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?
1 Answer
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.
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$\begingroup$ Thank you very much Terry Tao. If we suppose $f$ is absolutely continuous and satisfies above inequality, in this case, is it true that $f$ is differentiable and its derivative satisfies $$ |f'(x)-f'(y)|\leq\frac{1}{(\log\frac{1}{|h|})^{\beta}} $$Thanks $\endgroup$– RaviCommented Oct 2, 2013 at 5:36
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$\begingroup$ No, and one can see this by applying appropriate truncation to each of the terms in the previous example (note that absolute continuity is equivalent to the weak derivative of f being in L^1). It would probably be a good exercise for you to work out the details. $\endgroup$ Commented Oct 2, 2013 at 20:08