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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?

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

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  • $\begingroup$ How does this contradiction imply that the range of $\lambda I-T$ is dense? $\endgroup$ – user408856 Jan 31 '19 at 8:52
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    $\begingroup$ @James A linear subspace of a Hilbert space is dense iff the only vector orthogonal to the subspace is the zero vector. $\endgroup$ – Kavi Rama Murthy Jan 31 '19 at 8:55

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