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Let $\Omega$ be a bounded domain and let $u_n$ and $u \in L^2(\Omega)$.

Question: If $\int_{\Omega}u_nv \to \int_{\Omega}uv$ for all $v \in H^1(\Omega)$, does $\int_{\Omega}u_nh \to \int_{\Omega}uh$ for all $h \in L^2(\Omega)$?

I think so. Let $v_m \to h$ in $L^2$ where $v_m \in H^1$. We can find this sequence by density. We have $$|\int u_nv_m - \int uv_m| \to 0.$$ This is where I am stuck...

delimit
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No, this is not true. Weak convergence is sensitive to the space of test functions we use.

Let $d$ be the dimension of our space. Consider the sequence $$u_n = n^r \chi_{\{|x|\le 1/n\}} $$ where $r\in (0,d)$ is to be determined. The $L^p$ norm of $u_n$ is a constant multiple of $n^{r- d/p}$. Therefore, for $p<d/r$ we have $u_n\to 0$ strongly in $L^p$, whereas for $p>d/r$ the sequence is unbounded, and therefore does not converge even weakly in $L^p$.

Let's rephrase this in terms of dual spaces, using the Hölder conjugate exponent $(d/r)'=\frac{d}{d-r}$.

  1. If $v\in L^q$ with $q>\frac{d}{d-r}$, we have $\int u_n v \to 0$. (Hölder's inequality)
  2. There exists $v\in L^q$ with $q<\frac{d}{d-r}$ such that $\int u_n v \not\to 0$. (Weak convergence fails)

That's pretty sensitive.

Assume $0\in\Omega$. By the Sobolev embedding theorem, every function in $H^1(\Omega)$ belongs to $L^{2^*}(B)$ where $2^*=\frac{2d}{d-2}>2$ and $B$ is some ball around $0$. Choose $r$ so that $\frac{d}{d-r}$ is strictly between $2$ and $2^*$, and you have a counterexample.