0
$\begingroup$

I want to show that $$R_\Delta(1):=(1-\Delta)^{-1} $$ is not compact in $\mathbb{R}^3$.

I have found that for $\chi_{B}$ being the characteristic function for a set $B\subset\mathbb{R}^n$, $\chi_{B}(1-\Delta)^{-1}$ is compact since it has kernel $$K(x,y):=\chi_B(x)\frac{e^{-||x-y||}}{4\pi||x-y||},$$ which is clearly $K\in L^{2}(\mathbb{R}^3\times\mathbb{R}^3)$.

Theorem 9.14 in "Introduction to Spectral Theory" states that if an integral operator on $L^2(\mathbb{R}^n)$ has a Kernel in $L^2(\mathbb{R}^{2n})$, then the operator is compact. But shouldn't $(1-\Delta=^{-1}$ then not be compact too, since it has the same Kernel without the $\chi_B(x)$? Or am I too tired to see why this characteristic function decides upon $K\in L^{2}(\mathbb{R}^3\times\mathbb{R}^3$)?


Computing $\sigma((1-\Delta)^{-1})$ with Fourier transform gives me $$(F(1-\Delta)^{-1}F^{-1}\phi)(k)=(1+||k||^2)^{-1}\phi(k)$$ so the spectrum is the same as the multiplication with $(1+||k||^2)^{-1}$. This means the spectrum would be $(0,1]$, a contradiction to $(1-\Delta)^{-1}$ compact since $0\not\in \sigma((1-\Delta)^{-1})$ (and since it is bounded but not closed?). But I could still apply Theorem 9.14?

Edited for clarity.

$\endgroup$
  • 1
    $\begingroup$ 1) If you want to apply this Theorem 9.14, you need to figure out exactly what the kernel of $(1-\Delta)^{-1}$ itself is, and then check whether or not that kernel is square integrable. I can buy that the kernel of $(1-\Delta)^{-1}$ is $e^{\|x-y\|}(4\pi\|x-y\|)^{-1}$, but it isn't at all obvious to me that it is square integrable on $\mathbb{R}^6$. $\endgroup$ – Branimir Ćaćić Jan 14 '14 at 14:23
  • 1
    $\begingroup$ 2) The computation with the Fourier transform is the better way to approach this problem, because it gives you the spectrum, namely, $[0,1] = \overline{(0,1]}$ (remember, the spectrum of a bounded operator is closed). The way this contradicts compactness, however, is the fact that a self-adjoint compact operator has discrete spectrum, possibly with an accumulation point at $0$. $\endgroup$ – Branimir Ćaćić Jan 14 '14 at 14:24
1
$\begingroup$

Thanks to the comments of Branimir I got it now:

So $\Delta$ is locally compact, since we can write $(\chi_B(x)R_\Delta(1)f)(x)$ as$$(\chi_B(x)R_\Delta(1)f)(x)=\int\underbrace{\chi_{B}(x)\frac{e^{-||x-y||}}{4\pi||x-y||}}_{:=K(x,y)}f(y)dy,$$ where $K(x,y)\in L^2(\mathbb{R}^6)$. This can be seen by transforming $w=x-y$ and then going to spherical coordinates. In $\mathbb{R}^3$ this gives a factor $r^2$ in the integral, so the whole thing stays bounded.

With this we have shown with Theorem 9.14, that $\chi_B(x)R_\Delta(z)$ is compact for a $z\in\rho(\Delta)$ (and therefore for all $z\in \rho(\Delta)$ (first resolvent identity)). So $\Delta$ is locally compact.

Now if we look at $R_\Delta(1)$, the theorem cannot be applied, since the kernel here will be just $K(x,y)=\frac{e^{-||x-y||}}{4\pi||x-y||}$ which is not in $L^2(\mathbb{R}^6)$ (the characteristic function IS needed), the theorem can therefore not be applied.

One can even show that $R_\Delta(1)$ is not compact, since as seen in the post, $\sigma(R_\Delta(1))=[0,1]$ which is not discrete. This is a contradiction to $R_\Delta(1)$ being compact.

Thanks for the help!

$\endgroup$

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.