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Let $A$ be a bounded, nonnegative operator on a complex Hilbert space $H$. Prove that the spectrum $$\sigma(A)\subset[0,+\infty].$$ We say that an operator $A$ is nonnegative if it is self adjoint and $$ \langle Au,u\rangle \geq 0 \ \ \ \forall u \in H.$$ It is exercise 9.5 page 240 from https://www.math.ucdavis.edu/~hunter/book/ch9.pdf.

A more general question, there is a theorem that for a self-adjoint operator $$ \sigma(A)\subset\left[-\|A\|,\|A\|\,\right],$$ but is it true that $$ \sigma(A)\subset[\underset{\|u\|=1}{\inf}\langle Au,u\rangle,\underset{\|u\|=1}{\sup}\langle Au,u\rangle ]?$$


marked as duplicate by Alex M., GNUSupporter 8964民主女神 地下教會, Namaste, Xander Henderson, Parcly Taxel Mar 3 '18 at 4:52

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So we want to show that if $A-\lambda I$ is not invertible, then $\lambda\geq0$. There are three ways in which $A-\lambda I$ may fail to be invertible:

  1. $\ker(A-\lambda I)\ne\{0\}$. In this case $\lambda $ is an eigenvalue. So there exists a unit vector $v\in H$ with $Av=\lambda v$. Then $$\lambda=\langle\lambda v,v\rangle=\langle Av,v\rangle\geq0.$$
  2. $\ker(A-\lambda I)=\{0\}$, but $A-\lambda I$ is not bounded below. In this case, there exists a sequence $v_n$ of unit vectors with $(A-\lambda I)v_n\to0$. Then $\lambda=\langle\lambda v_n,v_n\rangle$, so $$\lambda=\lim\langle Av_n,v_n\rangle\geq0.$$
  3. $A-\lambda I$ is bounded below, but not surjective. This case has nothing to do with the sign of $\lambda$; it is just the fact that the residual spectrum of a selfadjoint operator is empty. As the image of a bounded below operator is closed, we get $$\ker (A-\lambda I)=\ker(A^*-\bar\lambda I)=\text{ran}\,(A-\lambda I)^\perp\ne\{0\};$$ this is a contradiction, since a bounded-below operator is injective. So no such $\lambda $ exists.

As for your last question, yes. For a selfadjoint operator (normal, actually), the convex hull of the spectrum is equal to the closure of the numerical range: $$ {\text{conv}}\,\sigma(A)=\overline{\{\langle Av,v\rangle:\ \|v\|=1\}.} $$

  • $\begingroup$ In 3. we have indeed $ker(A^*-\bar\lambda I)=\text{ran}\,(A-\lambda I)^\perp$. But why are you saying $\text{ran}\,(A-\lambda I)^\perp\ne\{0\}$?. In a Hilbert space the orthogonal of a subspace is 0 if and only if the subspace is dense, which is no contradiction for us. You are supposing $A-\lambda I$ is bounded below thus injective, and at the end you have $Av=\lambda v$ which means $A-\lambda I $ is not injective so i think you are contradicting yourself. $\endgroup$ – Emilian Dec 1 '15 at 11:33
  • $\begingroup$ However, the set $\{\lambda\in\mathbb{C}: \ \ A-\lambda I \ \text{is not bounded below}\}$ is called the approximate spectrum of $A$, and it can be proven that both the eignenvalues and the continous spectrum are included in the approximate spectrum which is included in the spectrum. And by another result in the book above, the residual spectrum of $A$ is empty because $A$ is selfadjoint. As you have proven in 2. that that approximate spectrum is nonnegative, the proof is finished. $\endgroup$ – Emilian Dec 1 '15 at 11:39
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    $\begingroup$ When an operator is bounded below, the range is closed. But yes, as you say, case 3 shows that the residual spectrum is empty. $\endgroup$ – Martin Argerami Dec 1 '15 at 12:16
  • $\begingroup$ Yes, it is clear now. Thank you. Do you know a book/link when I can find the theorem: ${\text{conv}}\,\sigma(A)=\overline{\{\langle Av,v\rangle:\ \|v\|=1\}}$ ? $\endgroup$ – Emilian Dec 1 '15 at 17:24
  • $\begingroup$ It follows in a straightforward way from the Spectral Theorem. if $A$ is normal, do it first in finite dimension, and in the infinite case approximate with linear combinations of projections. $\endgroup$ – Martin Argerami Dec 27 '15 at 15:14

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