If $u_{n} \rightarrow u$ in $L^{p}$ and $v_{n} \rightarrow v$ in $L^{q}$, can I conclude that $\int u_{n}^{p}v_{n}^{q}$ converges?

Question

Suppose that $u_{n} \rightarrow u$ in $L^{p}(\Omega)$ and $v_{n} \rightarrow v$ in $L^{q}(\Omega)$, where $\Omega \subset \mathbb{R}^{N}$ is a bounded set. Remark that this gives us the convergences: $$\displaystyle\int_{\Omega} |u_{n}|^{p} \rightarrow \displaystyle\int_{\Omega}|u|^{p} $$ and $$\displaystyle\int_{\Omega} |v_{n}|^{q} \rightarrow \displaystyle\int_{\Omega}|u|^{q}, $$

I want show that $$\displaystyle\int_{\Omega} |u_{n}|^{p}|v_{n}|^{q} \rightarrow \displaystyle\int_{\Omega}|u|^{p}|v|^{q}. $$

Are there some hypothesis which turn out the above statement true? If the converge is true in general, how I can show it? If not is true, what is the counterexample?

Answer 1

This is not true in general. Consider the counterexample:

$$ u_n = v_n = \frac{1}{n \sqrt{x}}, \qquad u=v=0, \qquad \Omega = [0,1], \qquad p=q=1 $$

We have that:

$$ \int \limits_{\Omega} |u_n - u| = \frac{1}{n} \int \limits_0^1 \frac{1}{\sqrt{x}} = \frac{2}{n} \rightarrow 0 $$

The same goes for $v_n$ and $v$. However:

$$ \int \limits_{\Omega} |u_n||v_n| = \frac{1}{n^2} \int \limits_0^1 \frac{1}{x} = \infty, \qquad \text{while} \qquad \int \limits_{\Omega} |u||v| = 0 $$

So we indeed don't have the convergence. Note that with a slight modification, in particular $u_n = v_n = (nx)^{-1/4}$, we can switch to $p=q=2$, which even satisfies $p^{-1}+q^{-1}=1$.