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 which is not real analytic, but such that all the differences are real analytic for every ?

The hypothesis implies that the Newton quotients are real analytic for every . If analyticity was preserved by smooth limits, this would imply that is real analytic, which would make 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 is periodic, say periodic with period , one can answer the question in the negative by Fourier series. Perform a Fourier expansion . If is not real analytic, then there is a sequence going to infinity such that as . From the Borel-Cantelli lemma one can then find a real number such that (say) for infinitely many , hence for infinitely many . Thus the Fourier coefficients of 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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4 May, 2021 at 1:58 pm

Is there a non-analytic function with all differences analytic? – Mathematics, for all Mankind![…] Is there a non-analytic function with all differences analytic? […]

4 May, 2021 at 2:16 pm

Real-userHi,

This is a very interesting question. What about posting it in math.stackexchange, or mathoverflow?

4 May, 2021 at 6:03 pm

sylvainjulienMaybe a detour via algebraic geometry considering germs of real analytic functions could work? I’m also thinking of drawing a parallel with differential geometry viewing the map that maps f to its finite Newtion quotient of parameter h as an analogue of a tangent vector, and the real analyticity of f emerging as a result of the existence of a connexion.

4 May, 2021 at 6:37 pm

Lior SilbermanLet be as given in the problem. Suppose we could find an analytic function so that . Then would be a periodic function with analytic difference quotients, hence analytic by the solution above and it would follow that is analytic as well.

It thus suffices to solve the following problem: given a real analytic function satisfying an appropriate hypothesis find an analytic function so that .

For a flavour of the hypothesis note that if and is integrable then vanishes on the integers, and is continuous so might possibly have an inverse Fourier transform.

4 May, 2021 at 8:00 pm

Xiaowei XuMaybe this is too elementary and I am sure you have considered this. But we could have worked with the telescoping sum

Then we have . By the hypothesis, each is analytic and have the radius of convergence in an open set of the form . Thus we just need some type of Baire-Category theorem to ensure that . Since is only smooth, we could not really restrict the size of , but shrinking to zero would imply behaves badly near (or positive ) for large enough . Since there is nothing particular about being an integer, this showed should be “almost” non-analytic almost everywhere on large enough values in . However, if is analytic somewhere a translation argument showed this cannot exist. So must be nowhere analytic and its translations also gradually “become worse”. But I am not sure if this is useful for anything.

4 May, 2021 at 8:58 pm

julien PUYDTThe indicator function for the rational numbers, denote it by , has the property that for each rational, is zero hence analytic, even though isn’t even continuous.

Granted, that doesn’t solve the question for all real, but it’s still a nice example, as it’s pretty trivial to explain to a young student and already shows how tricky things can become.

5 May, 2021 at 12:08 am

AnonymousI actually thought of this same example, but didn’t end up writing it because I’m an amateur. Glad to see it actually got some attention :)

I was also wondering about solutions to slight variations of the problem. For example, what if we remove the smoothness condition on , or what if we want that takes values in some set different than ? I was thinking something along the lines of taking values in some set , and then taking to be the indicator function of a set that is invariant by translations under elements of . The case of the above comment would then be .

Yet another interesting variation would be replacing the difference by some other function of and , for example there some or any arbitrary linear combination. I might be completely wrong, but something tells me that if we find an example for one linear combination, then we can easily modify it to an example for most (all?) linear combinations.

5 May, 2021 at 12:54 pm

AnonymousThis idea is applicable to any additive group of real numbers (which may be generated from any set of real numbers – as the smallest additive group containing ).

It is interesting to observe that if is Hamel basis, then the smallest additive group containing is .

7 May, 2021 at 11:32 am

Aditya Guha RoyIf you demand to be analytic for every where is an additive subgroup of but then taking to be the indicator function of suffices, since in this case is clearly not analytic, and also for every and every

5 May, 2021 at 2:16 am

Rafal FilipowThe question considered in the post is connected with the so called “the difference property in the sense of de Bruijn” (see [1], and a very good survey [2]).

We say that a family F of functions has the difference property if for each function f such that f(x+h)-f(x) belongs to F for each h, there are g in F and an additive function A such that f = g + A.

