Second order difference implies differentiability

Question

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 $x,h$?

Answer 1

As Gerald says, the answer is no without further hypotheses on f. But if one makes some minimal additional regularity hypotheses on f, such as continuity, then the answer is yes.

Write $D_h f(x)$ for the difference quotient $D_h f(x) := (f(x+h)-f(x))/h$, then the hypothesis is that $$ D_h f(x+h) = D_h f(x) + O( |h|^{1/2} )$$ for all $x, h$ (with $h$ nonzero), which implies $$ D_{2h} f(x) = D_h f(x + ih) + O( |h|^{1/2} )$$ for $i=0,1$. Iterating this we have $$ D_{2^j h} f(x) = D_h f(x + ih) + O( 2^{j/2} |h|^{1/2} ) \qquad (1)$$ for natural numbers $j$ and any integer $0 \leq i < 2^j$, which in particular implies $$ D_h f(x+ih) = D_h f(x) + O( |ih|^{1/2} ) \qquad (2)$$ for all integer $i$; in particular, we have $$ D_{(y-x)/n} f(y) = D_{(y-x)/n} f(x) + O( |y-x|^{1/2} )$$ for any distinct $x,y$ and natural number $n$. This already gives the derivative bound $|f'(y)-f'(x)| = O( |y-x|^{1/2} )$ if $f$ is differentiable.

To establish differentiability, we return to (1), which (in combination with (2)) gives $$ f(x+(i+2^j)h) - f(x+ih) = 2^j h D_h f(x) + O( |2^j h|^{3/2} )$$ whenever $i = O( 2^j )$. Telescoping this using binary expansion we see that $$ f(x+nh) - f(x) = n h D_h f(x) + O( |nh|^{3/2} )$$ or equivalently $$ D_{nh} f(x) = D_h f(x) + O( |nh|^{1/2} )$$ for any integer $n$ (not necessarily a power of two), and thus $$ D_{h} f(x) = D_{h/n} f(x) + O( |h|^{1/2} )$$ for any non-zero $h$ and nonzero integers $n$. In particular $$ D_{h} f(x) = D_{h'} f(x) + O( |h|^{1/2} + |h'|^{1/2} )$$ whenever $h,h'$ are nonzero rational (as then we can write $h = n h'', h' = n' h''$ for some nonzero integers $n,n'$ and some nonzero $h'$); by continuity of $f$, this is also true for nonzero real $h,h'$. Thus $D_h f$ is a Cauchy sequence as $h \to 0$, giving differentiability.

It is likely that one can also relax continuity to Lebesgue measurability (it seems that the above argument gives almost everywhere differentiability or something very close to this, in which case some version of the fundamental theorem of calculus should then finish the job).

Answer 2

I am assuming you mean $|f(x+2h)-2f(x+h)+f(x)|\le |h|^{3/2}$.

Well, what if $f$ satisfies $f(x+y)=f(x)+f(y)$ for all $x,y$? Certainly your inequality holds then. But (according to the Axiom of Choice) there are badly discontinuous functions like this.

Answer 3

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.