# How to show that fundamental solution of Laplace equation $\in L^2$given $f \in L^2$?

I need help with this homework question.

The question is :

Let $f:R^3\to R$ and $f\in L^2(R^3)$. $f$ is supported on a ball of radius 1/2 centred at origin. Let $u$ be the solution to $\Delta u=f$ , where $u$ is given by $u(x)= \frac{1}{4\pi}\int_{R^3}\frac{1}{|x-y|}f(y)\,dy$.

1. Show that $L^2$ norm of u in the unit ball of radius 1, centred at origin, is bounded by C$||f||_{L^2}$, where C is a constant independent of f.
2. Show that $u$ is $C^\infty$ outside the unit ball centred at origin.
3. Suppose that $\int_{R^3}f(y)dy = 0$ , show $u\in L^2(R^3)$. (Consider how an good approximation it is to replace $\frac{1}{|x-y|}$ by $\frac{1}{|x|}$ for $|x|$ large.
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I don't think you've copied question 1. properly. The inequality $||u||_{L^2} \leq C||f||_{L^2}$ doesn't depend on a variable, so what does it mean for the inequality to hold in unit ball centred at origin.? –  Byron Schmuland Aug 30 '12 at 2:04
@ByronSchmulandI've edited to make it clearer –  chris Aug 30 '12 at 8:09
Cross-posted to mathoverflow.net/questions/106000/… –  user16299 Aug 31 '12 at 9:50

The integrability of $u$ is a consequence of the following Lemma. It appears in Jost, Partial differential equations.
Lemma. For $\mu \in (0,1]$ and $f \in L^1(\Omega)$, $\Omega \subset \mathbb{R}^d$, put $$(V_\mu f)(x)=\int_\Omega |x-y|^{d(\mu-1)} f(y)\, dy.$$ Let $1 \leq p \leq q \leq \infty$, $$0 \leq \delta = \frac{1}{p}-\frac{1}{q} < \mu.$$ The $V_\mu$ maps continuously $L^p(\Omega)$ into $L^q(\Omega)$, and $$\|V_\mu f\|_q \leq \left( \frac{1-\delta}{\mu-\delta}\right)^{1-\delta} \omega_d^{1-\mu} |\Omega|^{\mu-\delta} \|f\|_p.$$ Here $\omega_d$ is the volume of the unit ball of $\mathbb{R}^d$ and $|\Omega|$ is the Lebesgue measure of $\Omega$.
The proof of this lemma is a repeated application of the Hoelder inequality. I finally suspect that 3. requires an estimate of the decay of $u$ at infinity. Probably you want to write $$\frac{1}{|x-y|} = \frac{1}{|x-y|}-\frac{1}{|x|}+\frac{1}{|x|}.$$ Hence $$\int \frac{f(y)}{|x-y|}dy = \int \frac{f(y)}{|x|}dy + \int f(y) \left( \frac{1}{|x-y|}-\frac{1}{|x|} \right) dy.$$ The first integral is zero because $\int f(y)dy=0$. Now you have to estimate the second integral when $|x|$ is large.