[Math] If $f_n \to f$ and $g_n \to g$ in $L^p$, and $g_n$ is uniformly bounded, then $f_ng_n \to fg$

lebesgue-integrallebesgue-measurelp-spacesreal-analysis

Problem: If $f_n \to f$ and $g_n \to g$ in $L^p$, and $g_n$ is
uniformly bounded, then $f_ng_n \to fg$ in $L^p$.

An official solution I saw for this problem looked very different. Here is my solution:

Also $$|fg_n|^p \leq |M f|^p = M^p |f|^p$$ with $M^p|f|^p$ integrable. So by the Lebesgue dominated convergence theorem, the fact that $|fg_n|^p$ converges to $|fg|^p$ pointwise means that $\int|fg_n|^p$ converges to $\int |fg|^p$. Then $\|fg_n\| = \sqrt[p]{ \int |fg_n|^p}$ converges to $\|fg\| = \sqrt[p]{\int |fg|^p}$. It can be shown that this is equivalent to saying that $\|fg – fg_n\|$ converges to $0$.

Best Answer

the fact that $|fg_n|^p$ converges to $|fg|^p$ pointwise

Where did you get this fact from? Convergence in $L^p$ does not imply convergence almost everywhere.

However, the proof can be fixed. Convergence in $L^p$ does imply convergence almost everywhere for a subsequence $g_{n_k}$. So, for this subsequence your argument works and gives $f_{n_k}g_{n_k}\to fg$ in $L^p$.

Moreover, the above can be applied to any subsequence of the original sequence $(g_n)$. Thus, every subsequence of $(f_ng_n)$ has a sub-subsequence converging to $fg$. It is a general topological fact that this implies the convergence of the whole sequence to $fg$. (Here this fact is proved for real valued sequences, but the same proof works in any topological space.)

Best Answer

Related Solutions

[Math] Prove $g_n f_n \rightarrow gf$ in $L_p([0,1])$. Where $f_n$ converges and $(g_n)_{n=1}^\infty $ is a sequence of bounded measurable functions

If $f_n \to f$ in $L^p$ and $g_n \xrightarrow{a.e.} g$ in $L^\infty$ then $f_ng_n \to fg$ in $L^p$

Related Question