If $(Tx)(t)=\alpha(t)x(t)$ is compact on $L^2[a,b]$, then $\alpha(t)=0$ a.e. on $[a,b]$

Question

I'm going to prove the following statement: Let $\alpha(t)$ be a bounded measurable function on $[a,b]$. If the operator $(Tx)(t)=\alpha(t)x(t)$ is compact on $L^2[a,b]$, then $\alpha(t)=0$ a.e. on $[a,b]$. Can anyone give some hints? And I wonder if this statement is true on $L^p[a,b]~ \text{for}~p\neq2$ as well.

Answer 1

Assume that $\alpha>\delta>0 $ on a set $E $ of positive measure. Show that the image of $T $ contains $L^p (E) $ and thus images of balls are not compact.

Answer 2

$T$ is a bounded linear operator and $T$ is normal. Furthermore

$||T||=||\alpha||_{\infty}$ .

Since $T$ is normal, $||T||=r(T)||$ ($r(*)$= spectral radius). Let $\sigma_p(T)$ denote the set of eigenvalues of $T$.

If $T$ is also compact, then $r(T)= \max \{|\mu|: \mu \in \sigma_p(T)\}$.

Now its your turn: what are the eigenvalues of $T$ ?

Answer 3

Hint: Let $E$ be a set of positive measure on which $|\alpha| > \epsilon$. Then $T$ is an isomorphism on $L^p(E)$ (considered as embedded in $L^p([a,b])$).