# Multiplier on a Sobolev space

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?

• 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? Dec 13, 2021 at 14:17
• @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. Dec 13, 2021 at 16:17
• 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 $L^2$). Dec 13, 2021 at 16:41
• @ChristianRemling ignoring issues at infinity more or less just means $W^{1,2}_{loc}$, doesn’t it? Dec 13, 2021 at 16:53
• @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.) Dec 13, 2021 at 17:14