Do the real-valued functions of bounded variation on $[0,1]$ belong to some Sobolev/Besov class?
What about a fractal, such as the Weierstrass function?
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asked Nov 24, 2020 at 17:37
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2$\begingroup$ BV functions are certainly in some Sobolev spaces: $f'$ (distributional derivative) is a (signed) measure, so $t\widehat{f}(t)\in L^{\infty}$. (In other words, $f\in H^s$ for $s< 1/2$.) $\endgroup$– Christian RemlingCommented Nov 24, 2020 at 18:57
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$\begingroup$ Thank you. What about fractals? $\endgroup$– Aryeh KontorovichCommented Nov 24, 2020 at 19:05
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$\begingroup$ @ChristianRemling: that's strict inclusion, right? So technically there is also the "trivial" answer that BV includes into L^p for all p. And by Sobolev embedding you also get that there are BV functions not in $W^{s,p}$ for any $s > 1/p$. $\endgroup$– Willie WongCommented Nov 24, 2020 at 19:12
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$\begingroup$ If you make this into an answer, I’ll be happy to accept. $\endgroup$– Aryeh KontorovichCommented Nov 24, 2020 at 19:25
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1 Answer
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(Summary post of comments)
BV functions are bounded, and hence trivially in any $L^p$. (Special case of Sobolev/Besov spaces.)
The distributional derivative $f'$ is a signed measure, so $t \hat{f}(t)\in L^\infty$. Also $f$ is bounded, so $\hat{f} \in L^\infty$. So we can in fact conclude that $f\in W^{s,p}$ for every $p \geq 2$ and $s < \frac{1}{p}$.
If $f\in W^{s,p}$ for $p\in [1,\infty]$ and $s > 1/p$ then by Sobolev embedding we have that $f$ is continuous, and thus there exists $f$ in BV that is not in $W^{s,p}$.
The usual definition of the Weierstrass function as $W(x) = \sum a^n \cos(b^n \pi x)$ where $b$ is an odd natural and $a\in (0,1)$ satisfy $ab > 1 +\frac32 \pi$, can be seen to be in $H^s$ for any $s$ satisfying $ab^s < 1$.
For the $b = 7$ case, this means that if you take any $s \in (0,1 - \log_7 (1 + \frac32 \pi))$ you can find an appropriate $a$ such that the Weierstrass function is in $H^s$.
answered Nov 24, 2020 at 20:34