Point spectrum of an integral operator

Question

Let we have $$Tu(x) = \cfrac{1}{x}\int_0^x u(y)dy$$ so that $u \in L^2(0,1)$. How can I show that $(0,2) \subset \sigma_p(T)$ and $T$ is not compact?

Answer 1

For $\alpha > -\frac{1}{2}$, we have $x^\alpha\in L^2(0,1)$ and

$$Tu(x)=\frac{1}{x}\int_0^x y^\alpha \, dy=\frac{1}{\alpha+1}u(x).$$

Hence $$(0,2)=\left\{\frac{1}{\alpha+1}: \alpha > -\frac{1}{2}\right\}\subset \sigma_p(T). $$

Non-compactness of $T$ follows immediately from the Riesz-Schauder theorem, as the spectrum of a compact operator on a Banach space can accumulate only at $0$.