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.

  • 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

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


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