Integral operator $\frac{g(s)}{1+(\cdot-s)^2}$ on $L^2(\mathbb{R})$

Show that

$(Ag)(t)=\int_{-\infty}^{\infty} \frac{g(s)}{1+(s-t)^2}ds, \ \ g \in L^{2}(\mathbb{R})$,

defines a bounded, self adjoint operator that is not compact.

Got the first part. As for the second part I want to apply Fubini's theorem but lacking a motivation to do so. Also like some on the compactness part if not shown to a suitable example.

Solution 1:

Let $F$ be the Fourier transform operator, defined on $L^2(\mathbb R)$as $$Fg(\xi)=\int_{\mathbb R} g(s)e^{-2i\pi \xi s}ds$$ Then $F$ is an isometry (by Plancherel/Parseval), which means that $F^{-1}=F^*$.

Note that $A$ is a convolution operator: $$Af = g \circledast h $$ where $$h(s)=\frac 1 {1+s^2}$$ The Fourier transform of that function is well-known (you can verify it by using the inverse Fourier transform formula): $$Fh(\xi)=\hat h(\xi)=\pi e^{-2\pi|\xi|}$$ Because the Fourier transform maps convolutions to pointwise products, it follows that: $$(FAg)(\xi)=F(h \circledast g)(\xi)=(Fh)(\xi).(Fg)(\xi)=\hat h(\xi).(Fg)(\xi)$$ In other words, for any $u\in L^2(\mathbb R)$: $$(FAF^{-1})u = (FAF^*)u = \hat h. u\tag{1}$$ Thus, this tells us everything we need to know about $A$ itself.

A is bounded

$F$ being an isometry, it suffices to verify that $FAF^{-1}$ is bounded, and that follows from $$\|FAF^{-1}u\|^2=\int_{\mathbb R}|\hat h(\xi)u(\xi)|^2 d\xi\leq \pi^2\|u\|^2$$

A is self-adjoint

Again, by $(1)$, it suffices to verify that the pointwise multiplication operator is self-adjoint. For $u$ and $v$ in $L^2(\mathbb R)$, $$\int_{\mathbb R} \overline{v(\xi)}\left(\hat h(\xi)u(\xi)\right)d\xi = 2\pi\int_{\mathbb R} \overline{v(\xi)}u(\xi)e^{-2\pi|xi|}d\xi = \overline{\int_{\mathbb R} \overline{u(\xi)}\left(\hat h(\xi)v(\xi)\right)d\xi}$$

A is not compact

If it were, then so would the pointwise multiplication operator $FAF^{-1}=FAF^*$, since $F$ is an isometry. We also know that self-adjoint compact operators have a countable set of eigenvalues. Let $\lambda$ be an eigenvalue with eigenfunction $u$. Then for almost all $\xi\in\mathbb R$, $$\hat h(\xi)u(\xi)=\pi e^{-2\pi |\xi|} u(\xi) = \lambda u(\xi)$$ Because the set of values of $\xi$ for which $u(\xi)\neq 0$ is of positive measure, this implies that $\lambda = 2\pi e^{-2\pi|\xi|}$ for that set of $\xi$, which is impossible.

Generalization to all convolution operators

There's nothing special about function $h$. The steps above can be repeated for any function $h$, and we conclude that all convolution operators (for the Lebesque measure) are bounded, self-adjoint and non-compact.