Elements of spectral theory - functional analysis I define: The operator $R( \lambda ; T)=(\lambda I-T)^{-1}$.
According to the chapter on elements of spectral theory, I must demonstrate the following:
$$\rho(T)=\rho(T^*)\quad\text{and}\quad R(\lambda;T^*)= R(\lambda;T)^*$$
Help please.
 A: I think it depends on what you're allowed to use to prove the claim.  First of all, the spectral radius $\rho(T)$ is
$$
\rho(T) \;\; =\;\; \sup_{\lambda \in \sigma(T)} |\lambda|.
$$
You can immediately see this if you know that $\sigma(T^*) = \overline{\sigma(T)}$, in other words the spectra of $T$ and $T^*$ are conjugate to one another. 
As for your other claim, I believe you're stating the result incorrectly.  The result should really be
$$
R\left (\lambda, T^*\right ) \;\; =\;\; R\left (\overline{\lambda}, T\right )^*.
$$
If you let $\varphi \in Dom(T)$ and $\psi \in Dom (T^*)$ you can see that 
\begin{eqnarray*}
\langle R(\lambda, T^*)\varphi, \psi\rangle & = & \langle \varphi, (\lambda I - T^*)\psi\rangle \\
& = & \left \langle \varphi, \left (\overline{\lambda}I - T\right )^*\psi \right \rangle \\
& = & \left \langle \left (\overline{\lambda}I - T\right )\varphi, \psi \right \rangle \\
& = & \left \langle \varphi, R\left (\overline{\lambda}, T\right )\psi \right \rangle
& = & \left \langle  R\left (\overline{\lambda}, T\right )^*\varphi, \psi \right \rangle.
\end{eqnarray*}
