Is there a continuous function $f$ satisfying the following Zygmund condition but not differentiable.

Question

Suppose that a continuous function $f$ on the line and satisfies $$ |f(x+2h)−2f(x+h)+f(x)|\leq const \frac{|h|}{(\log\frac{1}{|h|})^{\beta}}\,\,\,\,\,\,\text{where}\,\,\,\, \beta \in(0, 1] $$ for all $x,h$ real. Is it true that $f$ is differentiable? If not how can I prove it?

Answer 1

The Weierstrass-type function

$$ f(x) := \sum_{n=1}^\infty \frac{1}{2^n n^\beta} \sin(2^n x) $$

will obey the hypotheses, yet fails to be differentiable at the origin.

In general, one should look to Weierstrass-type functions (perhaps weighted by power weights $|x|^{-\alpha}$ or variants such as $|x|^{-\alpha} \log(\frac{1}{|x|})^{-\gamma}$, in case some $L^p$ norms are involved for a finite $p$) as key test cases for these sorts of endpoint functional embedding problems. It is also quite clarifying to reformulate the hypotheses and conclusion in terms of Littlewood-Paley theory (either the classical theory using harmonic extensions, or the modern theory using smooth partitions of unity in frequency space; see e.g. Stein's "Singular integrals" for the former, or the appendix to my PDE book for the latter). For instance, the hypothesis here is basically equivalent to the Besov-type bound

$$ \| P_N f \|_\infty \ll N^{-1} \log^{-\beta} N$$

for frequencies $N \gg 1$, where $P_N$ is a smooth Fourier projection to frequencies $|\xi| \sim N$, while the conclusion is roughly equivalent to the pointwise convergence of the series $\sum_N N P_N f(x)$ as $N$ ranges over dyadic integers. As $\sum_N \log^{-\beta} N$ diverges for $\beta \leq 1$, this indicates that the claim is false, and then guided by this analysis one can quickly come up with the aforementioned Weierstrass-type counterexample.