5
$\begingroup$

We say a non-constant function $f$ on $[0, 1]$ is singular if it is continuous, and in addition differentiable almost everywhere with $f' = 0$ a.e.

For every positive $\alpha < 1$, is the set of singular functions dense in the space of Holder continuous functions of order $\alpha$? Where we equip the space with the norm

$$\|f\|_{C^\alpha} := \sup|f| + \sup_{x, y \in [0, 1]} \frac{|f(y) - f(x)|}{|y - x|^\alpha}.$$

Comments: The prototypical example of a singular function is the Cantor function, which is Hölder continuous of order $\frac{\log 2}{\log 3}$.

$\endgroup$
3
  • 1
    $\begingroup$ If you consider the space of all Hölder functions, then that's not separable, so it's very unlikely that such a statement holds. If you take the completion of smooth functions under the Hölder norm, then the answer is yes and it's an easy approximation argument by pieces of Cantor function. $\endgroup$ Commented May 30 at 12:41
  • 1
    $\begingroup$ @MartinHairer Hm, why is seperability a necessary condition for the statement to hold? The space of singular functions is surely not countable. $\endgroup$
    – Nate River
    Commented May 30 at 12:42
  • $\begingroup$ It didn't say it's necessary, just very unlikely based on experience. I would be happy to learn something by being proven wrong... $\endgroup$ Commented May 30 at 19:39

1 Answer 1

5
$\begingroup$

No. Take $f(t) = \sum_{n=0}^\infty 2^{-\alpha 2^n}\cos(2^{2^n}t)$. This is $\alpha$-Hölder and there exists $c>0$ such that, for every point $t$, one has $\limsup_{s \to t} |t-s|^{-\alpha}|f(t)-f(s)| > c$. Take any "singular" functions $g$. Since there is some point $t$ such that $g'(t) = 0$, it follows from the property of $f$ that $\|f-g\|_\alpha \ge c$.

$\endgroup$
3
  • $\begingroup$ That’s an interesting looking counterexample… could you give a short sketch as to why the limsup is greater than $c$ uniformly? $\endgroup$
    – Nate River
    Commented May 30 at 21:00
  • $\begingroup$ … although it would suffice to show the limsup is bigger than $0$ on a set of nonzero measure too. $\endgroup$
    – Nate River
    Commented May 30 at 21:06
  • $\begingroup$ @NateRiver Just consider $|t-s|$ of order $2^{-2^n}$ for large $n$. $\endgroup$ Commented May 30 at 21:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.