Proving that weak limit in $L^p$ and strong limit in $H^{-1}$ are the same

Let $\Omega \subset \mathbb{R}^n$ be a bounded domain. Let $p \geq 1$ and suppose that $u_n \rightharpoonup u$ in $L^p(\Omega)$ and $u_n \to v$ in $H^{-1}(\Omega)$.

How to show that $u=v$?

I can do this for $p=2$ by using definition of weak convergence but for other $p$ I have no idea.

Both assumptions imply convergence in the sense of distributions: that is, for every smooth compactly supported function $\varphi$ we have $$\int \varphi u_n\to \int \varphi u,\qquad \int \varphi u_n\to \int \varphi v \tag1$$ So, $$\int \varphi u = \int \varphi v\tag2$$ for every such $\varphi$. This means exactly that $u=v$ in the sense of distributions. But $u$ is an $L^p$ function, so $v$ is a distribution represented by an $L^p$ function. By virtue of (2) it's the same function. Indeed, the elements of $L^p$ can be identified with bounded linear functionals on $L^q$ ($1/p+1/q=1$) and if two such functionals agree on a dense subspace, they are equal.
• Thank you. Can you explain one thing: why does $u_n \to u$ in $H^{-1}$ imply convergence in distribution? I was thinking: $(u_n, f)_{H^{-1}} \to (u, f)_{H^{-1}}$ for all $f \in H^{-1}$ (the brackets means the inner product on $H^{-1}$), and then we use Riesz map $\langle u_n, F \rangle_{H^{-1}, H^1} \to \langle u, F \rangle_{H^{-1}, H^1}$ for all $F \in H^1$. I think we want to pick $F \in C_c^\infty$, but the angled bracket only turns into an integral when $u_n$ and $u$ are in $L^2$ which is not necessary. So I guess this is not the way to prove it converges in distribution?
• @asda I was not thinking of inner product in $H^{-1}$, but rather of the definition of this space as the space of linear functionals on $H^1_0$. The way these functionals are evaluated is consistent with the evaluation of distributions on test functions. It's a matter of unwrapping the definition of this dual.