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Let $b$ be a function and $W^{1,2}$ the first-order Sobolev space on some Euclidean space (edit: so the whole space, but in arbitrary dimension, although $d=1$ would be interesting for me for the beginning as-well, no subsets or whatsoever). If I know that $b W^{1,2} \subseteq W^{1,2}$, what can I conclude about $b$? I also know that $b$ is bounded from above and below, if that helps.

What is clear is that $b$ is weakly differentiable (by multiplying bump-functions with it). I would like to conclude something in the direction (locally) Lipschitz. Any ideas?

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  • $\begingroup$ You also know that $b$ is in $W^{1,2}_{loc}$ also by bump functions. When you say "on some Euclidean space", do you mean all $\mathbb{R}^n$ or a bounded subset of it? $\endgroup$
    – Laithy
    Dec 13, 2021 at 14:17
  • $\begingroup$ @Laithy Yes, locally in $W^{1,2}$ is also immediate. With "on some" I meant that the dimension is arbitrary, though I would also be interested on results in $d=1$ if this is easier (which I could well imagine). Subsets are not interesting for me at the moment. $\endgroup$ Dec 13, 2021 at 16:17
  • $\begingroup$ I think in $d=1$ and ignoring issues at infinity, the answer should be exactly $H^1$ again (basically because the product rule suggest the existence of a derivative in $L^2$). $\endgroup$ Dec 13, 2021 at 16:41
  • $\begingroup$ @ChristianRemling ignoring issues at infinity more or less just means $W^{1,2}_{loc}$, doesn’t it? $\endgroup$ Dec 13, 2021 at 16:53
  • $\begingroup$ @SebastianBechtel: Yes, or assume that all functions are compactly supported. (This is not literally what you asked, but if my idea is correct, then it does refute Lipschitz continuity.) $\endgroup$ Dec 13, 2021 at 17:14

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