# 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?

## 1 Answer

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 see that there are several forms of the therem, so which one do we need here? – Ninja May 2 at 3:14
• For a compact operator on a Banach space, all nonzero elements of the spectrum are eigenvalues. Each eigenspace is finite dimensional and the eigenvalues can only accumulate at 0. We only require the last statement here, which can be shown by itself quite easily using the definition of compactness. – Goonfiend May 2 at 4:52
• I forgot to ask... how can we take $u(x)$ as $x^{\alpha}$? – Ninja May 2 at 20:59