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.

  • 1
    $\begingroup$ Typographical comment: $\|x\|$ looks better than $||x||$. Compare $\|x\|$ and $||x||$. $\endgroup$ – user147263 Nov 23 '14 at 23:09
  • $\begingroup$ 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 $\endgroup$ – Mike Nov 24 '14 at 0:23

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, $$ \|(A- \lambda I)x\| \ge |\langle (A- \lambda I)x, x \rangle| =|\langle Ax, x \rangle - \lambda| \ge |\operatorname{Im} \lambda| $$

  • $\begingroup$ 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? $\endgroup$ – PhoemueX Nov 23 '14 at 23:12
  • 1
    $\begingroup$ 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 $\ell^2$. $\endgroup$ – PhoemueX Nov 23 '14 at 23:16
  • $\begingroup$ Thanks; I edited the answer. $\endgroup$ – user147263 Nov 23 '14 at 23:19
  • $\begingroup$ 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 $\endgroup$ – Mike Nov 24 '14 at 0:22
  • $\begingroup$ @Mike This comment might help. $\endgroup$ – user147263 Nov 24 '14 at 0:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.