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?



BV spaces and fractals -- are they Sobolev? Besov?
(Summary post of comments)


1. BV functions are bounded, and hence trivially in any $L^p$. (Special case of Sobolev/Besov spaces.)
2. 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 $s < \frac{1}{p}$.
3. 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 $W^{s,p}$.
4. 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 $H^s$.


