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I'm following Kreyszig's "Introductory Functional Analysis with Applications" and am trying to follow the proof of the following Theorem (9.2-4 on p. 468)

For a bounded self-adjoint linear operator $T:H\to H$ on a complex Hilbert space $H$, $$\sigma_{r}\left(T\right)=\emptyset,$$ i.e. its residual spectrum is empty.

The proof refers to the following Lemma:

Lemma (projection theorem) Suppose that $Y$ is a closed subspace of a Hilbert space $H$. Then $$ H=Y\oplus Y^{\perp}. $$

Kreysig's begins his argument as follows.

"Suppose, for contradiction, that $\sigma_{r}\left(T\right)$ is non-empty; indeed, pick $\lambda\in\sigma_{r}\left(T\right)$. Then by definition of the residual spectrum, $R_{\lambda}:=T_{\lambda}^{-1}=(T-\lambda I)^{-1}$ exists but its domain is not dense in $H$."

I understand everything so far. He then goes to argue:

"Hence by the Projection Theorem, there is a $y \neq 0$ in $H$ which is orthogonal to the domain of $R_{\lambda}$."

I don't see why this is. I wonder if someone could very kindly explain?

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    $\begingroup$ Denote $Y=\operatorname{Im}(R_\lambda)$, then $Y$ is a proper subspace of $H$. By projection theorem $Y^\perp\neq\{0\}$, so you can choose $y\in Y^\perp\setminus\{0\}$ which is by definition of $Y^\perp$ orthogonal to $Y$, i.e. to $\operatorname{Im}(R_\lambda)$. $\endgroup$ – Norbert May 3 '13 at 10:13
  • $\begingroup$ But aren't we looking for $y \neq 0$ in $H$ which is orthogonal to the domain of $R_{\lambda}$, not its image? $\endgroup$ – Harry Williams May 3 '13 at 10:18
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    $\begingroup$ Sorry, in my first comment you should substitute $R_\lambda$ with $T_\lambda$ then $Y$ will be the domain of $R_\lambda$ $\endgroup$ – Norbert May 3 '13 at 10:23
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Denote $Y=\operatorname{Im}(T_\lambda)$, then $Y$ is a proper subspace of $H$. By projection theorem $Y^\perp\neq \{0\}$, so you can choose $y\in Y^\perp\setminus\{0\}$ which is by definition of $Y$ is orthogonal to $Y$, i.e. to $\operatorname{Im}(T_\lambda)$

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