# Arzela-Ascoli-type embedding: Is $H^1(0,T;X)$ compactly embedded in $C([0,T];X)$?

This is a variant of the question Compact Embedding of $W^{1,2}(0,T;ℝ^d)$ in $C(0,T;ℝ^d)$ where we had $X=\mathbb{R}^d$. Let now $X$ be some Banach space.

Question: Is $H^1(0,T;X)$ compactly embedded in $C([0,T];X)$?

We define $H^1(0,T;X)=\{u : (0,T) \to X ~|~ u \in L^2(0,T;X), u' \in L^2(0,T;X)\}$.

A proof for the continuous embedding can be found in a lot of books, e.g. Roubicek's book. Also, it is standard that $H^1(0,T,\mathbb{R})$ is compactly embedded in $C([0,T];\mathbb{R})$, e.g. Brezi's book. The main difficulty is that the classical Arzela-Ascoli theorem is only valid for handling something like $C(S;\mathbb{K}^d)$ and not for $C(S;Y)$ where $Y$ is some infinite-dimensional Banach space. But for this scenario there is the following version, which can be found in the book of Boyer.

Arzela-Ascoli: Let $E$ be a compact metric space and $F$ a metric space. Let $\mathcal{F}$ be a subset of $C(E,F)$. If $\{f(x) : f \in \mathcal{F} \}$ is relatively compact in $F$ for any $x\in E$ and $\mathcal{F}$ is equicontinuous, then $\mathcal{F}$ is relatively compact in $C(E,F)$.

I am translating the proof of Brezis to our setting.

Proof: Let $u \in B:=\{v \in H^1(0,T;X) : \|v\|_{L^2(0,T;X)}+\|v'\|_{L^2(0,T;X)} \leq 1\}$. In particular $u \in C([0,T];X)$. Then we have for all $t,s \in [0,T]$ $$\|u(t)-u(s)\|_X \leq \int_s^t \|u'(\tau)\|_X \, \text{d}\tau \leq \|u'\|_{L^2(0,T;X)} |t-s|^{1/2} \leq |t-s|^{1/2}$$ i.e. $u$ is equicontinuous. Further if we can show that $\{ u(t): u \in B\}$ is relatively compact in $X$ for any $t\in [0,T]$, then we would be finished.

Can someone help me with this last step written in bold font? I am not sure if this result is even correct. I think I have seen it in a article with $X=L^2$ or $X=H^1$, I don't remember clearly.

Before there was a picture of another Arzela-Ascoli theorem version which was written in an unclear/erroneous way. I've updated it with the version in Boyer's book which coincides with the notes provided in the comments.

• As for the compactness subquestion: continuous images of compact intervals are compact; I guess what you mention is only an issue for non compact $S$ – Bananach Apr 5 '18 at 12:24
• I believe his 'Warning' as written is actually incorrect, but there is a correct version of the warning: For any $x\in S$, the set $\{f(x) : f\in A\}$ must be precompact. This condition indeed follows from boundedness when $Y$ is finite-dimensional, but not when it is infinite-dimensional. See math.utk.edu/~freire/teaching/m447f16/AscoliArzelaNotes.pdf – Bananach Apr 5 '18 at 13:15
• You may mean the correct thing but you certainly don't write it: $u(t)$ for a given $u\in H^1([0,T];X)$ is a single element of $X$ and thus certainly is precompact. You need to show that $\{u(t) : u \in B\}$ is precompact in $X$ for all $t\in [0,T]$ – Bananach Apr 5 '18 at 14:44
• Now that we have clarified that, I don't think the statement to be proven actually holds: Let $B$ be the unit ball of $H^1([0,T];X)$, let $B_X$ be the unit ball of $X$, and let $\tilde{B}:=\{t\mapsto f : f \in B_X\}\subset B$ be the set of all constant functions in $H^1([0,T];X)$ whose constant value is in $B_X$. Then $\{u(0) : u \in B\}\supset \{u(0) : u \in \tilde{B}\} = B_X$ which is not precompact in $X$ unless $X$ is finite-dimensional – Bananach Apr 5 '18 at 15:04
• Actually, this disproves your original statement, and doesn't only show that your planned proof doesn't work: $B$ is not pre-compact in $C([0,T];X)$ because $\tilde{B}$ isn't, because $\phi(\tilde{B})= B_X \subset X$ isn't, where $\phi$ sends elements of $C([0,T];X)$ to their values at $0$ (a continuous mapping) – Bananach Apr 5 '18 at 16:34

Having infinite-dimensional $X$ is problematic, because even the space of constant $X$-valued functions is not compactly embedded in $C([0, 1];X)$. Indeed, the image of its unit ball under the embedding into $C([0,1];X)$ is an isometric copy of the unit ball of $X$; not a totally bounded set.
• Thank you for your answer. Too bad the result is not valid. I am aware of Aubin-Lions lemma and its special case indeed has a compactness result in $C([0,T];X)$; for a Gelfand triple $X \hookrightarrow \hookrightarrow Y \hookrightarrow Z$ we have $$L^\infty(0,T;X) \cap W^{1,p}(0,T;Z) \hookrightarrow \hookrightarrow C([0,T];Y).$$ I thought there could be a trade-off, since in my case the derivative is in a better space even though I have no $L^\infty$ estimate for the function itself. But thank you for clarifying this. – Fritz Apr 5 '18 at 21:35