Continuous but not Hölder continuous function on $[0,1]$

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.

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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
@AlexJ. There are several examples on the wikipedia page. – azarel Mar 28 '12 at 19:59

($1$-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.