point, and taking any approxmation $f_n\in L^1$ to the Dirac mass shows that even if $u_n=G*f_n$ is in some $L^p$ for fixed $n$ there cannot be continuity of $f\in L^1\mapsto u\in L^p$ (which is actually what I need, rather than "pointwise" $u\in L^p$ for given $f\in L^1$). Huuummm, I gues I have to find another way around. Thank you very much! $\endgroup$
    – leo monsaingeon
    Commented Jun 6, 2014 at 13:43
  • $\begingroup$ Actually let me step back for a while: the Newton potential it-self is $L^p(B_r)$ for all $p\in [1,\frac{d}{d-2})$ at the origin, and of course bounded far from the singularity. So the problem is what happens at infinity and I'm not so convinced anymore that the result fails if $f\in L^1$. $\endgroup$
    – leo monsaingeon
    Commented Jun 6, 2014 at 14:01
  • $\begingroup$ I think u belongs to the homogeneous Besseel space $\dot{H}^{2,1}$ here. Solutions can be distributions, not belong to any $L^p$. More about such spaces, you can see book like "Theory of function spaces" by Triebel, Hans. $\endgroup$
    – Tomas
    Commented Jun 8, 2014 at 21:31
  • $\begingroup$ @Shanlin: thank you, I will give it a look. Do you know by any chance if I can guarantee uniqueness for $-\Delta u=f$ within this class of solutions (still for $f\in L^1$)? Or does the decay at infinity deteriorate so badly that uniqueness may fail? $\endgroup$
    – leo monsaingeon
    Commented Jun 8, 2014 at 22:19

1 Answer 1

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The following result is well known. It is Theorem 5.1 in [1]. It is proved by the method of duality solutions sue to Stampacchia.

Theorem. Let $\mu$ be a signed Borel measure with the finite total variation in a bounded open set $\Omega\subset\mathbb{R}^{n}$. Then the Dirichlet problem $$ \left\{\begin{array}{ccc} \Delta u&=& \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \mu \\ u&\in& W^{1,1}_{0}(\Omega) \end{array} \right. $$ has a solution with the property that $u\in W_{0}^{1,p}(\Omega)$, $\Vert \nabla u\Vert_{L^{p}(\Omega)} \leq C\Vert\mu\Vert(\Omega)$ for all $1\leq p<n/(n-1)$. If the boundary of $\Omega$ is sufficently smooth, then the solution is unique.

Clearly, the result applies to $f\in L^1$, because such a function defines a measure of finite total variation $d\mu = f(x)\, dx$.

As was pointed put by YangMills in his comment, you can find a short proof of the result in the case when $d\mu=f(x)\, dx$, $f\in L^1$ in Lemma 14 in https://arxiv.org/abs/0809.2172.

In the case of solutions in $\mathbb{R}^n$ you obtain regularity $u\in W^{1,p}_{\rm loc}$ for all $1<p<n/(n-1)$.

[1] W. Littman, G. Stampacchia, H. F. Weinberger, Regular points for elliptic equations with discontinuous coefficients. Ann. Scuola Norm. Sup. Pisa (3) 17 1963 43–77.

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  • $\begingroup$ Thank you Piotr, this makes perfect sense. I'll try to get acquainted with the "duality method". If you ever come across the precise reference please, please let me know. (otherwise I guess I'll have to finally read Stampacchia's work, which sounds like a good idea anyway) $\endgroup$
    – leo monsaingeon
    Commented Apr 29, 2018 at 8:18
  • $\begingroup$ @leomonsaingeon I added a reference, but I might add a proof since it is not easy to read the paper [1]. $\endgroup$
    – Piotr Hajlasz
    Commented Apr 30, 2018 at 16:13
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    $\begingroup$ You will find a writeup of the proof in Lemma 14 here: arxiv.org/abs/0809.2172 $\endgroup$
    – YangMills
    Commented May 1, 2018 at 21:43

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