Closure of the image is equal to image of $u^\ast u$?

Question

Let $u \in B(H,H')$ where $H,H'$ are Hilbert spaces and let $u^\ast$ denote its adjoint.

How can I see that $\overline{u^\ast(H')} = u^\ast u (H)$?

Answer 1

If $u^*$ is a partial isometry, then your claim is correct, but in general $Im u^*u$ is not closed. To prove your claim in general we should show $\overline {Im u^*} = \overline {Im (u^*u)} $. Reduced it to show $\ker u = \ker u^*u$.

Clearly $\ker u\subset \ker u^*u$. Conversely we can decompose $H' = \overline{Im(u)} + \ker u^*$. Suppose $\xi \in \ker u^*u$ , $\eta\in \overline{Im(u)}$ , $\eta' \in \ker u^*$ then $$\langle u\xi , \eta+\eta'\rangle = \langle u\xi , \eta\rangle + \langle \xi , u^*\eta' \rangle = \lim \langle u\xi, u\eta_n\rangle = \lim \langle u^*u \xi , \eta_n\rangle =0 $$ where $\eta = \lim u\eta_n$.