Normal bounded operator in Hilbert space, whose spectrum is real, is self-adjoint Let $T$ be a bounded normal operator in Hilbert space such that the spectrum $\sigma(T)$ is contained in the real axis. By the Gelfand-Naimark theorem for commutative $C^*$-algebras the $C^*$-algebra generated by $T$ and $I$ is isometrically isomorphic to $C(\sigma(T)),$ the algebra of complex valued  continuous functions on $\sigma(T)\subset \mathbb{R}.$ The operator $T$ corresponds to multiplication by $x$ in $C(\sigma(T)),$ therefore $T$ is self-adjoint.
I would like to prove that fact  in a straightforward way, but I could not come up with any idea.
When $T$ is a compact operator the proof is relatively easy. Assume by contradiction that $T^*-T\neq 0.$ Then one of the numbers $\lambda:=\pm{1\over 2}\|T^*-T\|\neq 0$ is the eigenvalue of the self-adjoint operator $B:={i\over 2}(T^*-T).$ Let $V_\lambda$ denote the eigenspace of the operator $B$ corresponding to $\lambda.$ As $T$ and $T^*$ commute with $B,$ the subspace $V_\lambda$ is invariant for $A={1\over 2}(T+T^*).$ The operator $T=A+iB$ restricted to $V_\lambda$ is of the form $A+i\lambda I.$ Therefore $\sigma(T)\subsetneq \mathbb{R},$ which gives a contradiction.
 A: I know three ways, none really elementary.

*

*By the Spectral Theorem. You have
$$
T=\int_{\sigma(T)}\lambda\,dE(\lambda).
$$
Then, using the $\sigma(T)\subset\mathbb R$,
$$
T^*=\int_{\sigma(T)}\overline\lambda\,dE(\lambda)=\int_{\sigma(T)}\lambda\,dE(\lambda).
$$


*If you know that for normal $T$ you have $$\tag1\overline{W(T)}=\overline{\operatorname{conv}}\sigma(T),$$ where $W(T)$ is the numerical range, you get easily that $T=T^*$. The problem is that the only proof of $(1)$ that I know uses the Spectral Theorem, so it is easier to do the argument directly as in the previous case.


*If you have the Spectral Mapping Theorem (but again the only proof I know depends on the Spectral Theorem), you get that $T-T^*$ has real spectrum. So the selfadjoint operator $i(T-T^*)$ has imaginary spectrum, which implies that it is zero.
A: Let $T$ be a normal operator such that $\sigma(T)\subset \mathbb{R}.$ In particular the operator $iI-T$ is invertible. Consider $$U=(iI-T)(iI+T)^{-1}$$ Then $$\sigma(U)\subset \{z\in \mathbb{C}\,:\,|z|=1\}\setminus\{-1\}$$
Indeed, let $\lambda\notin  \{z\in \mathbb{C}\,:\,|z|=1\}\setminus\{-1\}.$ Then $$U-\lambda I =[(iI-T)-\lambda(iI+T)](iI+T)^{-1}\\
= [i(1-\lambda)-(1+\lambda)T](iI+T)^{-1}=(1+\lambda)\left [ i{1-\lambda\over 1+\lambda}- T\right  ](iI+T)^{-1}
$$
It can be verified easily that $i(1-\lambda)(1+\lambda)^{-1}$ is not real, hence the the operator $U-\lambda I$ is invertible.
Indeed assume  that $i(1-\lambda)(1+\lambda)^{-1}=t.$ Thus
$\lambda =(i-t)(i+t)^{-1}.$ If $t$ is a real number then $|\lambda|=1.$
The inverse operator also satisfies $\sigma(U^{-1})\subset \{z\in \mathbb{C}\,:\,|z|=1\}.$
The operator $U$ is normal, therefore its norm is equal to the spectral radius, i.e. $\|U\|= 1.$ Similarly $\|U^{-1}\|=1.$
Thus for any $x\in \mathcal{H}$ we have
$$\|x\|=\|U^{-1}Ux\|\le \|Ux\|\le \|x\|.$$ Therefore $\|Ux\|=\|x\|,$ which implies that $U$ is a unitary operator. We have
$$T=i(I-U)(I+U)^{-1}.$$ The formula is well defined as $-1\notin\sigma(U).$
The formula implies  that $T$ is selfadjoint as
$$T^*=-i(I+U^{-1})^{-1}(I-U^{-1}) =-i(I+U)^{-1}(U-I)=i(I-U)(I+U)^{-1}=T$$
A: In deference to the contentious nature of the sentence "to prove a fact in a
straightforward way", let me begin by listing the tools we will use:

*

*The spectral radius $r(T)$ of a normal operator $T$ coincides with its norm.


*For every operator $T$, one has that $\sigma (T+\lambda )=\sigma (T)+\lambda $.  Notice that this may be considered as a baby
version of the Spectral Mapping Theorem.


*An operator $T$ on a complex Hilbert space is self-adjoint if and only if
$
  \langle T(x),x\rangle \in {\mathbb  R},
  $
for every $x\in  H$.
If you consider these as "straigtforward facts", please keep reading!
In view of (3) above, let us  fix  $x\in H$, which we assume WLOG to be a
unit vector, and we write
$$
  \langle T(x),x\rangle =a+ib,
  $$
with $a$ and $b$ real.  We then have for every $c\in {\mathbb  R}$ that
$$
  a^2+b^2+2bc+c^2 =
  |a+ib+ic|^2= $$$$ =
  \big|\big \langle (T+ic)x,x\big \rangle \big|^2  \leq \|T+ic\|^2= r(T+ic)^2 = $$$$ =
  \sup \{|\lambda |^2: \lambda \in \sigma (T+ic)\} = \sup \{|\lambda +ic |^2: \lambda \in \sigma (T)\} = $$$$ =
  \sup \{\lambda ^2 +c^2: \lambda \in \sigma (T)\} =   c^2+\sup \{\lambda ^2: \lambda \in \sigma (T)\}.
  $$
Cancelling out the term $c^2$, we get
$$
  a^2+b^2+2bc \leq    \sup \{\lambda ^2: \lambda \in \sigma (T)\}.
  $$
Now,  since this holds for every $c\in {\mathbb  R}$, one necessarily has that $b=0$, as required.
