Let $U \subseteq \mathbb{R}^n$ be an open bounded set. Set $L^p=L^p(U)$.
Let $f \in L^p$, and let $g_n \in L^p$, $g_n \to 0$ in $L^p$. Is it true that $fg_n \to 0$ in $L^p$? Is it even true that $fg_n \in L^p$ for sufficiently large $n$?
I am quite sure the answer is negative. Any ideas for counter-examples?
Let $n=1,p=1,U=(0,1),f(x)=x^{q}, g_n(x)=\frac 1 n x^{q}$ where $-1<q<-\frac 1 2$. Then $fg_n \notin L^{1}$ for any $n$ even though $f,g_n \in L^{1}$ for all $n$ and $\|g_n\|_1 \to 0$ .
Generically $f g$ is not in $L^p(U)$. This can be guessed from Holder's, which says $$ \int |fg| \leq \|f\|_{L^p}\|g\|_{L^{p_*}}$$ and so we see even in the case that $p=p_*=2$, we should only expect $fg \in L^1\supset L^2$ in general. For other $L^p$ spaces, it also follows from Holder's that $$ \int |fg|^q \leq \|f\|_{L^{pq}}^q \|g\|_{L^{p_*q}}^q$$ where again $p_*$ is the Holder conjugate of $p$. So by taking $p=p_*=2$, we see that the product of $L^q$ functions is generically in $L^{q/2}$.
To obtain a counterexample for your specific statement, Similarly to $f \in L^1$, but $f \not\in L^p$ for all $p > 1$ , for any $p\in[1,\infty)$, we can find a function $F$ with $\int |F|^{p/2} < \infty$ but $F$ is not in $L^p$. Then let $f=g=\sqrt{|F|} \in L^p$.
To finish the construction just set $g_n = g\mathbb 1_{|g|>n}$. Then $g_n \to 0$ in $L^p$ and yet $f g_n \notin L^{p}$ for every $n$, as $$ \infty =\int_U f^pg^p = \int_U g^{2p} = \int_{|g|<n} g^{2p} + \int_U f^p g_n^p $$ and $0\le \int_{|g|<n} g^{2p} < n^{2p}|U| < \infty $.
