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This thread is only Q&A!

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

Consider operators: $$T:\mathcal{D}(T)\to\mathcal{H}$$

Denote for shorthand: $$\Omega\subseteq\mathbb{C}:\quad\langle\Omega\rangle:=\operatorname{conv}(\Omega)$$

Then one has: $$\langle\sigma(T)\rangle=\overline{\mathcal{W}(T)}$$

Does this hold true?

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  • $\begingroup$ When it comes to matters like this, it doesn't matter whether or not someone is a moderator. A moderator has some degree of power, but he should only exercise it to enforce the rules established by the community. I don't intend to say that you shouldn't answer your own question (the option is included in the interface for a reason), just that those links are not really relevant. $\endgroup$ – tomasz Dec 29 '14 at 3:14
  • $\begingroup$ @tomasz: I know but it just seems that some people don't think about it before downvoting or voting to close. I got the impression that explicitely mentioning it prevents people from doing something to overhasty. But yes it even looks so ugly. :( $\endgroup$ – C-Star-W-Star Dec 29 '14 at 3:20
  • $\begingroup$ @tomasz: But good point: I replaced moderator with community. $\endgroup$ – C-Star-W-Star Dec 29 '14 at 3:22
  • $\begingroup$ In that case, I think it would be better suited for a comment (because that is what it actually is, and it prevents it from distracting from the actual content). $\endgroup$ – tomasz Dec 29 '14 at 3:22
  • $\begingroup$ Putting in comments seemed not to help. $\endgroup$ – C-Star-W-Star Dec 29 '14 at 3:25
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Bounded Operators

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

For nilpotent operators: $$N^{K+1}=0:\quad\langle\sigma(N)\rangle=(0)\subseteq\overline{\mathcal{W}(N)}$$

As standard example: $$\left\langle\sigma\begin{pmatrix}0&1\\0&0\end{pmatrix}\right\rangle=(0)\subsetneq\frac12\mathbb{D}=\overline{\mathcal{W}\begin{pmatrix}0&1\\0&0\end{pmatrix}}$$

But inclusion always holds!

Unbounded Operators

For normal operators: $$N^*N=NN^*:\quad\langle\sigma(N)\rangle=\overline{\mathcal{W}(N)}\subseteq\mathbb{C}$$

For symmetric operators: $$S\subseteq\overline{S}\subsetneq S^*:\quad\langle\sigma(S)\rangle\nsubseteq\mathbb{R}\quad\overline{\mathcal{W}(S)}\subseteq\mathbb{R}$$

For unclosed operators: $$T\subsetneq\overline{T}=T^{**}:\quad\langle\sigma(T)\rangle=\mathbb{C}\supseteq\overline{\mathcal{W}(T)}$$

It may happen even:* $$S=\overline{S}\subsetneq S^*:\quad\langle\sigma(S)\rangle=\mathbb{C}\supsetneq\mathbb{R}\supseteq\overline{\mathcal{W}(S)}$$

And even worse also:* $$T=\overline{T}=T^{**}:\quad\langle\sigma(T)\rangle=\varnothing\subsetneq\overline{\mathcal{W}(T)}\subseteq\mathbb{C}$$

So inclusion mostly fails!

References

For examples see: Closed Operators

For the proof see: Normal Operators, Spectral Measures

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