Orthogonal complement of the kernel of $u\in B(H, H')$ Let $H,H'$ be Hilbert spaces and $u \in B(H,H')$. Let $u^\ast$ denote the adjoint. 
I know (and can show) that $(\mathrm{im} u)^\bot = \ker u^\ast$.
From this I would deduce that $(\ker u^\ast)^\bot = \mathrm{im} u$. But instead,
$$ (\ker u^\ast)^\bot = \overline{\mathrm{im} u}$$
and it is not intuitive for me. 

How to (geometrically, if possible) see that $ (\ker u^\ast)^\bot =
 \overline{\mathrm{im} u}$ and not $ (\ker u^\ast)^\bot = \mathrm{im}
 u$?


If $X\subseteq H$, then $X^\perp$ is closed.
Let $\{y_n\}$ be any convergent sequence in $X^\perp$ with $y_n\to y$. Then for any $x\in X$,
$$\langle x,y\rangle=\langle x,y-y_n\rangle+\langle x,y_n\rangle=\langle x,y-y_n\rangle.$$
Applying Cauchy-Schwarz, this gives
$$|\langle x,y\rangle|\leq \|x\|\|y-y_n\|$$
which tends to zero since $y_n\to y$. Therefore $\langle x,y\rangle=0$ for all $x\in X$, so $y\in X^\perp$.
Edit: I must have misread the original question, as I thought you were only asking why $(\ker u^*)^\perp$ must be closed. At this point we know that $\overline{\text{ran } u}\subseteq (\ker u^*)^\perp$. To see the opposite inclusion, let $x\in (\text{ran } u)^\perp$. For any $y\in H$, we then have
$$0=\langle uy,x\rangle=\langle y,u^*x\rangle,$$
so $u^*x=0$ and therefore $(\text{ran } u)^\perp\subseteq \ker u^*$. Taking orthogonal complements gives
$$(\ker u^*)^\perp\subseteq (\text{ran } u)^{\perp\perp}=\overline{\text{ran } u}.$$
Thus $\overline{\text{ran } u}= (\ker u^*)^\perp$ as desired.
