Excercise 4.3 (3) in Brezis. Convergence in $L_p$

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.

• 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. – Keen-ameteur Jan 9 at 20:02
• 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. – Alfdav Jan 9 at 20:16
• 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 $\vert fg_n\vert^p$. – Keen-ameteur Jan 9 at 20:24
• $f(g_n-g)=fg_n-fg$ 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 $||f(g_n-g)||_p=||fg_n-fg||_p<\epsilon$ – Alfdav Jan 9 at 20:41
• 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 – Alfdav Jan 9 at 20:43

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 $$\left\lvert h_n(x) \right\rvert\leqslant \left(2M\right)^p\left\lvert f(x) \right\rvert^p \mbox{ a.e.}$$