If $f_n \to f$ in $L^p$ and $g_n \xrightarrow{a.e.} g$ in $L^\infty$ then $f_ng_n \to fg$ in $L^p$

Question

Exercise :

Let $\Omega \subseteq \mathbb R^n$ be open and bounded, $\{f_n\}_{n \geq 1} \subseteq L^p(\Omega)$ with $1<p< \infty$ and $\{g_n\}_{n \geq 1} \subseteq L^\infty(\Omega)$. If it is $f_n \to f$ in $L^p(\Omega)$ and $g_n \xrightarrow{a.e.} g$ in $L^\infty(\Omega)$ while $\{g_n\}_{n \geq 1}$ is also bounded then show $f_ng_n \to fg$ in $L^p(\Omega)$.

Attempt :

Since $\{g_n\}_{n \geq 1}$ is bounded, then it would be $\|g_n\|_\infty \leq M$ for some $M>0$ and thus for the $p$-norm it would be $\|g_n\|_p \leq M$ as well. Now, it is :

\begin{align*}
\|fg – f_ng_n\|_p &= \|fg – fg_n + fg_n – f_ng_n\|_p \\ &\leq \|fg-fg_n\|_p + \|g_n(f-f_n)\|_p \\ &\leq \|fg-fg_n\|_p + M\|f-f_n\|_p
\end{align*}

The second term goes to $0$ as $f_n \to f$ in $L^p(\Omega)$.

Now, since $g_n \xrightarrow{a.e.} g$ we can also deduce that $\|g_n\|_\infty \xrightarrow{a.e.} \|g\|_\infty$.

For the first term :

\begin{align*}
\|fg-fg_n\|_p &= \left(\int_\Omega|fg-fg_n|^p\mathrm{d}x \right)^{1/p} \\ &\leq \left[ \int_\Omega \left(|fg| + |fg_n|\right)^p\mathrm{d}x\right]^{1/p} \\ &\leq \left[ \int_\Omega \left(|fg| + M|f|\right)^p\mathrm{d}x\right]^{1/p}
\end{align*}

I can't see how to continue on though to prove that this term can become arbitrarily small, thus that $\|fg-f_ng_n\|_p \to 0$ and thus the desired convergence.

Any hints or elaborations will be greatly appreciated !

Edit :

I worked it around myself as such :

\begin{align*}
\|f(g_n-g)\|_p &= \left(\int_\Omega |(g_n-g)f|^p\mathrm{d}x\right)^{1/p} \\ &\leq \left(\int_\Omega ||g_n-g\|_\infty^p|f|^p\mathrm{d}x\right)^p \\ &= \|g_n-g\|_\infty\left(\int_\Omega |f|^p\mathrm{d}x \right)^p \to 0
\end{align*}

So finally we get $\|fg-f_ng_n\|_p \to 0 \Leftrightarrow f_ng_n \to fg$ στον $L^p(\Omega)$.

Answer 1

Use the Dominated Convergence Theorem.

• $|f\,g-f\,g_n|^p$ converges to $0$ a.e.
• $|f\,g-f\,g_n|^p\le2^pM^p|f|^p\in L^1$