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A: How to get the $<Vf, g>$?

HermèsHint (to get you started): Well $x \mapsto e^{-|x|}$ is $C^\infty$ on $\mathbb{R}^3\setminus\{0\}$, so that $$ \Delta e^{-|x|} = \textrm{div}(\nabla e^{-|x|}) $$ and we trivially have that $\nabla e^{-|x|} = -\frac{x}{|x|}\cdot e^{-|x|}$. Moreover, and we have the well-known identity for $f(x) ...

Bob Oakley
Bob Oakley
well, I have got $\Delta e^{-|x|}$. But how to deal with $<(-1/|x|)f, g>$?
Hermès
Hermès
Well, what did you find for $\Delta e^{-|x|}$ ? Can't you go from there ?
Bob Oakley
Bob Oakley
I just get $<-\Delta e^{-|x|}, g>=<Ce^{-|x|}, g>$.
Hermès
Hermès
That's a mistake; you should do the computations again ;-)
Bob Oakley
Bob Oakley
Sorry, do you mean $\Delta (x/|x| )=1/|x|$? So term $(-1/|x|) fg$ in the integral will cancel?
Hermès
Hermès
07:45
Yes exactly, at some point the terms should cancel each others !
Bob Oakley
Bob Oakley
But why $\Delta(x/|x|)=1/|x|$? Is it not the delta function?
Hermès
Hermès
I was saying that at some point terms will canceled, but $\Delta(x/|x|)$ doesn't mean anything (since $x/|x| \in \mathbb{R}^3$). Do the computation of $\Delta e^{-|x|}$ slowly and you will see that a term in $1/|x|$ pops up :-) ! (NB : I think that there should be a $2$ in front of $1/|x|$ in your equation, but I'm not sure, I haven't done the computations myself)
Bob Oakley
Bob Oakley
Sorry, I mean $\nabla (x/|x|)$. I do not know how to compute it. Can you give me some advice?
Hermès
Hermès
Well, you can come back to the definition and compute $\partial_{x_k} x/|x|$ for every $k$ ! Try, and let me know if you cannot do it ;)
Bob Oakley
Bob Oakley
I guess the result is $1/|x|$ but I compute $\partial_{x_1} x/|x|$ which is $(|x|^2-x_1^2)/|x|^3$.
So here your $f=-x/|x|$ and we need to compute $div(-x/|x|)$, right? But this is zeor...
Hermès
Hermès
07:45
No, this is not zero; you should use the formula above : $\textrm{div}(x/|x|) = \nabla (1 / |x|) \cdot x + \text{div}(x) / |x|$, and $\textrm{div}(x) = 3$...
Bob Oakley
Bob Oakley
So that is $-|x|^{1/2}+3/|x|$?
well, there is still a wired term $|x|^{1/2}$.
Sorry, I see.. thx
Hi Hermes. I have another question about the Fourier transform. Can you have a glance?
I try to simplify the following singular integral:

>$$\int_{R^2}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi= A\int_{R^2}\int_{R^2}p(x) q(y) (-\log|x-y|)dxdy$$
where $p, q\in C_c^{\infty}$ and $\int q=0$.

My way is LHS=
$$\lim_{\epsilon\to 0}\int_{R^2\setminus B(\epsilon)}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi$$
But why the RHS appears $(-\log|x-y|)$ which is the log potential of Poisson equation.
Its something like the Fourier transforms of $|x|^{-\alpha}$ for $0<\alpha<n$.

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