If $E=Kernel T$ then $E^\perp=\overline{Im(T^*)}$. Let $H$ be a Hilbert space, and $T:H\to H$ be bounded, so in particular $E:=\text{Kernel}(T)$ is a closed subspace of $T$.  Show that $E^\perp$ equals the closure of the image of $T^*$.
Here is what I have so far.
If $v\in \overline{Im(T^*)}$ then either $v\in Im(T^*)$ or $v$ is a limit point of the set of elements in $Im(T^*).$
Supposing that $v\in Im(T^*)$ then there exists a $w\in H$ such that $T^*w=v$.
I want to show that $v\in E^\perp$.
So let $x\in E$ then by definition, $Tx=0$.
Thus, $\langle v,x\rangle=\langle T^*w,x\rangle=\langle w,Tx\rangle=\langle,w,0\rangle=0.$
Otherwise if $v$ is a limit point of $Im(T^*)$ then there exist a sequence $(v_n)_{n=1}^\infty$ such that there exists $w_n$ such that $T^*w_n=v_n$ and $\lim\limits_{v_n\to\infty}v_n=v$ and $\lim\limits_{n\to\infty}w_n=w$ for some $w\in H$.
Thus $\langle v,x\rangle=\langle\lim\limits_{n\to\infty} v_n,x\rangle=\langle\lim\limits_{n\to\infty}T^*w_n,x\rangle=\langle T^*\lim\limits_{n\to\infty}w_n,x\rangle=\langle\lim\limits_{n\to\infty}w_n,Tx\rangle=\langle w,0\rangle=0.$
Hence $\overline{Im(T^*)}\subseteq E^\perp.$
I'm a little unsure about this part as I'm not quite convinced about $(w_n)$ having to converge.
My though is that since $T$ is bounded it is continuous hence $T^*$ is bounded and continuous so we may pass the limit through $T^*$. 
ETA. Instead of taking the limit immediately as above,
Let $n$ be fixed, then
$\langle v_n,x\rangle=\langle T^*w_n,x\rangle=\langle w_n,Tx\rangle=\langle w_n,0\rangle=0$. But since this is true for arbitrary $n.$ we have that
$=0\langle v_n,x\rangle\to\langle v,x\rangle=0$ as $n\to\infty.$
What I'm really stuck on though is showing $E^\perp\subseteq\overline{Im(T^*)}$.
If $v\in E^\perp$ then I know that for all $x\in E$
$\langle v,x\rangle=0$
I thought perhaps I might show that  $Im(T^*)^\perp\subset E$ then we should have that $E^\perp\subset Im(T^*)^{\perp\perp}=\overline{Im(T^*)}$.
Does this seems reasonable?
How can I prove that $Im(T^*)^{\perp\perp}=\overline{Im(T^*)}$?
 A: You've basically solved it. The detailed steps are as follows:
1) $Im(T^{*}) \subseteq E^{\perp}$. You've shown this.
2) $Im(T^{*})^{\perp} \subseteq E$. To prove this, let $x \in (\text{Im}(T^{*}))^{\perp}$. Then for every $y \in H$, $\langle x, T^{*}y \rangle = 0$. So for every $y \in H$, $\langle Tx, y \rangle = 0$. This implies $Tx = 0$. 
3) $E^{\perp} \subseteq (\text{Im}(T^{*})^{\perp})^{\perp}$, follows from 2), and the fact that $A \subseteq B$ implies $B^{\perp} \subseteq A^{\perp}$.
4) For any subspace $V$, $(V^{\perp})^{\perp} = \overline{V}$. This is slightly tricky, so we break it into two steps:
4a) First assume $V$ is closed. Then $H = V \oplus V^{\perp}$. Clearly $V \subseteq (V^{\perp})^{\perp}$. To show the other inclusion, let $v = x + y \in (V^{\perp})^{\perp}$, with $x \in V$ and $y \in V^{\perp}$. Then $0 = \langle v,y \rangle = \langle x + y, y \rangle = \langle x, y \rangle + \langle y, y \rangle = \langle y, y \rangle$. So $y = 0$.
4b)  For general $V$, we have $V \subseteq (V^{\perp})^{\perp}$, and $(V^{\perp})^{\perp}$ is closed since it is the zero set of a collection of continuous functions, so $\overline{V} \subseteq (V^{\perp})^{\perp}$. For the other direction use 4a): $V \subseteq \overline{V}$, so $\overline{V}^{\perp} \subseteq V^{\perp}$, so $(V^{\perp})^{\perp} \subseteq (\overline{V}^{\perp})^{\perp} = \overline{V}$.
5) Now you have $Im(T^{*}) \subseteq E^{\perp} \subseteq \overline{Im(T^{*})}$, and since $E^{\perp}$ is closed we're done.
