Spectrum of self-adjoint operator on Hilbert space real 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!!!
 A: 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.
A: The representation of the elements of $H^*$ is not needed. 
Let $\lambda$ be in the point spectrum of $A$. Then there is $0\neq x\in H=(H,(\cdot,\cdot))$ such that $Ax=\lambda x$ and by the self-adjointness of $A$:
$\lambda(x,x)=(\lambda x,x)=(Ax,x)=(x,Ax)=(x,\lambda x)=\overline{\lambda}(x,x)$.
Hence $\lambda=\overline{\lambda}$, thus $\lambda\in\mathbb{R}$.
See e.g. http://en.wikipedia.org/wiki/Spectral_theorem for related results in greater generality.
A: If $\Im\lambda \ne 0$, and $x \in X$, then
$$
        \Im\lambda \|x\|^{2} = -\Im((A-\lambda I)x,x),\\
          |\Im\lambda|\|x\|^{2} \le |((A-\lambda I)x,x)|\le \|(A-\lambda I)x\|\|x\|,\\
                |\Im\lambda|\|x\| \le \|(A-\lambda I)x\|.
$$
So $A-\lambda I$ is injective for all $\lambda\notin\mathbb{R}$. The above inequality can be used to show that the range $\mathcal{R}(A-\lambda I)$ is closed for $\lambda\notin\mathbb{R}$. So $A-\lambda I$ is surjective for $\lambda\notin\mathbb{R}$ because
$$
\begin{align}
      \mathcal{R}(A-\lambda I)& =\overline{\mathcal{R}(A-\lambda I)} \\
         & =\mathcal{N}(A^{\star}-\overline{\lambda}I)^{\perp} \\
         & = \mathcal{N}(A-\overline{\lambda}I)^{\perp}=\{0\}^{\perp}=H.
\end{align}
$$
Therefore, $A-\lambda I$ is injective and surjective for $\lambda\notin\mathbb{R}$, which leaves $\sigma(A)\subseteq\mathbb{R}$.
