Hölder continuity of Weierstrass Function

Question

Let us fix some notations, we will call Weierstrass function a function such that:

$$f(x)=\sum_{i=0}^{n}a^{n}\cos(b^{n}x) $$ with $0<a<1, b>1 \text{and} \ ab>1$.

Wondering whether a function of this type is Hölder continuous or not, I came up with a wikipedia statement claiming that such a function is Hölder continuous with exponent $-\log_{b}a$. After trying to prove that using some Prostaphaeresis formulas and giving a good estimate of the tail, I found a solution here on this forum. The argument used its pretty basic, but there are some problems, from now on I will use the notation used in the second answer of the previous link.

The choice of $m$ is not right, this can be seen using $b=3$ and $h=10$, a good constraint is $b^{m-1} |h| < 1 < b^{m} |h|$ this leads us to the inequality $b^{m}|h| < b$ instead of having $2$ on the right and side, but that's not the problem.

The main problem is the inequality $a^{m} < |h|^{-\log_{b}a}$, in fact from $ab>1$ follows that $ab^{m}>1$ and $m>-\log_{b}a$, moreover $ab^{m-1}>ab>1>b^{m-1}|h|$ so that $a>|h|$ and I'm not able from this to deduce if the inequality $a^{m} < |h|^{-\log_{b}a}$ holds true or not.

Any suggestions?

Another way to approach this is to use the Littlewood-Paley characterization of Holder spaces. — Jose27, May 12 '18 at 06:37
Grafakos's "Modern Fourier Analysis" has this and is pretty accessible. — Jose27, May 18 '18 at 19:37

score 1 · Accepted Answer · edited May 12 '18 at 04:52

1

It does hold. Note that it suffices to show that $$m\log_ba<(-\log_ba)(\log_b|h|),$$ that is, $m>\log_b(1/|h|)$. This, however, is a direct consequence of the inequality $b^m|h|>1$, which holds, as you noted above.

edited May 12 '18 at 04:52

answered May 11 '18 at 16:55

George K

492

Hölder continuity of Weierstrass Function

1 Answers1