In [1], de Bruijn proved that the family of all analytic functions has the difference property. Consequently, if all differences f(x+h)-f(x) are analytic, there is an analytic g and an additive A with f=g+A. Now, if we assume that f is smooth, then both f and g are bounded on the unit interval [0,1]. It means that the additive function A is bounded on [0,1] as well, so A is in fact linear. All in all, f is analytic as a sum of 2 analytic functions.

Best,

Rafal Filipow

[1] N. G. de Bruijn, Functions whose differences belong to a given class. Nieuw Arch. Wiskunde (2) 23, (1951). 194–218.

[2] M. Laczkovich, The difference property. Paul Erdős and his mathematics, I (Budapest, 1999), 363–410, Bolyai Soc. Math. Stud., 11, János Bolyai Math. Soc., Budapest, 2002.

5 May, 2021 at 5:35 am

itaibnFirst of all, it would be nice to have a short name for the property in the question. Perhaps “difference-analytic”? I suggest that this term is not restricted to smooth functions, but perhaps to arbitrary functions whose differences is analytic, since we might want to prove partial results such as that if such a function is continuous then it is . On the other hand, Julien’s remark suggests that there might exist pretty wild non-continuous difference-analytic functions, so it might be worth excluding them.

I was hoping it might be possible to start with one difference-analytic function , and transform it into another difference-analytic function which is periodic. One approach I considered is by multiplying with an analytic function . Alas, this does not work:

The first term is analytic, but we couldn’t possibly know that the second term is analytic without already knowing that is analytic.

On the other hand, the convolution of with a function of compact support is difference-analytic, as

which is analytic since it is a convolution of an analytic function and a function of compact support, and such a convolution is analytic because a function is analytic iff it has a holomorphic extension. and need to be regular enough so that the convolution makes sense.

5 May, 2021 at 7:33 am

Terence TaoIt occurs to me that the Steinhaus lemma could be useful. For any , the function is analytic, and so obeys bounds for all , all (say), and some constant depending on . As I mentioned in the main post, there is no uniformity assumed on the analyticity and so could behave in a bad way with respect to . However, if for each we let be the set of such that , then the are measurable and exhaust , hence one of the must have positive measure. The Steinhaus lemma then gives us an interval around the origin that lies in . Thus if then there exists such that , hence for any and any we have

and hence by the triangle inequality we now have a uniform bound

whenever , , and . Not sure where to go from here (still not enough uniformity to pass to the limit in the Newton quotient) but it does seem one step closer to making analytic.

5 May, 2021 at 8:28 am

Terence TaoI took a look at de Bruijn’s paper mentioned in Rafal’s answer and it turns out he basically uses the same argument as above (but with Baire category instead of Steinhaus) and completes the argument as follows. For any , the divided difference is of the form for some , where . On the other hand, since is bounded on say , this divided difference is bounded by . From the previous bound and the triangle inequality we now have

for any and , and this gives the analyticity of in this interval, and similarly for the rest of the real line.

Alternatively, one can integrate the inequality

using the fundamental theorem of calculus to conclude that

for either choice of sign , and iterate to obtain a similar estimate.

7 May, 2021 at 8:01 pm

duck_masterI think this basically means that you answered (this high-schooler’s) question. One note: Since we assumed a-priori that $f$ can be essentially an arbitrary function, the notation $f^{(j)}$ doesn’t really make sense unless it’s referring to some kind of divided difference (because $f$ can be arbitrary far from being non-differentiable as many times as you want). I think this invalidates the proof on a surface level but there’s a slightly modified version of it that doesn’t suffer from this problem.

By the way, I’ve never understood how we can get bounds of the type $|f^j|\leq CD^j j!$ for some $C, D$ for an arbitrary true real-analytic function $f$. I understand that any sequence of numbers $(a_j)_{j = 1}^{\infty}$ with bounds $|a_j|\leq CD^j j!$ is trivially the derivative sequence of a real-analytic function near zero by Taylor’s theorem, but I haven’t been able to come up with a converse of this (which would be that the derivative sequence of a real-analytic function can *only* be such a sequence).

7 May, 2021 at 8:08 pm

duck_masterSorry, I don’t know how to write math-that-displays-properly on this comments section, so that’s why there are lots of dollar signs. I think it’s a math tag or something, but I don’t know. You should make it clearer somewhere.

