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A few months ago I posted a question about analytic functions that I received from a bright high school student, which turned out to be studied and resolved by de Bruijn. Based on this positive resolution, I thought I might try my luck again and list three further questions that this student asked which do not seem to be trivially resolvable.

  1. Does there exist a smooth function {f: {\bf R} \rightarrow {\bf R}} which is nowhere analytic, but is such that the Taylor series {\sum_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n} converges for every {x, x_0 \in {\bf R}}? (Of course, this series would not then converge to {f}, but instead to some analytic function {f_{x_0}(x)} for each {x_0}.) I have a vague feeling that perhaps the Baire category theorem should be able to resolve this question, but it seems to require a bit of effort. (Update: answered by Alexander Shaposhnikov in comments.)
  2. Is there a function {f: {\bf R} \rightarrow {\bf R}} which meets every polynomial {P: {\bf R} \rightarrow {\bf R}} to infinite order in the following sense: for every polynomial {P}, there exists {x_0} such that {f^{(n)}(x_0) = P^{(n)}(x_0)} for all {n=0,1,2,\dots}? Such a function would be rather pathological, perhaps resembling a space-filling curve. (Update: solved for smooth {f} by Aleksei Kulikov in comments. The situation currently remains unclear in the general case.)
  3. Is there a power series {\sum_{n=0}^\infty a_n x^n} that diverges everywhere (except at {x=0}), but which becomes pointwise convergent after dividing each of the monomials {a_n x^n} into pieces {a_n x^n = \sum_{j=1}^\infty a_{n,j} x^n} for some {a_{n,j}} summing absolutely to {a_n}, and then rearranging, i.e., there is some rearrangement {\sum_{m=1}^\infty a_{n_m, j_m} x^{n_m}} of {\sum_{n=0}^\infty \sum_{j=1}^\infty a_{n,j} x^n} that is pointwise convergent for every {x}?

Feel free to post answers or other thoughts on these questions in the comments.

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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