T be compact operator defined on $L^2(\Omega)$ Show that the null space of T-I satisfies

Question

Can anyone help me out on this one

Let $$ \Omega\subset\text{R^d be a domain K}\,\in\,L^2(\Omega X \Omega)$$ and T be compact operator defined on $L^2(\Omega)$ by Tf(x)=$\int_\Omega K( x,y)\text{f(y)}\ dy$

Show that the null space of T-I satisfies

dim N(T-I)$\le\||K||_{L^2(\Omega X\Omega)}^2$

Answer 1

That's interesting. The same holds for any measure space $(X,\mathcal A,\mu)$.

In the special case $||K||_2<1$ it's something everybody knows: It follows that $||T||<1$, hence $T-I$ is invertible.

Anyway, say $f_1,\dots,f_n$ is an orthonormal subset of the null space of $T-I$. It's enough to show that $n\le||K||_2^2$. Define $\phi_j\in L^2(X\times X)$ by $$\phi_j(x,y)=f_j(x)\overline{f_j(y)}.$$ Simple applications of Fubini show that $\phi_1,\dots,\phi_n$ is an orthonormal subset of $L^2(X\times X)$. And Fubini plus the fact that $\int_XK(x,y)f_j(y)\,d\mu(y)=f_j(x)$ shows that $$\int_{X\times X}K(x,y)\overline{\phi_j(x,y)}\,d\mu(x)\,d \mu(y)=1.$$So Bessel's inequality shows that $$||K||_2^2\le\sum_{j=1}^n1^2=n.$$