# If $f_n \rightarrow f$ is a sequence of $L^p$ and $g_n \rightarrow g$ a bounded sequence of $L^{\infty}$ then $f_ng_n \rightarrow fg$ in $L^p$

I want to verify is my proof is correct.

Let $p \in [1,\infty)$, $f_n$ a sequence of $L^p$ that converges to $f$ and $g_n$ a bounded sequence of $L^\infty$ that converges almost everywhere to $g$. Show that $f_ng_n$ converges to $fg$ in $L^p$

$|g_n| \leq M$ and now: $$||f_ng_n -fg ||_p = ||(f_n-f)g_n + (g_n-g)f||_p \leq ||(f_n-f)g_n||_p + ||(g_n -g)f||_p$$ $$\leq ||(f_n-f)||_pM + ||(g_n-g)f||_p$$ All is left to do is to proove that $||(g_n-g)f||_p \rightarrow 0.$ We have $$||(g_n-g)f||_p^p = \int |(g_n-g)f|^pd\mu \leq \int(||g_n-g||_\infty)^p|f|^p d\mu= ||g_n-g||_\infty^p\int|f|^p d\mu$$ As $||g_n -g||_\infty \rightarrow 0$ we get $$||g_n-g||_\infty^p\int|f|^p d\mu \rightarrow 0$$

Thus $$||f_ng_n -fg||_p \rightarrow 0.$$ Is my proof correct?

• Yes, it looks good. Boundedness does the job. May 12 '18 at 14:51

If you want to improve the proof even more, then there is one aspect that might be a bit suboptimal. Your proof suffers from a certain imbalance. On the one hand, you seem to be perfectly comfortable to claim that $$\| (f_n-f) g\|_p \leq \| (f-f_n)\|_p \|g\|_\infty,$$ but you seem to find it necessary to prove that $$\| f(g_n-g)\|_p \leq \| f\|_p \|(g_n-g)\|_\infty.$$ To me, both follow from the same argument, hence it is not clear why you prove one and not the other. I think it would be optimal to state or prove once that $\|f_1 f_2 \|_p \leq \|f_1\|_p \|f_2\|_\infty$, if $f_1 \in L^p$ and $f_2 \in L^\infty$ and then use this observation twice.