Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does there exist a continuous function $F$ on $[0,1]$ which is not Hölder continuous of order $\alpha$ at any point $X_{0}$ on $[0,1]$. $0 < \alpha \le 1$.

I am trying to prove that such a function does exist. also I couldn't find a good example.

share|improve this question
1  
Such functions certainly exist. In fact, almost all continuous functions in the Baire category sense (using sup norm on the space of continuous functions) fail to have a positive pointwise Holder condition at each point, a result that was proved by Auerbach and Banach in 1931 [Uber die Höldersche Bedingung, Studia Mathematica 3, pp. 180-184]. I don't know of a "formula" for such a function off-hand, however. –  Dave L. Renfro Mar 28 '12 at 19:56
2  
@AlexJ. There are several examples on the wikipedia page. –  azarel Mar 28 '12 at 19:59

1 Answer 1

($1$-dimensional) Brownian motion is almost surely continuous and nowhere Hölder continuous of order $\alpha$ if $\alpha > 1/2$. IIRC one can define random Fourier series that will be almost surely continuous but nowhere Hölder continuous for any $\alpha > 0$.

EDIT: OK, here's a construction. Note that $f$ is not Hölder continuous of order $\alpha$ at any point of $I = [0,1]$ if for every $C$ and every $x \in I$ there are $s,t \in I$ with $s \le x \le t$ and $|f(t)-f(s)|>C(t-s)^\alpha$.

I'll define $f(x) = \sum_{n=1}^\infty n^{-2} \sin(\pi g_n x)$, where $g_n$ is an increasing sequence of integers such that $2 g_n$ divides $g_{n+1}$. This series converges uniformly to a continuous function. Let $f_N(x)$ be the partial sum $\sum_{n=1}^N n^{-2} \sin(\pi g_n x)$. Note that $\max_{x \in [0,1]} |f_N'(x)| \le \sum_{n=1}^N n^{-2} \pi g_n \le B g_N$ for some constant $B$ (independent of $N$).

Now suppose $s = k/g_N$ and $t = (k+1/2)/g_N$ where $k \in \{0,1,\ldots,g_N-1\}$. We have $f(s) = f_{N-1}(s)$ and $f(t) = f_{N-1}(t) \pm N^{-2}$. Now $|f_{N-1}(t) - f_{N-1}(s)| \le B g_{N-1} (t-s) = B g_{N-1}/(2 g_N)$, so $|f(t) - f(s)| \ge N^{-2} - B g_{N-1}/(2 g_N)$. The same holds for $s = (k+1/2)/g_N$ and $t = (k+1)/g_N$. So let $g_n$ grow rapidly enough that $g_{n-1}/g_n = o(n^{-2})$ ($g_n = (3n)!$ will do, and also satisfies the requirement that $2g_n$ divides $g_{n+1}$). Then for every $\alpha > 0$, $(t-s)^\alpha = (2g_N)^{-\alpha} = o(N^{-2}) = o(|f(t) - f(s)|)$. Since for each $N$ the intervals $[s,t]$ cover all of $I$, we are done.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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