# Spectrum of a Self-Adjoint Operator is Real

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.

• Typographical comment: $\|x\|$ looks better than $||x||$. Compare $\|x\|$ and $||x||$.
– user147263
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 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, $$\|(A- \lambda I)x\| \ge |\langle (A- \lambda I)x, x \rangle| =|\langle Ax, x \rangle - \lambda| \ge |\operatorname{Im} \lambda|$$

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