Point spectrum of an integral operator

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?

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 $$0$$.
• I forgot to ask... how can we take $u(x)$ as $x^{\alpha}$? – Ninja May 2 at 20:59