Measure space and polar decomposition of a multiplicative operator This question was asked in my quiz of Functional Analysis and I was not able to solve this particular problem.

Question :Let $(\Omega , B, m)$ be a measure space and let $\phi \in L^{\infty} (\Omega ,m)$. Explain the polar decomposition of multiplicative operator $M_{\phi}$.

Attempt:  For each such $\phi \in L^{\infty} $ define: $M_{\phi}, L^{2} (\Omega ,m) \to L^2 (\Omega ,m)$ as $M_{\phi} : \phi \times f $  for every $f\in L^2 (\Omega ,m)$. Here $\phi  f= \phi (x) \times f(x)$ for every $x\in \Omega$.
But I have no idea on which direction I should work in to get the polar decomposition.
Can you please give a couple of hints?
I have been following walter rudin's Functional Analysis.
Thanks!
 A: Define $M:L^\infty(X,\mu)\to B(L^2(X,\mu))$ by $f\mapsto M_f$, where $M_f:L^2(X,\mu)\to L^2(X,\mu)$ is the linear operator defined as $M_f(g):=fg$.

*

*Not needed, but for completeness: $M_f$ is a bounded linear operator with $\|M_f\|\le\|f\|_\infty$. If the measure space is semi-finite, then $\|M_f\|=\|f\|_\infty$.


*You can easily show that $f\mapsto M_f$ preserves linearity, multiplication and adjoints i.e. $M_{\lambda f+g}=\lambda M_f+M_g$, $M_{fg}=M_fM_g$ and $M_f^*=M_{\bar{f}}$ for all $f,g\in L^\infty(X,\mu)$ and all $\lambda\in\mathbb{C}$.


*Let $f\in L^\infty(X,\mu)$. As nicely hinted in the comments, write $f:=|f|\cdot h$, where $h:=\frac{f}{|f|}\cdot\chi_E$, where $E:=\{x\in X:f(x)\ne0\}$.
First note that $|M_f|=M_{|f|}$. Indeed, by definition, $|M_f|=(M_f^*M_f)^{1/2}=(M_{\bar{f}\cdot f})^{1/2}=(M_{|f|^2})^{1/2}=M_{|f|}$. Now note that
$$M_f=M_{h}\cdot M_{|f|}=M_h\cdot |M_f|$$
so all we need to show is that $M_h$ is a partial isometry with the same kernel as $M_f$.
Note that for $g\in L^\infty(X,\mu)$, $\ker(M_g)=\{h\in L^2(X,\mu): gh=0\}=\{h\in L^2(X,\mu): \{h\ne0\}\subset\{g=0\}\}$, and therefore $\ker(M_h)=\ker(M_f)$, since $\{h=0\}=\{f=0\}$.
Finally, $M_h^*M_h=M_{|h|^2}=M_{\chi_E}$, which is a projection, since $\chi_E$ is a projection in $L^\infty(X,\mu)$ and thus $M_h$ is a partial isometry.
