$Tf = xf(x)$ is not compact in $L^2([0,1])$

Question

I want to prove, in a rather elementary way, that $Tf = xf(x)$ is not compact in $L^2([0,1])$. I cannot find the appropriate bounded sequence whose image has no Cauchy sub-sequences. I have tried variants of $f_n(x) = \sqrt n\cdot I_{[0,1/n]}$ to no avail. Any suggestions?

Answer 1

Start with an orthonormal set $\{ f_n \}_{n=1}^{\infty}$ in $L^2[1/2,1]$, and extend the functions to be $0$ on $[0,1/2]$ in order to obtain an orthonormal set $\{ \tilde{f}_n \}_{n=1}^{\infty}$ in $L^2[0,1]$. Then, for $n\ne m$, \begin{align} \|T\tilde{f}_n-T\tilde{f_m}\|^2 & =\|x\tilde{f}_{n}-x\tilde{f}_m\|^2 \\ & =\int_{1/2}^{1}x^2|f_n(x)-f_m(x)|^2dx \\ & \ge \frac{1}{4}\int_{1/2}^{1}|f_n(x)-f_m(x)|^2dx \\ & = \frac{1}{4}(\|f_n\|_{L^2[1/2,1]}^2+\|f_m\|^2_{L^2[1/2,1]})=\frac{1}{2}. \end{align}