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 $TI$ is bounded by $K_{L^{2}(\Omega\times\Omega)}^{2}$. In other word if $m = dim(N), N = ker(TI)$, 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 $m=f_1^{2}+\cdots + f_{m}^{2}\leq mK_{L^{2}(\Omega\times\Omega)}^{2}$. I want to get rid of the $m$ factor on the right. Any help much appreciated.
