TITLE: Second order difference implies differentiability
QUESTION [8 upvotes]: Suppose that a function $f$ on the line satisfies $|f(x+2h)-2f(x+h)+f(x)|\le |h|^{3/2}$ for all $x,h$ real. Is it true that $f$ is differentiable and its derivative satisfies
$|f'(x+h)-f'(x)|\le c |h|^{1/2}$ for all $x,h$?

REPLY [3 votes]: 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 $$|f'(y)-f'(x)|\le\frac{C}{2^\alpha -1 } |h|^{\alpha}\, .\,\qquad \qquad(2) $$

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