If $T$ is a closed, densely defined linear transformation on a Hilbert space $\mathcal{H}$, then $T$ has a polar decomposition $T=V(T^{*}T)^{\frac{1}{2}}$ where $V$ is the partial isometry defined by extending the map $(T^{*}T)^{\frac{1}{2}}x\mapsto Tx$ for $x\in D(T^{*}T)$ (where $D(T^{*}T)$ is the domain of $T^{*}T$ and a core for both $T$ and $(T^{*}T)^{\frac{1}{2}}$). Since $V$ is bounded it follows that $T^{*}=(T^{*}T)^{\frac{1}{2}}V^{*}$, and hence $TT^{*}=VT^{*}TV^{*}$.
What I do not understand is the following claim: $T^{*}T$ restricted to the closure of the range of $T^{*}$ is unitarily equivalent to $TT^{*}$ restricted to the closure of the range of $T$ and $V$ implements this equivalence.
Specifically, I do not see how it can possibly be true that if $Tx_{n}\rightarrow y$, then $y\in D(TT^{*})$ (or even that $y\in D(T^{*})$). Since $TT^{*}$ is closed, this amounts to showing that $TT^{*}Tx_{n}$ has a limit in $\mathcal{H}$. Since it is not assumed that $y$ is in the domain of any of these operators, it doesn't seem to me that this can possibly be shown.
Edit: I don't even think it's true that $Tx\in D(T^{*})$ for every $x\in D(T)$. This goes against everything I know about unbounded operators. If this were always true, then there would be no point in stipulating that $x\in D(T^{*}T)$ when you could just write $x\in D(T)$.