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This might look too an elementary question, but I am confined and is not able to find a textbook which answers the following question.

I have a function $f:{\mathbb R}\rightarrow{\mathbb R}$, such that $f\in L^3({\mathbb R})$ and $$\int\int\frac{|f(y)-f(x)|^3}{|y-x|^4}dydx<\infty.$$ May I conclude that $f\in W^{1,3}({\mathbb R})$ ?

This is a limit case of Sobolev-Slobodeckij space, as $4=1\cdot3+1$. Obviously, the same integral but with exponent $s\cdot3+1$ with $s<1$ is valid, hence $f\in W^{s,3}({\mathbb R})$

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    $\begingroup$ 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 $W^{1,3}(\mathbb{R})$. $\endgroup$
    – Hannes
    Commented Apr 8, 2020 at 9:40
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    $\begingroup$ 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. $\endgroup$
    – Michael Renardy
    Commented Apr 8, 2020 at 19:40
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    $\begingroup$ @MichaelRenardy. Indeed, you are right. I had an exchange, after posting the Q, with Petru Mironescu, who confirmed the fact. $\endgroup$
    – Denis Serre
    Commented Apr 8, 2020 at 20:19
  • $\begingroup$ This implies that (f) is constant (see the paper by Haïm Brezis, “How to recognize constant functions. Connections with Sobolev spaces”). $\endgroup$
    – Jean Van Schaftingen
    Commented Apr 24, 2020 at 13:04

1 Answer 1

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Let summarize the comments there: In order for the seminorm here to be finite, one needs at least $|f(x)-f(y)| = o(|x-y|)$ when $x-y\to 0$, and this is possible only for constants functions. Since $f∈L^3$, $f=0$ (and so $f∈W^{1,3}$ ...).

If to avoid that one uses the second order difference $f(2y-x)-2f(y)+f(x)$ instead of $f(y)-f(x)$, one could however not conclude that $f∈W^{1,3}$, and this is due to the misleading definition of fractional Sobolev(-Slobodeckij) spaces $W^{s,p}$ since when $s>0$ is not an integer $W^{s,p} = B^s_{p,p} = F^s_{p,p}$ (where the $F^s_{p,q}$ are the Triebel-Lizorkin spaces) while $W^{n,p} = F^n_{p,2}$ when $n$ is an integer. An other fractional extension of Sobolev spaces are the Bessel-Sobolev spaces $H^{s,p}$ where the seminorm is the $L^p$ norm of the fractional Laplacian. They verify $H^{s,p} = F^s_{p,2}$ (For every $s≥0$).

All these spaces are ordered in this way when $p≥ 2$ (with strict inclusion when $p>2$) $$ B^s_{p,1} ⊂ B^s_{p,2} ⊂ F^s_{p,2} (=H^{s,p}) ⊂ F^s_{p,p} = B^s_{p,p} ⊂ B^s_{p,\infty}. $$

This is well explained for example in the book of Hans Triebel, Theory of Function Spaces II. Springer Basel, 1992.

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    $\begingroup$ As a heads up, I did not claim hat the integral in the OP is exactly the $B^1_{3,3}$ seminorm. In the integer case, the usual equivalence theorem for the Besov norm (Thm. 2.5.1 in the "old Triebel") would give something like a "centered" term $f(2y-x)-2f(y)+f(x)$ in the numerator. But I figured that finiteness of the OP integral would imply finiteness of the integral with centered term, so the $B^1_{3,3}$ seminorm. $\endgroup$
    – Hannes
    Commented Apr 9, 2020 at 15:07
  • $\begingroup$ Yes you are absolutely right, there are second order differences when $n=1$ (and this is why with first order differences lead to constant functions). I edit my answer to summarize all that. $\endgroup$
    – LL 3.14
    Commented Apr 9, 2020 at 17:57

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