Let $\mu$ be a probability measure. Prove that, function $f$ is $L^1$ iff $f=g\bar{h}$ for some $g,h\in L^2$.

Question

Let $(X,M,\mu)$ be a probability space and $g,h\in L^2(\mu)$. Let $f=g\bar{h}$. Then by Holder's inequality, $\int_X |f| d\mu\le \lVert g\rVert_2 \lVert h\rVert_2<\infty$. So $f\in L^1(\mu)$.

But I am not getting any idea to prove the converse. I know that for finite measure $\mu$, $L^p\subseteq L^q$ for $q\ge p\ge 1$.

Can anyone give me wayout to prove the converse? Thanks for your help in advance.

Answer 1

For each $x$, choose a $0\leq\theta(x)<2\pi$ such that $e^{i\theta(x)}f(x)=|f(x)|$. Take $g(x)=\sqrt{|f(x)|}$ and $h(x)=e^{i\theta(x)}\sqrt{|f(x)|}$, then both $g,h\in L^{2}$ and that $f(x)=g(x)\overline{h(x)}$.