Spectrum of adjoint bounded linear operator on hilbert space I have been struggling to analyse the spectrum of the adjoint of a bounded linear operator on a hilbert space.
Throughout the internet I have found vague references that
$\sigma(T^*) = \sigma(T)$ but I fail to be able to prove that.
What I have succeeded in doing is to prove that $\rho(T^*) = \overline{\rho(T)}$ (where $\rho$ is the resolvent $\rho(T) = \mathbb{C} \backslash \sigma(T)$).
Thus imho $\sigma(T^*)$ should be equal to $\overline{\sigma(T)}$ and not the above... 
 A: You're correct that $\sigma(T^*) = \overline{\sigma(T)}$, where the overline denotes complex conjugation (and not closure as a subset of $\mathbb{C}$ - recall that the resolvent set of a bounded operator is open, and hence the spectrum is closed). In order to conclude that $\sigma(T^*) = \sigma(T)$, you need to impose the stronger condition that the eigenvalues be real (by specifying that $T$ be self-adjoint, e.g.).
A: The critical point is in complex Hilbert space you use a sesquilinear inner product to make adjoints but in Banach space you use a bilinear form. So in the Banach situation the conjugates disappear.
To reduce confusion I'll replace the two senses of adjoint with $T^T$ for Banach adjoint and $T^H$ for Hilbert adjoint. We borrow the rest of our notation from Pedersen: 

Adjoint $T^{H}$ for $T$ in (complex) Hilbert space $\mathfrak{H}$ is constructed through a sesqui-linear inner product $(\cdot|\cdot):\mathfrak{H}\times \mathfrak{H}^\ast→\mathbb{C}$ as follows:
  $$T^H:\mathfrak{H}^\ast\to \mathfrak{H}^\ast; \qquad (\cdot|T^H\phi)\triangleq (T \cdot |\phi) \in \mathfrak{H}^\ast $$
On the other hand, in a Banach space $\mathfrak{X}$, adjoint $T^T$ of $T$ is constructed using a bi-linear composition-with-dual operator $\langle \cdot,\cdot\rangle :\mathfrak{X}\times \mathfrak{X}^\ast\to \mathbb{F}$ in an identical looking way:
$$T^T:\mathfrak{X}^\ast\to \mathfrak{X}^\ast; \qquad \langle \cdot, T^T\phi\rangle\triangleq\langle T\cdot ,\phi\rangle \in \mathfrak{X}^\ast$$
Spectrum is:
  $$\sigma(T)=\left\{ \lambda : \lambda I - T \not\in \mathbf{GL} \right\}.$$
$\mathbf{GL}$ is the general linear group, 'operators with bounded inverse.'

With this notation, it is true that $\sigma(T)=\sigma(T^T)=\overline{\sigma(T^H)}$. 
We see $\sigma(T)=\overline{\sigma(T^H)}$ by computing:
\begin{align}
    \sigma(T^H)&=\{\lambda: \lambda I^H - T^H \not\in \mathbf{GL}(\mathfrak{H^\ast})\} \\
&= \{\overline\lambda: \phi\mapsto\phi((\lambda I - T)(\cdot)) \not\in\mathbf{GL}(\mathfrak{H}^\ast)\}\\
&= \{\overline\lambda: (\lambda I - T) \not\in\mathbf{GL}(\mathfrak{H})\} \tag{A} \\
&= \overline{\sigma(T)}
\end{align}
where $(\text{A})$ follows because, letting $S^H=\phi \mapsto \phi((\lambda I-T)(\cdot))$, if $(\lambda I -T)\in\mathbf{GL}(\mathfrak{H})$, then by computation $\phi\mapsto \phi((\lambda I-T)^{-1}(\cdot))\in \mathbf{GL}(\mathfrak{H}^\ast)$ is an inverse of $(S^H)^{-1}$ so $S^H \in\mathbf{GL}(\mathfrak{H}^\ast)$. On the other hand $S^H\in\mathbf{GL}(\mathfrak{H}^\ast)\Rightarrow (S^H)^{-1}\in \mathbf{GL}(\mathfrak{H}^\ast)$ and by computation, $((S^H)^{-1})^H\in \mathbf{B}(\mathfrak{H})$ must behave as an inverse for $\lambda I - T.$ (We have just shown the complements of the two sets are subsets of one another)
The Banach part follows by the same argument, replacing $(\cdot|\cdot)$ with $\langle \cdot,\cdot\rangle$, $\mathfrak{H}$ with $\mathfrak{X}$, $\cdot^H$ with $\cdot ^T$, and getting rid of conjugation.
This is part of E 4.1.11 in Analysis Now by Pedersen.
