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Sep 24, 2021 at 2:06 comment added sharpend This is why it would be great if somebody wrote an up-to-date and friendly version of Ladyzhenskaya-Solonnikov-Uraltseva...
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Sep 4, 2019 at 22:13 history edited YCor CC BY-SA 4.0
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Nov 8, 2018 at 19:53 answer added DCM timeline score: 1
Nov 8, 2018 at 0:31 comment added DCM The $C^{2+\alpha,\beta}(Q_T)$ in my last comment should have been a $C^{2+\alpha,1+\beta}(Q_T)$, and what I'm really interested in is whether you agree that, for $u$ in this space, $u$ and its first order $x$ and $t$ derivatives are Lipschitz continuous with respect to the metric given by $d((x,t),(x',t')) = |x-x'|^\alpha + |t-t'|^\beta$. I suspect that'll suffice, but will check and post an answer once I'm clearer on your definitions.
Nov 8, 2018 at 0:01 comment added DCM Hi again. I'm not sure I'm quite clear on the definitions of your spaces. Could you perhaps confirm (or give an indication to the contrary) that all the functions in your first space $C^{2+\alpha,\beta}(Q_T)$ have $u, \dfrac{\partial u}{\partial x}, \dfrac{\partial u}{\partial t}\in C^{\alpha,\beta}(\bar Q_T)$? I think this is likely to be all you need (provided $\Omega$ is an interval or similar) to get a compact embedding here, but I'm reluctant to work on an interpolation inequality until I'm clear on what spaces you're considering. Feel free to refer me to a reference if you like :)
Nov 6, 2018 at 22:23 history edited YCor
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Nov 6, 2018 at 21:14 comment added DCM I think you should be ok if $\Omega \subset \mathbb{R}$ is a finite union of bounded open intervals (I think you may want some sort of boundary smoothness for compactness of this sort of embedding in general). I'm not sure you'll find the statement you want explicitly, but you might be able to derive it from suitable interpolation estimates. Might be worth looking what google turns up on interpolation inequalities in parabolic Holder spaces.
Nov 6, 2018 at 19:59 comment added VirgoMath @DCM Yes, $\Omega$ is an open bounded interval. Do you mean that if $Q_T$ be a bounded domain, then the above embedding is compact?
Nov 6, 2018 at 19:47 comment added DCM Do you mean to say that $\Omega$ is a bounded interval here? If so, isn't it the case that all the domains you're considering here are bounded rectangles (where you'll almost certainly have a compact embedding)? Unless I'm missing something, I think you should be able to get what in that setting using something like the interpolation inequality trick described here: books.google.co.uk/…
Nov 6, 2018 at 14:45 review First posts
Nov 6, 2018 at 14:53
Nov 6, 2018 at 14:41 history asked VirgoMath CC BY-SA 4.0