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Apr 24, 2020 at 13:04 comment added Jean Van Schaftingen This implies that (f) is constant (see the paper by Haïm Brezis, “How to recognize constant functions. Connections with Sobolev spaces”).
Apr 8, 2020 at 20:19 comment added Denis Serre @MichaelRenardy. Indeed, you are right. I had an exchange, after posting the Q, with Petru Mironescu, who confirmed the fact.
Apr 8, 2020 at 19:40 comment added Michael Renardy If f is smooth with nonzero derivative, the integrand is of order $1/|x-y|$, which is not integrable. I suspect the condition implies that f is constant.
Apr 8, 2020 at 17:24 answer added LL 3.14 timeline score: 3
Apr 8, 2020 at 9:40 comment added Hannes If I see it correctly, then by the "modulus of continuity-characterization" of Besov spaces, $f$ will lie in $B^1_{3,3}(\mathbb{R})$. Unfortunately, this a slightly larger space than $W^{1,3}(\mathbb{R})$.
Apr 7, 2020 at 10:08 history edited YCor
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Apr 7, 2020 at 8:41 history asked Denis Serre CC BY-SA 4.0