Some remarks on the issue of weakening the continuity assumptions on $f$ to locally integrable, and about generalizing to other exponents of $|h|$.

Keeping track of the big O term in the first part of Terry Tao's proof, we may state it as an a priori bound on the Hölder norm of $f'$: Let $f\in C^1(\mathrm{R})$: 

> If $f$ satisfies, for some $0< \alpha
> \le 1$ and for all $x$ and $h$
> $$|f(x+2h)-2f(x+h)+f(x)|\le
> C|h|^{1+\alpha} \qquad \qquad(1)$$
> 
> then its derivative is
> $\alpha$-Hölder, and in fact
> 
> $$  |f'(y)-f'(x)|\le
> \frac{2^\alpha}{2^\alpha -1 }
> C|h|^{\alpha}\\ .\\ \qquad \qquad(2) $$

The same conclusion holds if we only assume $f\in L^1_{loc}(\mathbb{R})$. Indeed, we may consider the standard approximation of $f$ by convolution, $f_\epsilon:=f*\phi_\epsilon$ with $\phi_\epsilon(x):=(1/\epsilon)\phi(x/\epsilon)$, for  $\phi\in C^\infty_c(\mathbb{R}) _ + $ with $\int_\mathbb{R}\phi\\ dx=1\\ .$  Then the $f_\epsilon$ are in $C^{1,\alpha}(\mathbb{R})$ and satisfy the above hypothesis (1) with the same $C$, so the $f _\epsilon'$ are equicontinuous. By the Ascoli-Arzelà theorem, since $f _\epsilon \to f$ locally uniformly, this is sufficient to conclude that $f$ is also in $C^{1,\alpha}(\mathbb{R})$, with the same bound on the Hölder norm.


For continuous (or just $ L^1_{loc}$, as before) functions $f$, the above hypotesis with $\alpha=0$ defines the Zygmund class. It is well known that a function in this class is "quasi-Lipschitz", i.e. it has modulus of continuity of the form $Kt(\log |t| +1)$, but   may fail to be of bounded variation, and may be differentiable at no point. An example is the [Hardy-Weierstrass function][1], or also the Takagi or blancmange function (see e.g. this [MO question][2]). Note that, for $\sigma=1$ and $C=1$, the Takagi function is the largest $f\in C^0$ in the pointwise order, with $\mathrm{supp}(f)\subset[0,1]$ satisfing (1). Hence, it  produces a (local) modulus of continuity for functions in the Zygmund class, which implies   the "quasi-Lipschitz" property of these functions, since the Takagi function itself is  dominated by a function $Kt (\log|t| +1)$. 

For completeness: the condition (1) for $\alpha > 1$ became trivial: a (locally integrable) $f$ satisfying it is then linear.


[1]:http://en.wikipedia.org/wiki/Weierstrass_function
[2]:http://mathoverflow.net/questions/38751/a-holder-continuous-function-which-does-not-belong-to-any-sobolev-space/38791#38791