Polar decomposition of Bounded Normal Operator on Hilbert Space

Let $T$ be normal. By the spectral theorem, there is a unitary map $V : H \to L^2(\mu)$ for a suitable measure $\mu$ on a measure space $X$ such that we have $$ T = V^\ast M_f V $$ for some bounded function $f : X \to \Bbb{C}$, where $M_f : L^2(\mu) \to L^2 (\mu), g \mapsto f\cdot g$ is the multiplication operator which multiplies by $f$.

By the usual properties of the spectral calculus, we have $\varphi(T) = V^\ast M_{\varphi \circ f} V$ for every bounded measurable $\varphi : \Bbb{C} \to \Bbb{C}$. In particular, we have $|T|= V^\ast M_{|f|} V$.

Now, it is easy to see that there is a measurable $g : X \to \Bbb{C}$ with $|g(x)| = 1$ for all $x \in X$ and $|f(x)| \cdot g(x) = f(x)$. In fact, the function $$ g : X \to S^1 \subset \Bbb{C}, x \mapsto \begin{cases} \frac{f(x)}{|f(x)|}, & f(x) \neq 0,\\ 1, & f(x) = 0\end{cases} $$ does the job.

Since $g$ has modulus one, the multiplication operator $M_g$ is unitary. Hence, so is $U = V^\ast M_g V$ and we have $$ T = V^\ast M_{f} V = V^\ast M_g M_{|f|} V = V^\ast M_g VV^\ast M_{|f|} V = U\cdot |T| $$ as desired.

EDIT: As I wrote in the comments, the unitary $U$ is not unique in general. For example for $T=0$, we have $|T|=0$, so that any unitary $U : H \to H$ will do the job.

For any bounded operator $T$, the operator $T^{\star}T$ is a positive selfadjoint operator with unique positive square root $|T|=(T^{\star}T)^{1/2}$. And, $$ \|Tx\|^{2}=(T^{\star}Tx,x)=(|T|^2x,x)=(|T|x,|T|x)=\||T|x\|^{2},\;\;\; x \in H. $$ The operators $T$ and $|T|$ have the same null spaces, and both maps induce injective linear maps from $H/\mathcal{N}(T)$ into $X$. Therefore, there is a linear map $$ U : \mathcal{R}(|T|) \rightarrow \mathcal{R}(T) $$ such that $U|T|=T$. $U$ is isometric because $\|U|T|x\|=\|Tx\|=\||T|x\|$. So it extends uniquely by continuity to an isometric map from $\overline{\mathcal{R}(|T|)}$ onto $\overline{\mathcal{R}(T)}$.

Now assume that $T$ is normal. Then $\|Tx\|^{2}=(T^{\star}Tx,x)=(TT^{\star}x,x)=\|T^{\star}x\|^2$, which gives $\mathcal{N}(T)=\mathcal{N}(T^{\star})$. Hence, $$ \mathcal{R}(|T|)^{\perp} =\mathcal{N}(|T|) =\mathcal{N}(T) =\mathcal{N}(T^{\star}) = \mathcal{R}(T)^{\perp}. $$ So $U$ has a unitary extension to all of $H$ obtained by setting $U=I$ on $\mathcal{N}(T)$. $U$ is not unique because $U$ may be chosen to be any unitary map on $\mathcal{N}(T)$. For example, in the most extreme case where $T=0$, the operator $U$ can be any unitary map of $H$.