# Compact integral operator

I have a question regarding compact integral operators on $L^{2}({\Omega})$ with $\Omega$ a bounded domain in $\mathbb{R^{n}}$ Suppose we are given $T$ from $L^{2}(\Omega)$ to $L^{2}(\Omega)$ as $Tf(x) = \int_a^b K(x,y)f(y)\,dy$ with $K(x,y) \in L^{2}(\Omega\times\Omega)$ and $T$ is compact. I would like to know how the dimension of the kernel of $T-I$ is bounded by $||K||_{L^{2}(\Omega\times\Omega)}^{2}$. In other word if $m = dim(N), N = ker(T-I)$, then I would like to have $m \leq ||K||_{L^{2}(\Omega\times\Omega)}^{2}$. Here's what I did so far. Since $T$ is compact, $m$ is finite, and so we can construct an orthonormal basis for $f_{i}, i = 1,\cdots, m$ for $N$. We also have $<f_{i}, f_{j}> = <Tf_{i}, Tf_{j}>$ and so $m=||f_1||^{2}+\cdots + ||f_{m}||^{2}\leq m||K||_{L^{2}(\Omega\times\Omega)}^{2}$. I want to get rid of the $m$ factor on the right. Any help much appreciated.

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I think you are also assuming Hermitian symmetry $K(x,y)=\overline{K(y,x)}$ (or just $K(x,y)=K(y,x)$ if $K$ is real) otherwise the eigenfunctions need not to be othogonal. Then $K(x,y)=\sum_{j=1}^\infty\lambda_jf_j(x)f_j(y)$ and by Parseval $\|K\|_2^2= \sum_{j=1}^\infty|\lambda_j| ^2\ge m$. – Pietro Majer Oct 24 '13 at 0:20
No symmetry is needed, Pietro. The Hilbert-Schmidt norm of $T$ is equal to the $L_2$ norm of $K$. – Bill Johnson Oct 24 '13 at 0:51
I don't say more because this looks like a homework problem. – Bill Johnson Oct 24 '13 at 0:54