Is this operator T compact on $L^2([0,1])$?

Question

Consider $L^2([0,1])$ and the operator $$Tu(t) = tu(t)$$for a.e. $t \in [0,1]$

How can I prove or disprove that T is compact? Thanks!

Answer 1

Hint: $T$ is self-adjoint and has no eigenvectors. Use the spectral theorem.

Answer 2

Hint

Let $f_n(x)=n^{3/2}\boldsymbol 1_{[0,\frac{1}{n}]}(x)$ for all $n\in\mathbb N$. Set $K=\{f_n\mid n\in\mathbb N\}\subset L^2(0,1)$. Is $K$ bounded ? Is $T(K)$ relatively compact ?

Answer 3

On the infinite dimensional invariant subspace $\{f\in L_2([0,1]): f_{[0,1/2]}=0\}$ the operator is invertible.