Consider and Hilbert Space $X$, $T\in B(X)$ and a scalar $\mu$ s.t. $|\mu|=||T||$
By a simple argument I deduced that $\ker(\mu I- T)=\ker(\bar\mu I-T^*$) where $*$ denotes the adjoint.
I am then asked to deduce that $ \text{Im} (\mu I - T) + \ker (\mu I - T)$ is dense in $X$.
It does not seem to me that this follows, I think we need, in addition, the image to be closed.
If anybody could indicate the reason why this holds true I would be extremely grateful!
Generally, we have for any subspaces $A, B$ of a topological vector space $E$ the relation
$$\overline{A} + \overline{B} \subset \overline{A+B}.\tag{1}$$
Further, for any $S \in B(X)$, where $X$ is a Hilbert space, we have
$$\ker S^\ast = (\operatorname{im} S)^\perp.\tag{2}$$
Apply $(1)$ and $(2)$ to the right operator and subspaces to reach the conclusion.