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Hi, I am really struggling with this 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.

$(1)$ seems straightforward since $f$ is compactly supported. For (2), I managed to show $\frac{1}{|x-y|}$ can be approximated by a sequence of smooth function say $\delta_n$ such that $u(x)= \int \delta_n(x-y)f(y)dy$ . But no success.

Could anyone here help me with this problem? Thanks in advance.

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This question was asked here: – B R Aug 30 '12 at 22:49
Are you both sarah and chris? – Steven Gubkin Aug 30 '12 at 23:34
I'm surprised you're not able to get more help from You could also try the appropriate forum on The Art of Problem Solving web site. Or get help from your classmates or professor? But you should also keep struggling with it yourself. You appear to be making progress, and even if it takes you a long time, it's always best to somehow do it yourself. Don't worry about how long it takes, even if it all seems too easy and obvious afterwards. Even some of us now successful research mathematicians suffered like that as undergraduates. – Deane Yang Aug 30 '12 at 23:34
You don't need to know anything. Just use your bare hands and struggle. – Deane Yang Aug 30 '12 at 23:59
This is obviously homework, voting to close. – Igor Rivin Aug 31 '12 at 15:49
up vote 1 down vote accepted

Your question text suggests that yo know how to do (1). (Interchange the order of integration). Now (2) is easy: outside the support of $f$ we have $\Delta u=0$, that is $u$ is harmonic outside the support of $u$. Harmonic functions are surely $C^\infty$. For (3), expand the kernel $1/|x-y|$ in a series of spherical harmonics outside of some ball, and integrate this series term by term (it is absolutely and uniformly convergent). You obtain $u(x)=O(1/|x|^2)$ which is more than enough for $L^2$.

References: 1. Landkof, Introduction to modern potential theory, 2. Hormander, Analysis of linear partial differential operators, vol. 1, 3. Stein, Weiss, Introduction to harmonic analysis on Euclidean spaces, 4. Any book with "Potential theory" in the title.

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Let $E$ be the fundamental solution of the Laplace equation in $\mathbb R^n$ for $n\ge 3$: $E=c_n\vert x\vert^{2-n}$ so that $E$ is $L^1_{loc}$. For $f$ compactly supported, $\chi\in C^\infty_c$, $$ \chi u=\chi (E\ast f)=\chi(x)\int_{y\in supp f} E(x-y) f(y) dy. $$ Since $x$ and $y$ are moving in compact sets, it is also the case of $x-y\in K=supp \chi-supp f$: as a result $$ \chi u=\chi \bigl((1_KE)\ast f\bigr)\Longrightarrow \Vert \chi u\Vert_{L^2}\le \Vert 1_K E \Vert_{L^1}\Vert f\Vert_{L^2},\quad\text{proving (1)}. $$ Moreover since $\Delta u=0$ on the open set $(supp f)^c$, the function $u$ is harmonic there, thus in fact real-analytic on that open set, proving more than (2). Going back to $ u(x)=\int_{y\in supp f} E(x-y) f(y) dy, $ we already know that $u\in L^2_{loc}$. We have also from the vanishing of the integral of $f$, $$ u(x)=\int_{y\in supp f} \bigl(E(x-y)-E(x)\bigr) f(y) dy, $$ so that for $\vert x\vert\gg diam supp f$, $y\in suppf$, $ \vert E(x-y)-E(x)\vert\lesssim \vert x\vert^{1-n}\in L^2(\mathbb R^n\cap \vert x\vert \ge 1) $ since we have $$ 2n-2>n,\quad \text{from $n>2$.} $$

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