I need to prove (if true) that the space $W^{1,2}(0,T;\mathbb{R}^d)$ is compactly embedded in $C(0,T;\mathbb{R}^d)$. The proof for the continuous embedding part is straightforward and is given in PDE book by Evans. However could someone give me some ideas on how to prove the compact part.

This question pops up in proving certain bounds for existence of solutions to a simple GENERIC problem. Is there some extension of Azrela-Ascoli theorem to say Bochner spaces which can be used here.

  • $\begingroup$ I think you can find some hints here: hrcak.srce.hr/file/7912 $\endgroup$ – Siminore Jan 24 '13 at 12:38
  • $\begingroup$ I am a little new to the PDE world, so I would be grateful if you could let me know what exactly am I supposed to be looking for. $\endgroup$ – UPS Jan 24 '13 at 13:04

I guess this is the solution. Would be grateful if someone could verify the arguments.

The proof of the continuous embedding $W^{1,2}(0,T;\mathbb{R}^d)\hookrightarrow \mathbb{L}^2(0,T;\mathbb{R}^d)$ follows from Evans 5.9.2.

So let's begin with the assumption that we have a bounded sequence $(x_n)$ in $X:=W^{1,2}(0,T;\mathbb{R}^d)$, $\sup\limits_{n}\|x_n\|_X<c$ (i.e. we have uniform bounds).

(This assumption may be incorrect (is it??). But it actually holds in the original problem.)

For proving the compact embedding we make use of Azrela-Ascoli theorem which requires

(1) Uniform bounds on $(x_n)$ in the space $X$ which we have.

(2) Equicontinuity: Let $0\leq s\leq t\leq T$. \begin{equation} \|x_n(t)-x_n(s)\|_{\mathbb{R}^d}\leq\int\limits_{s}^t\|\dot{x}_n(\sigma)\|_{\mathbb{R}^d}d\sigma\leq \|\dot{x}_n\|_{\mathbb{L}^2(0,T;\mathbb{R}^d)}(t-s)^{\frac{1}{2}}\leq C(t-s)^{\frac{1}{2}}, \end{equation} The first inequality is due to fundamental theorem of calculus (and existence of weak derivative).

By Azrela-Ascoli we have managed to show the compactness of $(x_n)$ in the Holder space $C^{\frac{1}{2}}(0,T;\mathbb{R}^d)$. By similar arguments we can show that Holder space $C^\alpha$ is compactly embedded in $C^\beta$ where $\beta>\alpha$. Hence we are done.

Note: There is a subtlety involved in the usage of Azrela-Ascoli theorem. We require that the space $\overline{\{x_n(t): t\in[0,T]\}}$ is compact in $\mathbb{R}^d$. Therefore this embedding would have not been possible for any range space $X$ instead of $\mathbb{R}^d$.

  • $\begingroup$ You said in the question that the continuous embedding of $W^{1,2}(0,T;\mathbb{R}^d)$ into $C(0,T; \mathbb{R}^d)$ was "straightforward". If you unwind that statement, you'll find it says that any sequence $\{x_n\}$ which is bounded in $W^{1,2}$-norm is also bounded in sup norm. $\endgroup$ – Nate Eldredge Jan 27 '13 at 0:38
  • $\begingroup$ I think you may be confused by "uniform bounds" which is a vague term to begin with, and is used in two ways here. You want to prove that any bounded sequence in $X$ has a convergent subsequence in $C$. So, $\sup_n \|x_n\|\le c$ is our starting point: we have a bounded sequence in $X$. You could call $c$ a "uniform bound" since it's uniform in $n$. It is also true (on the basis of FTC) that there is $M$ such that $|x_n(t)|\le M$ for all $n$ and all $t\in [0,1]$. You could also call $M$ a "uniform bound" since it's uniform in $n$ and $t$. In the application of Arzelà–Ascoli theorem you use $M$. $\endgroup$ – user53153 Jan 27 '13 at 0:42
  • $\begingroup$ @NateEldredge I do get that. However it would seem that the more important part of the continuous embedding is that every $W^{1,2}$ function is a C function as well and therefore we can use the Fundamental Thm. of Calculus. Is my understanding correct? $\endgroup$ – UPS Jan 27 '13 at 1:07
  • $\begingroup$ @5PM I would think the bound $\sup_n\|x_n\|_X<c$ would automatically also give the bound $|x_n(t)|<M$ for all n,t as $\sup_n\|x_n\|=\sup_n\sup_{t\in[0,T]}\|x_n\|_{\mathbb{R}^d}$. $\endgroup$ – UPS Jan 27 '13 at 1:09
  • $\begingroup$ @UPS You wrote $\|x_n\|$ without a subscript; I don't know what that means. By definition of Sobolev space $X=W^{1,2}$, $$\|x_n\|_X= \sqrt{\int_0^1 \|x_n'(t)\|^2_{\mathbb R^d}\,dt}$$ which is not exactly the same as the supremum of $\|x_n(t)\|_{\mathbb R^d}$ over $t$. $\endgroup$ – user53153 Jan 27 '13 at 1:15

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