6
$\begingroup$

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?

$\endgroup$
1
3
$\begingroup$

The proof is correct. I think you could add periods at the end of your equations if they end a sentence, but that is just personal taste.

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.

$\endgroup$

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.