What's the name of this theorem: $\int_a^b|u_n-u|^pdt\rightarrow0\wedge\int_a^b|u_n'-v|^pdt\rightarrow0\Rightarrow u\in W_p^1(a,b) \ : \ u'=v$?

Question

My Professor gave me this as an exercise to prove, that:

$$\int_a^b|u_n(t) -u(t)|^pdt\rightarrow0\wedge\int_a^b|u_n(t)'-v(t)|^pdt\rightarrow0\Rightarrow u(t)\in W_p^1(a,b) \ : \ u'(t)=v(t)$$

given, that we're in $L_p \left[a,b\right]$ space for $1 \le p < \infty$, and our sequence $(u_n)$ is taken so, that $\forall n \ u_n \in W_p^1 (a,b)$ (Sobolev Space).

He didn't give me the name for that theorem though, he just said that it's "one of the most fundamental ones in functional analysis" and that "its proof is not that trivial".

Answer 1

I don't believe this statement has its own name. However, one can rephrase it in a more descriptive language:

Thm. 1. The subspace $\{ (u,u') : u \in W^{1,p} \}$ is a closed subspace of $L^p \oplus L^p$.

As mentioned by daw, this is related to:

Thm. 2. The space $W^{1,p}$ is complete.

I see four things that could be said about these two statements, and none of them is really hard.

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Proof of Thm. 1

Given $u_n$, $u$ and $v$ as in your question, it's enough to justify that $u'=v$. To this end, fix a test function $\varphi \in C_c^\infty((a,b))$. Given that $$ 0 = \int_a^b u_n' \varphi + u_n \varphi' $$ for each $n$, we can pass to the limit and obtain $$ 0 = \int_a^b v \varphi + u \varphi'. $$ This is nothing else than saying $u' = v$ in the weak sense.

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Proof of Thm. 2 given Thm. 1

Since $V = \{ (u,u') : u \in W^{1,p} \}$ is a closed subspace of the complete space $L^p \oplus L^p$, then $V$ has to be complete as well. Note that $V$ is isometric to $W^{1,p}$ by the map $$ W^{1,p} \ni u \mapsto (u, u') \in V, $$ so $W^{1,p}$ is complete too.

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Proof of Thm. 1 given Thm. 2

This is a very strange thing to do, but since OP asked for it in a comment, here it goes.

If $W^{1,p}$ is known to be complete, then $V$ is also complete as its isometric copy. Then $V \subseteq L^p \otimes L^p$ has to be a closed subspace. This last implication doesn't even use completeness of $L^p$.

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Reflexivity of $W^{1,p}$

OK, nobody asked about it, but it's important. It is a consequence of Hahn-Banach that any closed subspace of a reflexive space is also reflexive. In the case $p > 1$ the space $L^p \oplus L^p$ is reflexive, and thus so is $V$, and finally also $W^{1,p}$.