$T$ is self adjoint iff $\sigma(T)\subset R$ Let $H$ be a complex Hilbert space and  $T\in B(H)$ a normal operator.
Prove that:
A.$T$ is self adjoint ($T^*=T$) iff $\sigma(T) \subset R$.
B.$TT^*=T^*T=I$ iff $\sigma(T) \subset T=${$z\in C$ : |z|=1}.
For a.
=> Let T be self adjoint i.e. $T^*=T$.
We know that the spectrum of T is
$\sigma(T)$={$\lambda$ : $\lambda I-T$ is not invertible} so
$\lambda I-T$ is not invertible iff $\overline{\lambda} I-T^*$ is not invertible iff $\overline{\lambda} I-T$ is not invertible and that happen iff $\lambda = \overline{\lambda}$.
Therefore $\sigma(T)\subset R$.
<= similarly, assume that the spectrum $\sigma(T) \subset R$ so
If $\lambda \in \sigma(T)$ then
$\lambda I-T$ is not invertible iff $\overline{\lambda} I-T^*$ is not invertible iff $\lambda I-T^*$ is not invertible and that heppen iff $T=T^*$.
However I don't see where the assume that T is normal is used in my proof.
*How the solution would be changed using the spectral theorem instead?
 A: You say

"$\lambda I - T$ not invertible if and only if $\overline\lambda I-T$ not invertible" implies $\lambda=\overline\lambda$.

No, it doesn't. The unilateral shift is not even normal, and it doesn satisfy your double implication, as any operator whose spectrum is symmetric along the $x$-axis will be.
Similarly, you say

"$\lambda I - T$ not invertible if and only if $\lambda I-T^*$ not invertible" implies $T=T^*$.

No, it doesn't. As far as I can tell, neither of these implications can be proven directly just using the definitions. The alternative to the arguments I provide below is to use functional calculus.
For the first part, you would usually use that a selfadjoint operator has no residual spectrum, and that $\langle Tx,x\rangle\in\mathbb R$ for all $x$. Then every $\lambda\in\sigma(T)$ is in the approximate point spectrum. So there exists a sequence $\{x_n\}$ with $\|x_n\|=1$ for all $n$ and $\lambda x_n-Tx_n\to0$. Then
$$
\lambda=\lim_n\langle \lambda x_n,x_n\rangle=\lim_n\langle Tx_n,x_n\rangle\in\mathbb R.
$$
So $\sigma(T)\subset\mathbb R$. For the converse, one has to use that because $T$ is normal, the closure of its numerical range is the closed convex hull of the spectrum. So the numerical range of $T$ is real. Then, for any $x$,
$$
\langle Tx,x\rangle=\langle x,Tx\rangle=\langle T^*x,x\rangle. 
$$
So $\langle (T-T^*)x,x\rangle=0$ for all $x$, which by polarization implies that $T-T^*=0$.
For part b, if $T$ is a unitary  note that $\|T\|=\|T^*T\|^{1/2}=1$, so $\sigma(T)\subset\overline{\mathbb D}$. Also
$$
\sigma(T)=\{\lambda^{-1}:\ \lambda\in\sigma(T^{-1})\}. 
$$
But $T^{-1}=T^*$ is a unitary, so $\sigma(T^{-1})$ is in the unit disk. Thus
$$
\sigma(T)\subset\{\lambda:\ |\lambda|\leq1\ \text{ and } |\lambda|\geq1\}=\{\lambda:\ |\lambda|=1\}. 
$$
For the converse, if $T$ is normal and $\sigma(T)\subset\mathbb T$, there might be a trick, but the most natural way is to use functional calculus. So the Gelfand transform $\Gamma$ maps $T$ to the function $f\in C(\mathbb T)$ with $f(t)=t$. Then
$$
T^*T=\Gamma^{-1}(f^*f)=\Gamma^{-1}(1)=I. 
$$
