# 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 $L^p$

Exercise :

Let $$\Omega \subseteq \mathbb R^n$$ be open and bounded, $$\{f_n\}_{n \geq 1} \subseteq L^p(\Omega)$$ with $$1 and $$\{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 $$L^p(\Omega)$$.

• It is redundant to say $g_n\to g$ "a.e." and "in $L^\infty$". Convergence in $L^\infty$ implies convergence a.e. . – Giuseppe Negro May 18 at 10:51
• @GiuseppeNegro It's given as such in the exercise body. – Rebellos May 18 at 10:53
• That's not important. Use your own mind. – Giuseppe Negro May 18 at 11:09

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