$ker(T)^{\bot} = \overline{im(T^*)}$ if $T$ is a linear operator between Hilbert spaces

Question

Let $T$ be a linear operator. For any underlying normed spaces it holds that $$ker(T)^{\bot} \subset \overline{im(T^*)},$$ but if they are both Hilbert spaces we get $$ker(T)^{\bot} = \overline{im(T^*)}.$$ Now my question is: How to prove the inclusion from right to left?

Answer 1

For a Hilbert space, and any vectors $x,y$, $$\langle T^*x,y\rangle = \langle x,Ty\rangle$$ Hence $x\in(im T)^\perp\iff x\in\ker T^*$, i.e., $(im T)^\perp=\ker T^*$. Taking a second perp gives $$(\ker T^*)^\perp=(im T)^{\perp\perp}=\overline{im T}$$ Apply this identity with $T^*$ instead of $T$ to get $$(\ker T)^\perp=\overline{im T^*}$$

Edit: This works in Hilbert spaces but not Banach spaces because $T^{**}=T$ is valid in the former but not the latter in general.

Edit 2: Direct proof. Let $x\in\overline{im(T^*)}$ and $y\in\ker T$, then there are vectors $z_n$ such that $T^*z_n\to x$, and $Ty=0$. So $$\langle y,x\rangle=\lim_{n\to\infty}\langle y,T^*z_n\rangle=\lim_{n\to\infty}\langle Ty,z_n\rangle=0$$