TITLE: Spectrum of self-adjoint operator on Hilbert space real
QUESTION [5 upvotes]: My book says that a self-adjoint bounded linear operator $A:H\to H$ on a complex Hilbert (not sure if separability is needed) space has a real spectrum.
I guess that the key is in the fact that any $f\in H^{\ast}$ can be represented by a functional of the form $\langle-,x_0\rangle$ for some $x_0\in H$, but I am not able to use that fact. I also see that $\forall x,y\in H\quad\langle(A-\lambda I)x,y\rangle=\langle x,(A-\bar{\lambda} I)y\rangle$ but I'm not sure that can be useful to prove the statement...
Thank you very much for any help!!!

REPLY [3 votes]: The fact that the spectrum of $A$ is real when $A = A^*$ is a general fact about commutative $C^*$ algebras; the discussion of $A$ as an operator on Hilbert space is not needed.  The main point is that in such a $C^*$ algebra, we have $||xx^*|| = ||x||^2$ for all $x$, and the spectral radius $\rho(x) = ||x||$ for all $x$.  Thus if $Y$ is a large real number (large multiple of the identity), then $$\rho(A+iY)^2 = ||A+iY||^2 = ||A^2+Y^2|| = Y^2 + O(1).$$  Now if $x+iy$ is in the spectrum of $A$, say with $y>0$, then $\rho(A+iY)^2 > |x+iy+iY|^2 > (y+Y)^2$ and hence $y=0$.  Thus the spectrum of $A$ is real.