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.
 A: ($1$-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.
