Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.
share|cite|improve this question
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… –  user16299 Aug 31 '12 at 9:50

1 Answer 1

up vote 0 down vote accepted

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.

share|cite|improve this answer
It is in chapter 8 of my book, but I think that it moved somewhere to chapter 9 in the second edition. There is no pde theory in that proof, only the Hoelder inequality applied twice or so. –  Siminore Aug 30 '12 at 10:28
But any idea about 3? –  chris Aug 30 '12 at 12:59
@chris I have added a possible idea about 3. –  Siminore Aug 30 '12 at 13:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.