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I was asked the following interesting question from a bright high school student I am working with, to which I did not immediately know the answer:

Question 1 Does there exist a smooth function {f: {\bf R} \rightarrow {\bf R}} which is not real analytic, but such that all the differences {x \mapsto f(x+h) - f(x)} are real analytic for every {h \in {\bf R}}?

The hypothesis implies that the Newton quotients {\frac{f(x+h)-f(x)}{h}} are real analytic for every {h \neq 0}. If analyticity was preserved by smooth limits, this would imply that {f'} is real analytic, which would make {f} real analytic. However, we are not assuming any uniformity in the analyticity of the Newton quotients, so this simple argument does not seem to resolve the question immediately.

In the case that {f} is periodic, say periodic with period {1}, one can answer the question in the negative by Fourier series. Perform a Fourier expansion {f(x) = \sum_{n \in {\bf Z}} c_n e^{2\pi i nx}}. If {f} is not real analytic, then there is a sequence {n_j} going to infinity such that {|c_{n_j}| = e^{-o(n_j)}} as {j \rightarrow \infty}. From the Borel-Cantelli lemma one can then find a real number {h} such that {|e^{2\pi i h n_j} - 1| \gg \frac{1}{n^2_j}} (say) for infinitely many {j}, hence {|(e^{2\pi i h n_j} - 1) c_{n_j}| \gg n_j^2 e^{-o(n_j)}} for infinitely many {j}. Thus the Fourier coefficients of {x \mapsto f(x+h) - f(x)} do not decay exponentially and hence this function is not analytic, a contradiction.

I was not able to quickly resolve the non-periodic case, but I thought perhaps this might be a good problem to crowdsource, so I invite readers to contribute their thoughts on this problem here. In the spirit of the polymath projects, I would encourage comments that contain thoughts that fall short of a complete solution, in the event that some other reader may be able to take the thought further.

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