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Preparing for an exam in functional analysis, I'm trying to show that for a self-adjoint operator $A$, $\sigma(A) \subset \mathbb{R}$. I came across the following proof in the book (or rather, lecture notes) we're using for the course. The proof is even stronger, giving bounds for the spectrum. However, I have issue with the proof. Here is the proof:

Let $m = \inf_{||x||=1}{\langle Ax, x \rangle}$ and $M= \sup_{||x|=1}{\langle Ax, x \rangle}$. Let $\lambda \in \mathbb{C} \backslash [m,M]$. Define $d = \mathrm{dist}(\lambda, [m,M])$. By the Cauchy Schwarz inequality we have $||(A- \lambda I)x|| \ge |\langle (A- \lambda I)x, x \rangle| =|\langle Ax, x \rangle - \lambda| \ge d$. Thus $||A-\lambda I||$ is bounded below and is hence invertible.

My problem wiht this proof is 2-fold. First, this seems to assume that (if $||x||=1$, then) $\langle Ax,x \rangle \in \mathbb{R}$. Further, I don't see how this uses the fact that $A$ is self-adjoint! If I had to guess, $A$ self-adjoint implies that $\langle Ax,x \rangle \in \mathbb{R}$ (when $||x|| = 1$), but I don't seem to be able to figure out how to make that connection.

Any help is appreciated. Thanks in advance.

Mike
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    Typographical comment: $\|x\|$ looks better than $||x||$. Compare $|x|$ and $||x||$. –  Nov 23 '14 at 23:09
  • Thanks. I have a command (\norm{}) that I got from someone else, and couldn't actually remember what the long-hand for it was offhand – Mike Nov 24 '14 at 00:23

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Self-adjointness is used to show that $\langle Ax,x\rangle$ is always real. Recall that $\langle u,v\rangle = \overline{\langle v,u\rangle} $ and use the definition of self-adjointness: $$ \langle Ax,x\rangle=\langle x,Ax\rangle = \overline{\langle Ax,x\rangle} $$

The boundedness of $A$ implies that the set $\{\langle Ax,x\rangle:\|x\|=1\}$ is bounded. So, it's a bounded subset of $\mathbb R$. Now $m$ and $M$ make sense and the proof goes as before.

By the way, there is a gap in is bounded below and is hence invertible: for example, the shift operator $(x_1,x_2,\dots)\mapsto (0,x_1,x_2,\dots)$ is bounded from below but is not invertible. This is another place where self-adjointness will be invoked: re-read the proof to see what goes on there.

If you don't need bounds for the spectrum, you could simplify the proof by dropping $M$ and $m$. Since the modulus of a complex number is at least the modulus of its imaginary part, $$ \|(A- \lambda I)x\| \ge |\langle (A- \lambda I)x, x \rangle| =|\langle Ax, x \rangle - \lambda| \ge |\operatorname{Im} \lambda| $$

  • Self adjointness is also used to conclude from the "bounded below" property (which is equivalent to $A-\lambda$ being injective with closed range) to $A-\lambda$ being invertible (which adds surjectivity to the preceding properties), isn't it? – PhoemueX Nov 23 '14 at 23:12
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    Yes, but $\Vert Ax \Vert \geq \Vert x \Vert$ does not imply in general that $A$ is invertible, take $A x = (x_1,0,x_2,0,x_3,\dots)$ on $\ell^2$. – PhoemueX Nov 23 '14 at 23:16
  • Thanks; I edited the answer. –  Nov 23 '14 at 23:19
  • Hmm... These lecture notes do have something about for A self-adjoint, invertible if and only if bounded below... He only has a brief proof, though. Guess I'll have to try and dissect it – Mike Nov 24 '14 at 00:22
  • @Mike This comment might help. –  Nov 24 '14 at 00:24
  • It didn't really, but I think I understand it now anyway – Mike Nov 24 '14 at 02:38
  • Not to de-rail my own question, but I actually have a slightly new question: For your alternative (when I don't need $m$ and $M$), does that work generally if I don't assume $|x| =1$? I have another problem to prove that fact for general $x$, and that's a pretty easy proof if so. But I'm concerned because I think a $|x|^2$ would show up, wouldn't it? – Mike Nov 24 '14 at 02:46
  • @Mike I am still assuming $|x|=1$ there. When proving an upper or lower bound on a linear operator, it's pretty common to make this assumption. Otherwise the argument is the same, but with a few factors of $|x|$ thrown in. –  Nov 24 '14 at 02:48
  • Well, my thinking is where you have $|\langle Ax, x \rangle - \lambda|$, if $|x| \neq 1$ then we would get $|\langle Ax, x \rangle - \lambda|x|^2|$, right? How do I pull out a $|x|$? – Mike Nov 24 '14 at 02:54
  • @Mike It would be $$|(A- \lambda I)x| \ge |\langle (A- \lambda I)x, x \rangle|/|x| =|\langle Ax, x \rangle - \lambda |x|^2|/|x| \ge |\operatorname{Im} \lambda||x|^2/|x| $$ –  Nov 24 '14 at 02:55
  • @Mike When working with linear operators, it is usual to use their homogeneity. Every nonzero vector $x$ can be written as $x=ty$ with $|y|=1$, just by letting $t=|x|$ and $y=x/t$. With this substitution, the inequality $|(A-\lambda I)x| \ge |\mathrm{Im} \ \lambda| |x|$ that you want becomes $|(A-\lambda I)y| \ge |\mathrm{Im} \ \lambda|$. –  Nov 24 '14 at 02:57
  • Wait, I see it now. For some reason I couldn't see how it was working in the longer version before, but now I see it. I knew you generally assume $|x| = 1$, but for some reason I was having problems with it before because I couldn't see how something worked. But now I do. Thanks for all the help. – Mike Nov 24 '14 at 03:02