Tests: \frac{1}{2}, \(\frac{1}{2}\), \[\frac{1}{2}\], \begin{equation}\frac{1}{2}\end{equation}, $$\frac{1}{2}$$

Again but with spaces for posterity: \ f r a c { 1 } { 2 } , \ ( \ f r a c { 1 } { 2 } \ ) , \ [ \ f r a c { 1 } { 2 } \ ] , \ b e g i n { e q u a t i o n } \ f r a c { 1 } { 2 } \ e n d { e q u a t i o n } , $ $ \ f r a c { 1 } { 2 } $ $

[Instructions for using LaTeX are at the bottom left of the blog – T.]5 May, 2021 at 8:21 am

Márton ElekesA function class C has the so called difference property, if the following holds: if f(x+h)-f(x) is in C for every h then f = g + A, where g is in C and A is an additive function (i.e. A(x + y) = A(x) + A(y) for every x,y).

In the following paper it was shown that the class of real analytic functions has the difference property:

de Bruijn, N. G.

Functions whose differences belong to a given class.

Nieuw Arch. Wiskunde (2) 23, (1951). 194–218.

This implies that if f is, say, measurable, and the differences f(x+h)-f(x) are all real analytic, then f itself is real analytic.

For more information on the difference property see e.g.

Laczkovich, M. (H-EOTVO-AN)

The difference property. Paul Erdős and his mathematics, I (Budapest, 1999), 363–410,

Bolyai Soc. Math. Stud., 11, János Bolyai Math. Soc., Budapest, 2002.

5 May, 2021 at 10:25 am

Helloif we replace f differentiable or continuous instead of smooth what is the answer?

7 May, 2021 at 12:22 pm

Aditya Guha RoySee section 1.6 from the measure theory textbook by prof. Tao.

5 May, 2021 at 10:37 am

Johan AspegrenThe sums and negations of analytic functions are analytic. From $A- B = A + (-B)$ you can just calculate $f(x+h) – f(x) – (f(x +h) – f(x)) = -2f(x)$ is analytic, thus f(x) is analytic?

5 May, 2021 at 11:29 am

Johan AspegrenSorry, a calculation mistake. $f(x+h) – f(x) – (f(x +h) – f(x)) = 0. $ I don’t know much about analytic functions, but I remember that if it holds on a point it holds in some interval. So $f(0+y) – f(0) +f(0) – f(0-y)= 2f(y) $ is analytic if $f$ is odd. $f(0+y) – f(0) +f(0) – f(0-y)= 0$ is if $f$ is even. Every analytic function is a sum of it’s odd and even parts so $f$ should be analytic because this holds for any $y$.

7 May, 2021 at 9:37 am

Johan AspegrenIt seems that for even part I need this $f(0+y) – f(0) – f(0) + f(0-y)= -2f(0) +2f(y)$. However, W.L.O.G we can assume that $f(0) = 0.$ Thus, $f(0 + h) – f(0) = f(h)$ is analytic is enough.

7 May, 2021 at 7:44 am

DonaldTake a hypothetical such function f. It must clearly be nonanalytic in every open neighbourhood. let k(x)=exp(-1/x) (0 for x<=0). let p(x) be formed by taking the talor expansion of f(x)/k(x) at 1. let P(x)=k(x)p(x). This is an analytic function at every point except 0. So f(x)-P(x) is not analytic, and has all derivatives 0 at 1.

(with the taylor series at point 0 not being effected.) flip this round, swaping 0 and 1. You now have a function F(x) such that F(x) is not analytic, and F(x) has a taylor series of 0 at points 0 and 1.

If you make this cyclic, the analytic difference property should hold. Even at the endpoints, its analytic at both sides, and the taylor series at the sides is the same, so analytic. Thus we have constructed a periodic function with this property. Contradiction.

(Probably an error somewhere here, don't know where.)

7 May, 2021 at 11:03 am

Anonymouswas claimed to be “formed” by taking the Taylor expansion of , the problem is that is not necessarily analytic (i.e. its Taylor series has radius of convergence = 0) – so p(x) is not necessarily well defined on any open interval by its Taylor series.

8 May, 2021 at 4:59 am

Aditya Guha RoyReblogged this on Aditya Guha Roy's weblog.