Show that $T f(x) = \frac{1}{x^2}\int\limits_0^x t f(t) dt$ is not compact.

Question

Problem

Let $T : C([0,1]) \to C([0,1]),$ where

$T f(x) = \frac{1}{x^2}\int\limits_0^x t f(t) dt, \ x\neq 0$

$T f(0) = f(0)/2$

Show that $T$ is not compact.

Solution (wrong)

I'm using the following theorem:

Let $X$ be Banach and $T$ a compact bounded linear operator. If $dim(X) = \infty$, then $0 \in \sigma(T)$.

$T$ is clearly linear and I have shown that $T$ is bounded. Since $C([0,1])$ is Banach and infinite dimensional, if I show that $0 \notin \sigma(T)$, then I have shown that $T$ is not compact.

Assume that $0\in \sigma(T)$. Then $T f = 0$ for some $f\neq 0$. But that is not possible, since if $\forall x :\int\limits_0^x t f(t) dt = 0 $, implies that $f = 0$. So $0 \notin \sigma(T)$, hence $T$ is not compact.

Answer 1

$x^{1/(n-1)}$ is an eigen function corresponding to the eigenvalue $\frac {n-1} {2n-1}$. If $T$ is compact then the only possible limit point for the set of eigen values is $0$. Since $\frac 1 2$ is a limit point in this case it follows that $T$ is not compact.