Showing $u'=v$ a.e. given $u_k \to u$ and $u'_k \to v$ in $L_2(\mathbb{R})$. Suppose $(u_k)$ is a sequence of differentiable functions in $L_2(\mathbb{R})$ satisfying
(1) There is a $u \in L_2(\mathbb{R})$ so that $\| u_k - u\|_2 \to 0$.
(2) There is a $v \in L_2(\mathbb{R})$ so that $\|u'_k - v \|_2 \to 0$.
If $u$ is differentiable, show that $u'=v$ almost everywhere.
This is a problem on a practice qualifying exam that I am having trouble with.

Define $w(x) := \int_0^x v(t)\,dt$. Then $w$ is differentiable a.e. and $w' = v$ a.e. Let $\varphi\in C_0^\infty(\mathbb R)$. Then
$$
\langle u'-w',\varphi\rangle = \langle w-u,\varphi'\rangle = \langle w-u_k,\varphi'\rangle + \langle u_k-u,\varphi'\rangle.
$$
Now,
$$
\langle w-u_k,\varphi'\rangle = \langle u_k'-v,\varphi\rangle\to 0
$$
as $k\to\infty$. Therefore, and since $u_k\to u$ in $L^2$, we have $\langle u'-w',\varphi\rangle = 0$ for all $\varphi\in C_0^\infty(\mathbb R)$. Since $C_0^\infty(\mathbb R)$ is dense in $L^2(\mathbb R)$, this shows $u' = w' = v$ a.e., which we had to prove.
