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Problem

Given a Hilbert space $\mathcal{H}$.

Denote for readability: $$\Omega\subseteq\mathbb{C}:\quad\|\Omega\|:=\|\omega\|_{\omega\in\Omega}:=\sup_{\omega\in\Omega}|\omega|$$

Denote for shorthand: $$\Omega\subseteq\mathbb{R}:\quad\Omega_+:=\sup_{\omega\in\Omega}\omega\quad \Omega_-:=\inf_{\omega\in\Omega}\omega$$

For bounded operators: $$A\in\mathcal{B}(\mathcal{H}):\quad\sigma(A)\subseteq\overline{\mathcal{W}(A)}$$

For normal operators: $$N^*N=NN^*:\quad\|\sigma(N)\|=\|\mathcal{W}(N)\|$$

But one has even: $$H=H^*:\quad\sigma(H)_\pm=\mathcal{W}(H)_\pm$$

How can I prove this?

Attempt

The argument goes as: $$|\langle N\hat{\varphi},\hat{\varphi}\rangle|\leq\|N\|\cdot\|\hat{\varphi}\|^2=\|\sigma(N)\|$$ (It exploits normality!)

Example

As standard example: $$\left\langle\sigma\begin{pmatrix}0&1\\0&0\end{pmatrix}\right\rangle=(0)\quad\overline{\mathcal{W}\begin{pmatrix}0&1\\0&0\end{pmatrix}}=\tfrac12\mathbb{D}$$

(It exploits nilpotence!)

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Let $H=H^{\star}$ be a bounded selfadjoint operator and let $\rho=\inf_{\|x\|=1}(Hx,x)$. Then $\rho_{\epsilon}(x,y)=((H-\rho I+\epsilon I)x,y)$ is an inner product on $\mathcal{H}$ for any $\epsilon > 0$. So, $$ |\rho_{\epsilon}(x,y)| \le \rho_{\epsilon}(x,x)^{1/2}\rho_{\epsilon}(y,y)^{1/2}. $$ Letting $\epsilon \downarrow 0$ gives $$ |((H-\rho I)x,y)| \le ((H-\rho I)x,x)^{1/2}((H-\rho I)y,y)^{1/2}. $$ Let $y = (H-\rho I)x$ in order to obtain \begin{align} \|(H-\rho I)x\|^{2} & \le ((H-\rho I)x,x)^{1/2}((H-\rho I)(H-\rho I)x,(H-\rho I)x)^{1/2} \\ & \le ((H-\rho I)x,x)^{1/2}\|H-\rho I\|^{1/2}\|(H-\rho I)x\| \end{align} Hence, $$ \|(H-\rho I)x\| \le \|H-\rho I\|^{1/2}((H-\rho I)x,x)^{1/2}. $$ By the definition of $\rho$, there is a sequence of unit vectors $\{ x_{n} \}$ such that $((H-\rho I)x_{n},x_{n})\rightarrow 0$. Therefore, $$ \lim_{n}\|(H-\rho I)x_{n} \|=0, $$ which means $\rho \in \sigma(H)$ because it is in the point spectrum or approximate point spectrum of $H$. By the way, you also see that a unit vector $x$ minimizes $(Hx,x)$ over all unit vectors iff $Hx = \rho x$.

The opposite direction is not so difficult, which is to say that every $\rho'$ such that $\rho ' < \rho=\inf_{\|x\|=1}(Hx,x)$ is in the resolvent of $H$. I've shown that to you before.

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  • $\begingroup$ Man I must admit again and again: You're really awesome!!! :D $\endgroup$ – C-Star-W-Star Jul 9 '15 at 17:09
  • $\begingroup$ I found an extremely elementary proof. xD (See my answer.) $\endgroup$ – C-Star-W-Star Jul 10 '15 at 0:42
  • $\begingroup$ @Freeze_S : You are assuming facts I'm proving from scratch. The assumption $\|\sigma(N)\|=\|\mathcal{W}(N)\|$ is what I've proved here for selfadjoint, and I proved $\mathcal{W}(H)_{\pm}=\sigma(H)_{\pm}$. My assumptions: $H=H^{\star}$ and the CS inequality. $\endgroup$ – DisintegratingByParts Jul 10 '15 at 2:07
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    $\begingroup$ @Freeze_S : It's at a much higher level than Cauchy-Schwarz. :) $\endgroup$ – DisintegratingByParts Jul 10 '15 at 15:24
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    $\begingroup$ Ya kind of true - you got me. ^^ $\endgroup$ – C-Star-W-Star Jul 10 '15 at 18:05
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Define the value: $$\omega:=\omega_-:=\mathcal{W}(H)_-$$

By linearity of range: $$\mathcal{W}(H-\omega)=\mathcal{W}(H)-\omega\geq0$$

For bounded operators: $$\mathcal{W}(H-\omega)\geq0\implies\sigma(H-\omega)\geq0$$

For normal operators: $$\sigma(H-\omega)_+=\|\sigma(H-\omega)\|=\|\mathcal{W}(H-\omega)\|=\mathcal{W}(H-\omega)_+$$

By linearity of both: $$\sigma(H)_+=\sigma(H-\omega)_++\omega=\mathcal{W}(H-\omega)_++\omega=\mathcal{W}(H)_+$$ $$\sigma(H)_-=-\sigma(-H)_+=-\mathcal{W}(-H)_+=\mathcal{W}(H)_-$$

Concluding the assertion.

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