Polar Decomposition: Let $v$ be a continuous linear operator on a Hilbert space $H$. Then there is a unique partial isometry $u\in B(H)$ such thats $v=u|v|$ and $\ker(u) = \ker(v)$ more over, $u^*v=|v|$.
Assume that $X$ is a compact set, $\mu$ measure on it and $\varphi \in L^{\infty}(X,\mu)$. We define $M_{\varphi}: L^{\infty}(X,\mu)\longrightarrow L^{\infty}(X,\mu)$ such that $M_{\varphi}(f)=\varphi f$.
I know that $M_{\varphi}^*=M_{\overline{\varphi}}$.
so I can say $$| M_{\varphi}|=\left(M_{\varphi}^{*}M_{\varphi}\right)^{1/2}=M_{|\varphi|} $$
I want to find $u$ in polar decomposition.
Q:What is $u$? What is the form of it?
We have (by uniqueness) that $u = M_{{\mathrm sgn}\,\varphi}$, where $$ {\mathrm sgn}\,\varphi(x) = \begin{cases}0, &\varphi(x) =0,\\ \frac{\varphi(x)}{|\varphi(x)|}, &\varphi(x)\neq 0.\end{cases} $$