Prove that $T$ is not compact

Question

(a) Prove that the operator $T:L^2([0,1])\to L^2([0,1])$ defined by setting $T[f](x)=xf(x)$ is continuous and symmetric(self-disjoint).

(b) Prove that $T$ is not compact

I have proved question(a) which is easy (just use definitions), but I am stuck in part(b), can someone tell me how to prove part(b)?

Answer 1

Let $$ f_n(x) = \frac{1}{\sqrt{n(n+1)}}\chi_{[1-\frac{1}{n},1-\frac{1}{n+1}]}(x). $$ This is an orthonormal set of vectors in $L^2[0,1]$ because $f_nf_m=0$ a.e. for $n\ne m$. And $\{ Tf_n \}$ is an orthogonal set of vectors with $$ \|Tf_n\|^2 = \int_{1-1/n}^{1-1/(n+1)}x|f_n(x)|^2dx \ge (1-1/n)\int_{0}^{1}|f_n(x)|^2 = 1-1/n. $$ So there cannot be a convergent subsequence of $\{ Tf_n \}_{n=1}^{\infty}$, because $$ \|Tf_n-Tf_{n+n}\|^2 = \|Tf_n\|^2+\|Tf_{n+m}\|^2 \ge 1-1/n,\;\;\; m > 0, n \ge 1. $$

Answer 2

Hint:

Suppose, on contrary that $T$ is compact. Then, since $T$ is self-adjoint, either $\|T\|$ or $-\|T\|$ is an eigenvalue of $T$. But you can prove that $T$ has no eigenvalues, giving a contradiction.