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