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No. It is not true.

However, I was wondering under what conditions it is true.

I think that if $s>n/2$ and $1/f$ is bounded the result holds. Is it true?

Edit. If the domain is unbounded the set described may be empty as shown by TZakrevskiy's answer. Therefore, I am focusing on compact or bounded domains.

Adrian Keister
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yess
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  • Yes, the composition $\phi\circ f$ should be as good as $f$ is, provided that $\phi$ is smooth with bounded derivatives on the range of $f$ (which is the case for $\phi(t)=1/t$ under your assumption). –  May 16 '18 at 22:23

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The obvious big question - can you find $f\in H^s$ such that $1/f$ is bounded?

Consider $\Bbb R$: if $1/f \in L^\infty(\Bbb R)$, then $|f|$ is bounded from below by $\frac{1}{vrai\sup |1/f|}>0$, therefore $f \notin L^s(\Bbb R)$ for $s\ge 1$.

Unless we answer these considerations rigorously, I suspect that your hypothesis describes an empty set of functions.

TZakrevskiy
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    Thanks, I would be happy to consider bounded domains. I will edit the question accordingly. – yess May 16 '18 at 13:02