1
$\begingroup$

Let $(f_n)$ in $L_p(\Omega)$ $1\leq p< \infty$ and $(g_n)$ bounded in $L_{\infty}(\Omega)$ assume that $f_n \rightarrow f$ in $L_p(\Omega)$ and $g_n \rightarrow g$ a.e. Prove that $$f_ng_n\rightarrow fg$$ in $L_p(\Omega)$.

I have done the following:

Note that $f_ng_n-fg=(f_n-f)g_n+f(g_n-g)$ therefore $$||f_ng_n-fg||_p=||(f_n-f)g_n+f(g_n-g)||_p\leq||f_n-f||_p||g_n||_{\infty}+||f(g_n-g)||_p$$ I know the first term is less than $\epsilon$ (because $f_n$ converges in $L_p(\Omega)$ to $f$) however there is a hint in the book that says $f(g_n-g)\rightarrow 0$ in $L_p(\Omega)$ by the dominated convergence theorem. Is the first part of my reasoning correct? how can I see what the hint is telling me? Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ Are you asking how the $L_p$ convergence is shown from the dominated convergence theorem? Because if you show that the hint is true then the second term will also be smaller then $\epsilon$ for $n$ large. $\endgroup$ – Keen-ameteur Jan 9 at 20:02
  • $\begingroup$ yes but i made another advance, since $g_n\rightarrow g$ a.e. can i say that $fg_n\rightarrow fg$ a.e. ???? Because if so, since i have that $|fg_n|^p\leq M^p|f|^p$ then I can use dominated convergence. $\endgroup$ – Alfdav Jan 9 at 20:16
  • $\begingroup$ The first thing you wrote is true, but I'm not sure you can say the latter simply follows from dominated convergence theorem, because I think you need to bound $\vert f(g-g_n) \vert^p $ instead of $\vert fg_n\vert^p$. $\endgroup$ – Keen-ameteur Jan 9 at 20:24
  • $\begingroup$ $f(g_n-g)=fg_n-fg$ if the first one is bounded that is $|fg_n|^p\leq M^p|f|^p$ and converges a.e. to the second one that is $fg$ then by dominated convergence $fg$ belongs in $L_p$ and $||f(g_n-g)||_p=||fg_n-fg||_p<\epsilon$ $\endgroup$ – Alfdav Jan 9 at 20:41
  • $\begingroup$ I only needed to know if the a.e. convergence was correct, because I think the dominated convergence theorem applies the way i did it $\endgroup$ – Alfdav Jan 9 at 20:43
1
$\begingroup$

The only remaining thing to is show that $\int_\Omega h_n\to 0$, where $h_n(x)=\left\lvert f(x)\left(g_n(x)-g(x)\right)\right\rvert^p$.

Since $f$ belongs to $\mathbb L^p$, the quantity $\left\lvert f(x) \right\rvert^p$ is finite for almost every $c$ hence the fact that $h_n(x)\to 0$ follows from the implication (if $c\geqslant 0$ and $a_n\to 0$) then $c\cdot a_n\to 0$.

For the domination condition, let $M:=\sup_{n\geqslant 1}\left\lVert g_n\right\rVert_\infty$. Then $\left\lvert g(x)\right\rvert\leqslant M$ for almost every $x$ hence $$\left\lvert h_n(x) \right\rvert\leqslant \left(2M\right)^p\left\lvert f(x) \right\rvert^p \mbox{ a.e.}$$

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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