Can a function be continuous but not Hölder on a compact set? Is it possible to construct a function $f: K \to \mathbb{R}$, where $K \subset \mathbb{R}$ is compact, such that  $f$ is continuous but not Hölder continuous of any order?  It seems like there should be such a function--it would probably oscillate wildly, like the Weierstrass-Mandlebrot function.  However, the W-M function itself doesn't work, since it is Hölder.
Edit: I guess I did have in mind for the function to not be Hölder anywhere, even though I didn't explicitly say so.
 A: Quoting Wikipedia: 

The function $f(x) = x^β$ (with $β ≤ 1$) defined on $[0, 1]$ serves as a prototypical example of a function that is $C^{0,α}$ Hölder continuous for $0 < α ≤ β$, but not for $α > β$. 

Of course only the part of $x \mapsto x^\beta$ near $0$ is responsible for the failure of it being $C^{0,\alpha}$ when $\alpha > \beta$.
Now take $f_n: [0,1] \to \mathbb R$ to be suitably scaled, translated and truncated version of $x^{1/n}$ such that it's supported on $[\frac1{n+1},\frac1n]$.  Now consider $f = \sum_{n\geq2} f_n$.
A: Consider the function $f\colon[-\pi,\pi]\to\mathbb R$, $f(0)=0$ and $f(x)=1/(\ln\lvert x\rvert-\ln2\pi)$. It's continuous, but not Hölder of any order since the Dini integral
$$\int_0^h\frac{\lvert f(t)+f(-t)-2f(0)\rvert}tdt=2\int_0^h\frac{dt}{t(\ln t-\ln2\pi)}$$
diverges.
The preceding example is extracted from an example of the fact that Dini criterion and Dirichlet-Jordan criterion for convergence of Fourier series are incomparable.
I hope if someone could give a continuous function $f\colon[-1,1]\to\mathbb R$ such that the Dini integral $\int_0^\epsilon\lvert\phi_x(t)\rvert dt/t$ diverges for some small $\epsilon>0$ and all $x\in(-1,1)$, where $\phi_x(t)=f(x+t)+f(x-t)-2f(x)$.
Maybe the totality of continuous functions whose Dini integral converges at some $x$ is meager, i.e. a countable union of nowhere dense subsets of $C[-1,1]$.
A: Take a look at the functon
$$f(x) = \sum_{n=0}^\infty {\sin(2^n x)\over n^2}.$$
This function is continuous on the line.  Much ugliness ensues when you try to show it's Hölder continuous.
A: No oscillation needed: the function $$f(x) = \begin{cases} 1/\log x,\quad &x\in (0,1/2], \\ 0,\quad & x=0\end{cases}$$
fits the bill. 
Indeed, $f(x)/x^\alpha\to \infty$ as $x\to0$, for any positive $\alpha$.
