Let $\mathcal H$ be a Hilbert space and $T \in \mathcal L (\mathcal H).$ Suppose that $\text {ran} (T)$ is closed. Does it always mean that $\text {ran} (T^*)$ is also closed?
I am trying to prove that by sequential convergence but I couldn't quite do that. Can anybody please help me in this regard? Thanks for your time.
Yes, it is. Let $H_1=\text{Ker}(T)^\perp$ and $H_2=\text{Ran}(T)$. Also let $\iota _1$ and $\iota _2$ be the inclusions of $H_1$ and $H_2$ into $H$, respectively, and observe that $\iota _1^*$ and $\iota _2^*$ are the corresponding projections.
If $T$ has closed range then $\iota _2^*T\iota _1$ is an invertible operator from $H_1$ to $H_2$, whence its adjoint, namely $\iota _1^*T^*\iota _2$ is an invertible operator from $H_2$ to $H_1$.
Observing that the range of $T^*$ is contained in $H_1$, we deduce that it is precisely $H_1$, hence closed.