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Suppose I have $f \in W^{s, p}(B_R)$ for some ball $B_R\subset \mathbb{R}^d$, and for some $s > 0$ and $p > 1$. Is it true that the function ${\rm sgn}(x_1)f$ belongs to some $W^{s', p'}(B_R)$ for some $s' > 0$ and $p' > 1$? (note that ${\rm sgn}(x_1)$ represents the sign function, a one dimensional step function).

I already know that the result is true if $f \in W^{s, p}(\mathbb{R}^d)$ and $g \in W^{1, 1}(\mathbb{R}^d)\cap L^\infty(\mathbb{R}^d)$, but I am not sure about this local case.

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    Can you take $f=1$? – 2017-01-21
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    It obviously holds in this case, I wonder if itis true in the general case. – 2017-01-22
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    If it would hold for $f=1$, the sign function itself would be in some Sobolev space, what is not true, at least for $s>1$, see e.g. the discussion in this question http://math.stackexchange.com/questions/1540688/weak-derivative-in-sobolev-spaces – 2017-01-23
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    Precisely, I am taking fractional Sobolev spaces, I am actually only interested in the case $s \in (0,1)$. – 2017-01-25
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    Do you have any lower bound on the dimension? Can $d$ be equal to one? – 2017-01-25
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    Well, in my case I would like it for $d\geq 3$, but it would also be nice to have it for 1 and 2. – 2017-01-26
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    I don't understand how your statement could be true for $f=1$. Do you have a proof? It would be nice if you could provide at least a sketch for this case in your question (in the spirit of "I have tried this example and it made me think that the statement I am interested in might be true in general")? – 2017-01-26

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