I was asked the following interesting question by a bright student I work with that I didn’t know the answer to right away:

Question 1 Is it working properly? {f: { bf R}  rightarrow { bf R}} This is not really analytical, but such that there are all differences {x  mapsto f (x + h) - f (x)} are real analytical for everyone {h  in { bf R}}?

The hypothesis implies that the Newton’s quotient { frac {f (x + h) -f (x)} {h}} are real analytical for everyone {h  neq 0}. If the analyticity were preserved by smooth boundaries, this would imply {f '} is really analytical of what would do {f} real analytical. However, we are not assuming any uniformity in the analysis of Newton’s quotients, so this simple argument does not seem to solve the question immediately.

In the event that {f} is periodic, let’s say periodic with a point {1}one can answer the question in the negative by Fourier series. Do a Fourier expansion {f (x) =  sum_ {n  in { bf Z}} c_n e ^ {2  pi i nx}}. If {f} is not really analytical, then there is a sequence {NJ} go to infinity, so that {|  c_ {n_j} |  = e ^ {- o (n_j)}} how {j  rightarrow  infty}. A real number can then be found from the Borel-Cantelli lemma {H} so that {|  e ^ {2  pi ih ​​n_j} - 1 |   gg  frac {1} {n ^ 2_j}} (say) for an infinite number {j}therefore {|  (e ^ {2  pi ih ​​n_j} - 1) c_ {n_j} |   gg n_j ^ 2 e ^ {- o (n_j)}} for an infinite number {j}. Thus the Fourier coefficients are of {x  mapsto f (x + h) - f (x)} Do not decay exponentially and so this function is not analytical, a contradiction.

I was unable to quickly resolve the non-periodic case, but I thought this might be a good problem for Crowdsource, so I invite readers to share their thoughts on this issue here. In the spirit of the Polymath projects, I encourage comments that contain thoughts that do not correspond to a complete solution, in case another reader may be able to advance the thought.


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