Let $H$ be a complex Hilbert space and $T:H\to H$ a self-adjoint, bounded operator. Let $\lambda\in\sigma(T)\setminus\sigma_{pt}(T)$ where we define $\sigma_{pt}(T)=\{\lambda\in\mathbb{C}\,\big|\,\lambda I-T\,\text{is not injective}\}.$
Then, the image of $\lambda I-T$ is dense.
We can deduce that $\lambda I-T$ is injective but not surjective.
I wanted to show that given $\epsilon>0$, for all $x\in H$, there exists $y\in H$ such that $||x-(\lambda I-T)y||<\epsilon.$ However, I get stuck when I want to how this. How can I proceed?
Best Answer
Suppose $y$ is orthogonal to the range of $\lambda I -T$. Then $\langle \lambda x-Tx, y \rangle=0$ for all $x$. Since $T$ is self adjoint this gives $\lambda\langle x, y \rangle -\langle x, Ty \rangle=0$ for al $x$ which implies $Tx=\overline {\lambda} x$, a contradiction unless $y=0$. [ Eigen values of $T$ are real. So if $\overline {\lambda}$ is an eigen value the so is $\lambda$, but the hypothesis says this is not true]. Hence range of $\lambda I -T$ is dense.