Let $T$ be a self adjoint operator in a Hilbert space $H$. Let $I$ be the identity operator on $H$ and $z\in \mathbb{R}$. Why does it hold that the range of $$T-izI$$ is $H$? Thanks in advance!
Range of self-adjoint operator
functional-analysisoperator-theory
Related Solutions
Hint:
$\ker(A)=R(A^*)^\perp$
For a subspace $W\subset H$ we have that $W^\perp =\{0\}$ if and only if $W$ is dense.
The facts you state do not depend on $K$ being compact, but only on $K$ being selfadjoint. For example, suppose $K^2x=0$. Then $0=\langle K^2x,x\rangle=\langle Kx,Kx\rangle=\|Kx\|^2$ implies $Kx=0$. So, $$ \mathcal{N}(K^2)=\mathcal{N}(K). $$ Therefore, if $K^2=K^3$, then $K^2(I-K)=0$ implies $K(I-K)=0$ or $K^2=K$.
There are several ways to prove that either $\|K\|$ or $-\|K\|$ is in the spectrum of $K$. One way is the prove that the norm of $K$ is the same as the spectral radius, which can be done by showing that $\|K\|=\|K^2\|^{1/2}$ for any selfadjoint operator $K$. This is because the spectral radius is $r_{\sigma}(K)=\lim_{n}\|K^n\|^{1/n}$, and $$ \|K\|=\|K^2\|^{1/2}=\|K^4\|^{1/4}=\cdots=\lim_{n}\|K^{2^n}\|^{1/2^n}=r_{\sigma}(K). $$ To see that $\|K\|=\|K^2\|^{1/2}$ for any selfadjoint $K$, note that $$ \|K^2\| \le \|K\|\|K\|=\|K\|^2, $$ and $$ \|K\|^2=\sup_{\|x\|=1}\|Kx\|^2 = \sup_{\|x\|=1}(K^2x,x) \le \sup_{\|x\|=1}\|K^2x\|\|x\|=\|K^2\|. $$ Every non-zero point of the spectrum $\sigma(K)$ of a selfadjoint compact operator is an eigenvalue. So, either $\|K\|=0$ or at least one of $\|K\|,-\|K\|$ is such a non-zero point of the spectrum because $\|K\|=r_{\sigma}(K)$.
An alternative method for showing that either $\|K\|$ or $-\|K\|$ is in the spectrum relies on the following operator norm equality which holds for selfadjoint operators $K$: $$ \sup_{\|x\|} |\langle Kx,x\rangle| = \|K\|. $$ Either $\sup_{\|x\|=1}\langle Kx,x\rangle =\|K\|$ or $\inf_{\|x\|=1}\langle Kx,x\rangle =-\|K\|$. The first case may be assumed to hold by replacing $K$ with $-K$ if necessary. Let $\{ x_n \}$ be a sequence of unit vectors chosen so that $\langle Kx_n,x_n\rangle$ converges to $\|K\|$. Then $$ \|\|K\|x_n-Kx_n\|^2=\|K\|^2-2\|K\|\langle Kx_n,x_n\rangle+\langle Kx_n,x_n\rangle^2\rightarrow 0. $$ Because $K$ is compact, there is a subsequence $\{ x_{n_k} \}$ such that $\{ Kx_{n_k}\}$ converges to some $y\in \mathcal{H}$. Then $\{ x_{n_k} \}$ also converges: $$ x_{n_k} = \frac{1}{\|K\|}(\|K\|x_{n_k}-Kx_{n_k})+\frac{1}{\|K\|}Kx_{n_k}\rightarrow 0+\frac{1}{\|K\|}y. $$ Hence $\frac{1}{\|K\|}y$ is a unit vector and, by the continuity of $K$, one obtains $K\left(\frac{1}{\|K\|}y\right)=y$, which proves that $\|K\|$ is an eigenvalue.
Best Answer
This is because $\sigma(T) \subseteq \mathbb{R}$, see here.
Now $iz \notin \mathbb{R}$ so $T-izI$ is invertible. In particular, $T-izI$ is surjective.