Compact operator in $L^2(0,2\pi)$

Question

I have to prove that an operator is compact. The operator is $A:L^2[0,2\pi]\to L^2[0,2\pi]$ given by $$ Au(x)=\cos x\,u(x). $$ I have showed that $A$ is bounded and symmetric, I have to prove only the compactness. I had an idea: if I prove that this operator is an integral operator, it's ok, because an integral operator is always compact. Is it true? Is $A$ an integral operator? Thank you!

Answer 1

Your operator is not compact. You can easily show that $\sigma(A)=[-1,1]$, which by being uncountable (and/or by having accumulation points other than $0$) prevents $A$ from being compact.