# On the spectrum of a self-adjoint operator

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?

• $\mathcal{R}(T-\lambda I)^{\perp}=\mathcal{N}(T^*-\overline{\lambda}I)$ In your case, $T^*=T$ and $\overline{\lambda}=\lambda$ and $\mathcal{N}(T-\lambda I)=\{0\}$. – DisintegratingByParts Jan 31 '19 at 15:53
• What are $\mathcal{R}$ and $\mathcal{N}$? – user408856 Jan 31 '19 at 16:13
• Range and Null Space, respectively. – DisintegratingByParts Jan 31 '19 at 16:14

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.
• How does this contradiction imply that the range of $\lambda I-T$ is dense? – user408856 Jan 31 '19 at 8:52