compactness of the operator $Tx (t) = t x(t)$ ($0< t < 1 $)

Question

Is the operator $T$ defined by $Tx (t) = t x(t)$ ($0< t < 1 $) compact in $L^2(0,1)$?

It is conjectured that it is indeed compact. But I cannot find a proof. Can anyone give one?

This is a problem in the book by Friedman.

Answer 1

This is not true.

Take the sequence $u_n = \chi_{(1-\frac1n,1)}\sqrt n$ with $\|u_n\|_{L^2}=1$. Then $$ \|Tu_n\|_{L^2}^2 = n \int_{(1-\frac1n,1)} t^2 dt = \frac n3( 1 - (1-\frac1n)^3) = 1-\frac1n + \frac1{3n^2} \to1, $$ and $(Tu_n)(t) \to 0$ almost everywhere. Hence $(Tu_n)$ cannot have a converging subsequence in $L^2$, since for strongly converging sequences the $L^2$-limit and the pointwise limit coincide.