# Resolvent R(1) of the Laplace operator not compact

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.

• 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 $\mathbb{R}^6$. – Branimir Ćaćić Jan 14 '14 at 14:23
• 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 $0$. – Branimir Ćaćić Jan 14 '14 at 14:24

## 1 Answer

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!