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.
Summary: both modes of convergence imply distributional convergence, and distributional limit is unique.