Let $I=(l_0,l_1)$ and $g(x)= f(x)\chi_{(\alpha,\beta)}$ where $(\alpha,\beta)\subset I$ and $f \in L^\infty(I)$ with $f(x)\geq c>0$ for all $x\in I$.
Now, if $h \in L^2(I)$, is the product $g(x)h(x) \in L^2(I)$ if $f(x)$ is also bounded?
My attemps:
Since $g(x)$ is measurable in $I$, then $g(x)h(x)$ is measurable. Now, I need to show that ${|gh|}^2 \in L^1(I)$. let $K\geq f(x)$, then
\begin{equation*} \begin{aligned} \int_{I} (g(x)h(x))^2\,dx&=\int_{I} (f(x)\chi_{(\alpha,\beta)}\,h(x))^2\,dx\\ &\leq K^2\int_{I} h(x)^2\,dx \end{aligned} \end{equation*}
This proves that $g(x)h(x) \in L^2(I)$. Am I right? What about only requiring $f(x) \in L^\infty(I)$?
I really appreciate your help!
