TITLE: Exercise in Evans, Partial Differential Equation
QUESTION [3 upvotes]: I want to solve exercise 4 in chapter 7.5 Problems in Evans book, i.e.
$u_k$ converges weakly to $u$ in $L^2(0,T;H^1_0(U))$ and $u_k'$ converges weakly to $v$ in $L^2(0,T;H^{-1}(U))$, then $v=u'$. There is a hint: let $\phi\in C_c^1(0,T)$ and $w\in H^1_0(U)$. Then $\int_0^T \langle u_k',\phi w\rangle dt = -\int_0^T \langle u_k,\phi ' w\rangle dt$. How should I use this hint? 
My plan was to use the fundamental lemma of calculus of variations. For that, I wanted to show,
$$\int_0^T (u'(t)-v(t))\phi(t) dt=0$$
for all $\phi \in C^\infty_c((0,T))$. By definition of weak derivative and weak convergence I can write
$$\int_0^T u_k'(t)\phi(t) dt= -\int_0^T u_k(t) \phi'(t) dt$$ 
The LHS converges to $\int_0^T v(t)\phi(t)dt$ and the RHS to $-\int_0^T u(t)\phi'(t)dt $. Therefore
$$\int_0^T v(t)\phi(t)+u(t)\phi'(t)dt=0$$
I guess the hint motivates something similar. But how could I proceed and prove the statement? Or is my start completely wrong?

REPLY [3 votes]: I think some care needs be taken at the point, where you write 

[...] The LHS converges to $\int_0^T v(t) \phi(t)dt$ and the RHS to $-\int_0^T u(t) \phi'(t)dt$. Therefore [...]

This is true, but actually not given in the exercise statement, and I believe its deduction is the main part of the problem. Otherwise you'd already be done, having shown that $\int_0^T u(t)\phi'(t)\, dt = -\int_0^T v(t)\phi(t)\, dt$ for all $\phi\in C_c^1(0,T)$. A function $v$ (by definition) is a weak derivative of $u$ if this holds.
Maybe we should first recall what it means for a sequence $w_n$ to converge to $w$ weakly in $L^2(0,T; X^\ast)$, where $X$ is some Banach space. By definition (at least if I'm not misunderstanding something myself) this means that
$$\int_0^T \langle w_n, g\rangle \, dt \to \int_0^T \langle w,g\rangle\, dt \qquad \forall \, g \in L^2(0,T; X)$$
where $\langle\; , \;\rangle $ is the natural pairing $X^\ast \times X \to \mathbb R$. 
Now following the hint: If we set $X = H^1_0(U)$, $X^\ast= H^{-1}(U)$ and $g = \phi(t) w$ for $\phi\in C_c^1(0,T)$, $w\in H^1_0(U)$, then $g, g'\in L^2(0,T;H^1_0(U))$. So by assumption on $u$ and $v$:
\begin{align}
\int_0^T \langle u_n, \phi' w\rangle \, dt &\to \int_0^T \langle u,\phi' w\rangle\, dt \\
\int_0^T \langle u_n', \phi w\rangle \, dt &\to \int_0^T \langle v,\phi w\rangle\, dt 
\end{align}
We can rewrite 
$$\int_0^T \langle u, \phi' w\rangle \, dt = \int_0^T \phi' \langle u,  w\rangle \, dt = \int_0^T \langle u\phi' , w\rangle \, dt = \left\langle \left(\int_0^T u(t)\phi'(t) \, dt\right) , w\right \rangle$$
where I have made use of Fubini in the last equality. A similar equality is true for $v$. Using this together with $\int_0^T \langle u_n, \phi' w\rangle \, dt = -\int_0^T \langle u_n', \phi w\rangle \, dt$, we obtain 
$$\left\langle \left(\int_0^T u(t)\phi'(t) \, dt\right) , w\right \rangle = \left\langle \left(\int_0^T -v(t)\phi(t) \, dt\right) , w\right \rangle \qquad \forall\, w\in H^1_0(U)$$
Therefore $\int_0^T u(t)\phi'(t) \, dt = -\int_0^T v(t)\phi(t) \, dt$ for all $\phi \in C_c^1(0,T)$. So $v$ is a weak derivative of $u$.
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