polar decomposition of multiplicative operator on L^2 induced by identity function.

Question

We know that every operator in B(H) has a polar decomposition. $T=VP$ that $P=|T|$ and V is a partial isometry with initial space closure of ImP and final space ImT.

How can i obtain polar decomposition of multiplicative operator on $L^2(\bar D)$ that induced by the identity function?

Answer 1

Hint: For any $f,g,h$ we have $f(x)=|f|(x)e^{i\theta(x)}$, and we know that $M_gM_h=M_{gh}$.

So, $M_f=M_{|f|}M_{f_1}$ where $f_1=f(x)/|f(x)|$ whenever $f(x) \neq 0$.

This is your Polar decomposition: Check that this works.

For $f(x)=1$ we have that both the positive part and the partial isometry is actually $M_1$