Is the regularity of $u$ necessary to deduce this result? (Evans PDE)

Question

One of the exercises in Evans book on PDEs (at the end of chapter 7) is given as follows: Assume $$u_k\rightharpoonup u\quad\mbox{in}\quad L^2(0,T;H^1_0(U)),$$ $$u_k'\rightharpoonup v\quad\mbox{in}\quad L^2(0,T;H^{-1}(U)),$$ Prove that $v=u'$.

Disregarding the following hint:

Let $\phi\in C^1_c(0,T)$, $w\in H^1_0(U)$. Then $$\int\langle u_k',\phi w\rangle\,dt=-\int_0^T\langle u_k,\phi'w\rangle\,dt.$$ The reason I choose to disregard the hint is because it relies on the duality pairing $\langle,\rangle$ of $u_k$ and $w\in H^1_0(U)$. Instead I believe we can prove this simply using the definition of weak derivative: Take $\varphi\in C^\infty_c(0,T)$, then $$\int_0^T u\varphi'\,dt=\lim_{k\to\infty}\int_0^T u_k\varphi'\,dt$$ $$=-\lim_{k\to\infty}\int u_k'\varphi\,dt$$ $$=-\int_0^Tv\varphi\,dt,$$ by the definition of the weak derivative, we see that $v=u'$.

This proof does not seem to require convergence in a particular space, as in it seems to me that it would work, so long as we had convergence into two Banach spaces $X$ and $Y$, rather that $H^1_0(U)$ and $H^{-1}(U)$.

Answer 1

• Is the regularity of u necessary to deduce this result?

No.

• This proof does not seem to require convergence in a particular space, as in it seems to me that it would work, so long as we had convergence into two Banach spaces [...]

I agree with you provided that the Banach space $Y$ is separable reflexive and the Banach space $X$ is continuously embedded in $Y$.

Furthermore, $L^2$ convergence for both, the functions and the derivatives, is not needed. We can assume $L^p$ convergence for the functions and $L^q$ convergence for the derivatives for any $1\leq p,q\leq\infty$. Explicitly, we have the following result:

Theorem: Let $Y$ be a separable reflexive Banach space, $X$ a Banach space continuously embeded in $Y$ and $1\leq p,q\leq\infty$. If $$\left\{\begin{align}u_k\rightharpoonup u\quad&\mbox{in}\quad L^p(0,T;X),\\ u_k'\rightharpoonup v\quad&\mbox{in}\quad L^q(0,T;Y),\end{align}\right.$$ then $u'=v$.

The proof follows as in your post. Let me tell where the hypotheses are used:

Since $X\hookrightarrow Y$, we get $L^p(0,T;X)\hookrightarrow L^1(0,T;Y)$ and $L^q(0,T;Y)\hookrightarrow L^1(0,T;Y)$. Thus,

$$\left\{\begin{align}u_k\rightharpoonup u\quad&\mbox{in}\quad L^1(0,T;Y),\\ u_k'\rightharpoonup v\quad&\mbox{in}\quad L^1(0,T;Y).\end{align}\right.$$

It follows that

$$\lim_{k\to\infty}\int_0^T u_k\varphi'\,dt=\int_0^T u\varphi'\,dt,\qquad \lim_{k\to\infty}\int_0^T u_k'\varphi\,dt=\int_0^Tv\varphi\,dt$$ because $Y$ is separable reflexive (where the limits are understood in the weak sense).