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Let $T$ be an integral operator on $L^2([0,1])$, such that: $$ (Tf)(x) = \int_0^1K(x,y)f(y)dy, $$ with $K(x,y): [0,1]^2 \rightarrow \mathbb{R}$ continuous and $K(x,y) = K(y,x)$, $K(x,y)\geq0$ $ \forall(x,y)\in[0,1]^2 $. I have to show that $T$ has a countable collection of positive real eigenvalues. Furthermore I have to show that the collection normed eigenfunctions form a complete orthonormal system of $L^2([0,1])$.

It is fairly easy to proof that this integral operator is self-adjoint: $$ \langle Tf\mid g\rangle = \int_0^1\int_0^1K(x,y)f(y)dy g(x)dx = \int_0^1\int_0^1K(y,x)f(x)dx g(y)dy = \langle f\mid Tg\rangle. $$ So I would say this determines that the eigenvalues must be real. Furthermore I know that $T$ is compact (lengthy but easy proof with Arzela Ascoli). But from this point I am stuck. Can anyone help me?

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Are you allowed to use spectral theorem? – Davide Giraudo Jan 17 '13 at 10:50
@DavideGiraudo if you mean $ \int_0^1\int_0^1 |k(x,y)^2dxdy = \sum_{\lambda\in\sigma(T)}\lambda^2\mathrm{dim}(E_{\lambda})$, then yes we have proven this before during my lectures. – Funzies Jan 17 '13 at 19:36
up vote 2 down vote accepted

Just notice that $L^{2}$ is separable. And if you have eigenvectors $\{f_i;i<n\}$, then you can retrict T to the orthogonal complement $M_n$ of the subspace spanned by $\{f_i;i<n\}$, because T is self-adjoint, it maps $M_n$ to $M_n$ itself. At last you only have to show that the restriction of T must have a eigenvector(which follows from the compactness). For the same eigenvalue you can always choose the eigenvectors orthogonal to each other(because the eigenspace would be finite dimensional if the eigenvalue is not 0). If the closed linear span S of eigenvectors of T doesn't contain $L^2$ then by considering the orthogonal complement of S you will get a contradiction.

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