some inclusions regarding linear operators Let $H$ be a Hilbert Space and $T:H\rightarrow H$ a linear operator.
Let $T^*$ be the adjoint operator of $T$ and let $\operatorname{Cl}(X)$ be the topological closure of the set X and $X^{\perp}$ denote the orthogonal complement of the subset $X$.
I'd like to see a proof of the following inclusions :
$ \operatorname{Cl}(\operatorname{im}(T)) \subset \operatorname{Cl}(\operatorname{im}(TT^{*}))$ and $\ker((T^{*}))^{\perp} \subset \operatorname{im}(T) $ ?
I guess it's pretty simple, but somehow I got stuck.
Thank you :)
 A: $\def\im{\mbox{im}\,}$ First, we show that $\ker T=(\im T)^\perp$. Indeed,
$$
Tx=0 \iff \langle Tx,y\rangle=0\,\forall y\ \iff \langle x,T^*y\rangle=0\,\forall y\ \iff x\in(\im T)^\perp
$$
Taking orthogonals we get $(\ker T)^\perp=\overline{\im T^*}$. So we get an orthogonal decomposition $H=\ker T \oplus \overline{\im T^*}$. 
So, if $x\in H$, then $x=y+z$, with $y\in\ker T$ and $z\in\overline{\im T^*}$, i.e. $z=\lim T^*x_n$ for some sequence $x_n$. Then
$$
Tx=T(y+z)=Tz=T(\lim T^*x_n)=\lim TT^*x_n\in\overline{\im TT^*}.
$$
This shows that $\im T\subset\overline{\im TT^*}$, and so $\overline{\im T}=\overline{\im TT^*}$. 
As for your last inclusion, we have the equality $(\ker T^*)^\perp=\overline{\im T}$. But the inclusion as you wrote is not true. It is enough to take any operator with non-closed image, and the inclusion will fail, as the left hand side is $\overline{\im T}$.
To see an easy example of such operator, fix an orthonormal basis $\{e_j\}$ on a separable Hilbert space, and define
$$
Te_j=\frac1j\,e_j.
$$
Then $\im T$ is dense, because it contains every element in the basis ($e_j=T(je_j)$). But it is not all of $H$, as one can check that the vector $\sum_j\frac1j\,e_j$ is not in $\im T$.
