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In the Wikipedia article, the general Sobolev inequalities are stated without saying to which spaces the parameters $k$, $p$ and $q$ belong. For the case $k<n/p$ the choose $q$ such that $$\frac1q=\frac1p-\frac kn\;.$$ I guess there is a reason why they don't simply write $$q=\frac{dp}{d-kp}\;.$$ However, the only reason would be that $p$ might be $\infty$; but in that case $k<n/p=0$ would be negative.

So, the simple quesiton is: To which domains belong $k$, $p$ and $q$?

(It's obvious that the authors copied the statements from the book of Evans. Unfortunately, Evans also doesn't specify these domains.)

0xbadf00d
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1 Answers1

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They talk about this earlier in the wikipedia article. See the first paragraph in the Sobolev embedding theorem section.

  • $k$ is a non-negative integer ($0,1,2,3,\dots$) specifying the number of weak derivatives that exist.

  • $q$ and $p$ are a real numbers in $[1,\infty)$ specifying the $L^p$ space that the weak derivatives reside in.

  • $n$ is a positive integer ($1,2,3,\dots$) specifing the spatial dimension of the domain of the function.

Also,

  • Sometimes you can take $p=\infty$ and $q=1$ or vice versa, though there are some subtleties here.

  • Depending on the properties of the domain, fractional derivatives can be defined via the Fourier transform (or wavelet transform, or other techniques). Then typically $k$ can be taken as a real number in the range $[0,\infty)$, rather than an integer in that range.

  • I guess $n=0$ also works (the domain is a single point), though I've never seen this in books.

If it turns out that some of the quantities are not in the appropriate ranges, then there would be no embedding for that choice of parameters.

Nick Alger
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  • I'm only interested in $k\in\mathbb N$ and the domain to be a bounded open subset (not necessarily connected, i.e. not really a domain) $\Lambda$ of $\mathbb R^d$, $d\in\mathbb N$. I just want to know for which $p,q$ is $$W_0^{k,:p}(\Lambda)\subseteq L^q(\Lambda)$$ and for which $p,q$ is $$W_0^{k,:p}(\Lambda)\subseteq C^0(\overline\Lambda);.$$ – 0xbadf00d Jan 26 '17 at 10:04
  • As I said before, I don't understand why they use this strange definition for $q$, since it's obvious that $p$ cannot be $\infty$ in the case where this $q$ is used (otherwise $k<d/p=0$). – 0xbadf00d Jan 26 '17 at 10:21
  • Oh, and what is $\left[;\cdot;\right]$ in the article? It usually denotes the nearest integer function but I guess they mean the floor function here. – 0xbadf00d Jan 26 '17 at 13:37