3
$\begingroup$

Let's consider in dimension $d\geq 3$ the Newton/riesz potential $f=I_2[g]$ $$ f(x)=\int_{R^d}\frac{1}{|x-y|^{d-2}}g(y)dy, $$ which solves $-\Delta f=g$ (up to positive normalizing constants, which I shall ignore), and assume that $g\in L^q$ for all $q\in[1,2d/(d+2)]$. By the Hardy-Littlewood-Sobolev inequality (or any other variation from [Stein, singular inegrals], [Lieb-Loss, Analysis] etc...) we know that $f\in L^{2d/(d-2)}$, and also $\nabla f\in L^2$ (this is one way to prove the Sobolev embedding $H^1\subset L^{2d/(d-2)}$). If $p=2d/(d+2)$ the conjugated Holder exponent is exactly $p'=\frac{2d}{d-2}$, thus $g\in L^p$, $f\in L^{p'}$ and $fg\in L^1$. Since by definition $-\Delta f=g$ we would expect that $$ \int \underbrace{f}_{\in L^{p'}}\underbrace{g}_{\in L^p}=\int f(-\Delta f)\overset{?}{=}\int |\underbrace{\nabla f}_{\in L^2}|^2. $$

When is the last integration by parts legitimate? With my hypotheses all the above terms are well defined, but is this enough? I seem to remember that there are "exotic" counterexamples...

It seems to me that an approximation argument works fine: if $g_n$ is a sequence of smooth compactly supported functions such that $g_n\to g$ in $L^{2d/(d+2)}$ then by continuity (HLS inequality) we have that $f_n\to f$ in $L^{2d/(d-2)}$ and $\nabla f_n\to \nabla f$ in $L^2$. Since the integration by parts is legitimate for smooth decaying functions then it should pass to the limit... I don't think anything is wrong here, did I miss something or is it really just that easy?

Edit: I just added the approximation argument

$\endgroup$
3
$\begingroup$

This "integration by parts" is true in larger generality: Let $\mu$ be a measure, $$u(x)=\int\frac{1}{|x-y|^{n-2}}d\mu(y).$$ Then $$\int u(x)d\mu=\int\frac{1}{|x-y|^{n-2}}d\mu(x)d\mu(y)$$ is called the energy of $\mu$. If this is finite, the measure is said of finite energy. Gradient of the potential of such measure is in $L^2$ and your formula holds. Then it is extended to differences of measures (charges) of finite energy. Measures and potentials of finite energy form a Hilbert space with respect to the scalar product $$(\mu,\nu)=\int u(x)d\nu(x).$$

Reference: Landkof, Introduction to modern potential theory, Ch. I, section 4.

$\endgroup$
  • $\begingroup$ OK, thank you Alexandre. What if the dimension is now $d=2$ and the Poisson Kernel $-\log|x-y|$? $\endgroup$ – leo monsaingeon Mar 4 '14 at 14:52
  • $\begingroup$ The case $n=2$ is somewhat special because the kernel is not positive definite. Roughly speaking it becomes positive definite if you restrict to charges of total charge 0 supported on the unit disc. For the details see the same book of Landkof. $\endgroup$ – Alexandre Eremenko Mar 5 '14 at 4:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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