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 $L^p(\Omega)$.

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

Use the Dominated Convergence Theorem.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.