# If $f \in L_p$ and $g \in L_\infty$, show that $fg \in L_p$.

If $f \in L_p, 1\le p\le \infty$, and $g \in L_\infty$, then the product $fg \in L_p$ and $\|fg\|_p \le \|f\|_p\|g\|_\infty$.

I am still trying to show the first part, that $fg \in L_p$. What I have tried so far is the following:

We must try and show that $$\int |fg|^p \, d\mu < \infty.$$ Notice that $$|fg|=|f||g|\le |f|\|g\|_\infty,$$ since $|g(x)|\le \|g\|_\infty$ for every $x$. Now \begin{align}\int |fg|^p d\mu &= \int \left(|f|~|g|\right)^p \, d\mu \\ &= \int |f|^p|g|^p d\mu \\ &\le \int |f|^p \|g\|_\infty^p \, d\mu \\ &= \|g\|_\infty^p \int |f|^p\, d\mu <\infty,\end{align} i.e. $fg \in L_p$.

Is this correct?

EDIT: The inequality in question then follows directly from seeing that $||fg||_p=(\int |fg|^p d\mu)^{\frac{1}{p}}$ and applying the inequality we obtained above.

• Replace every by "almost every" and it's fine. – user296602 Sep 3 '16 at 21:10
• @T.Bongers - Oh yes of course! On $L_\infty$ we have "almost everywhere" boundedness :) – DJS Sep 3 '16 at 21:11