Compact Operator on Hilbert Space

Question

How do I show that the range of $\lambda I-T$ is all of $H$ (Hilbert Space) if and only if the null-space $\bar\lambda I-T^{\ast}$ is trivial? Thanks!

Answer 1

The question reduces to the following: if $S\colon H\to H$ has a closed range, then the following are equivalent:

1. $\mathrm{Range}(S)=H$;
2. $\ker(S^*)=\{0\}$.

Assume that 1. is satisfied. If $S^*x=0$, then $\langle x,Sy\rangle=0$ for any $y\in H$ and by 1., there is some $y$ for which $Sy=x$ hence $\langle x,x\rangle =0$ and 2. is satisfied.

Assume that 2. is satisfied. Since $\mathrm{Range}(S)$ is a closed subspace of $H$, we are reduced to prove that its orthogonal is trivial. If $\langle x,Su\rangle =0$ for each $u\in H$, then $\langle S^*x,u\rangle=0$ for each $u$ hence $x=0$.