8
$\begingroup$

Let $\Omega \subset \mathbb{R}^n$ be a domain. There are two different types of Sobolev spaces on $\Omega \times \mathbb{R}$ that are used in PDE Theory: one is the Bochner-Sobolev space $W^{1,p}(\mathbb{R}, W^{1,q}(\Omega))$ of Banach-space valued functions; the other is the usual Sobolev space $W^{1,p}(\Omega \times \mathbb{R})$ on the product space.

It seems to me that there should be some kind of simple relation between these two kinds of spaces coming from Fubini-Tonelli. However, none of the standard references that I've looked at seem to state any kind of result along these lines, which makes me somewhat concerned. Can someone give a reference for a theorem that relates these two?

Background: I'm reading a PDE paper which states all of its results in terms of Bochner-Sobolev spaces and I'd like to understand how these results translate back into ordinary Sobolev spaces.

$\endgroup$
1
  • $\begingroup$ Is there some way I can move this to MO? $\endgroup$
    – Maxim
    Aug 28, 2016 at 0:47

2 Answers 2

3
$\begingroup$

The space $W^{1,p}(\mathbb{R};W^{1,q}(\Omega))$ involves derivatives up to order $2$, so this is not the right space to compare with $W^{1,p}(\mathbb{R}\times \Omega)$. But you have that an equivalent norm on $W^{1,p}(\mathbb{R}\times \Omega)$ is given by $\|\cdot\|_{W^{1,p}(\mathbb{R};L^p(\Omega))} + \|\cdot\|_{L^p(\mathbb{R};W^{1,p}(\Omega))}$. The first norm gives information on the time derivative, the latter of the spatial derivatives. If $\Omega$ is bounded, you can get embedings involving an-isotropic spaces as in you question (i.e. with $q$ instead of $p$ for suitable $q$).

$\endgroup$
-1
$\begingroup$

I may be late to give my feedback, But for example, if you have an element $$u \in W^{1,p}( (0,T), W^{1,p}(\Omega)) \Leftrightarrow \tilde{u} \in W^{1,p}( (0,T)\times\Omega)$$ where $\tilde{u}$ is defined as $$ \tilde{u}(t,x)=u(t)(x) $$ The function $\tilde{u}$ is defined almost everywhere in $(0,T) \times \Omega$. I hope this would enlighten you.

$\endgroup$
1
  • $\begingroup$ The equivalence is false, since in general $u'$ has values in $L^2$ if $\tilde{u}\in W^{1,p}$. $\endgroup$
    – No-one
    Jul 9, 2022 at 19:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .