Let $u_n, u \in L^2$. If $\int u_nv \to \int uv$ for all $v \in H^1$, does $\int_{}u_nh \to \int_{}uh$ for all $h \in L^2$? 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...
 A: 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 $(d/r)'=\frac{d}{d-r}$.


*

*If $v\in L^q$ with $q>\frac{d}{d-r}$, we have  $\int u_n v \to 0$. (Hölder's inequality)

*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.
