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Suppose we have two real-valued functions $f(x)$ and $g(x)$, both equal to their Newton series expansion:

$$f(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k f\left (0\right)$$

$$g(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k g\left (0\right)$$

Is their composition $F(x)=f(g(x))$ also equal to its Newton series expansion (if it converges)?

$$F(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k F\left (0\right)$$

Anixx
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    Please use at least one top-level tag (those with a two-letter prefix). –  Jun 15 '14 at 12:45
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    This question is cross-posted from M.SE - please add the link to that question. –  Jun 15 '14 at 13:20
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    Really, what's the point of asking all questions of the type "Does the class of functions representable by their Newton series possess natural property P?" in some random order? The answer to this one is "No, as usual". Take any entire $f$ of order less than $1$ that vanishes at every square of an integer and compose with $g(x)=x^2$. – fedja Jun 15 '14 at 18:13
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    @fedja interesting, but at what conditions it holds then? – Anixx Jun 15 '14 at 18:40
  • @fedja by the way, another question is, will a non-zero function which vanishes at any square of integer have its Newton series converging? – Anixx Jun 15 '14 at 18:46
  • I think another example should be :$g(z)=-z$, and a (non identically zero) $f$ everywhere represented by its Newton series, and vanishing at all negative integers. So f(g(z))=f(-z) has a Newton series with all zero coefficients. – Pietro Majer Jun 15 '14 at 18:58
  • Any entire function of order less than $1$ has convergent Newton series, hasn't it? @Pietro Majer It should if you can produce such a function. ;) – fedja Jun 15 '14 at 19:16
  • @fedja $\sin (\frac{\pi x}{2})$ is entire function? Yes. I suppose its order is less than 1. It does not have convergent Newton series. See here: http://mathoverflow.net/questions/99166/convergence-of-newton-series-for-sin-ax – Anixx Jun 15 '14 at 19:37
  • The order of $\sin$ is $1$. Don't confuse with type! Order less than $1$ means $|f(z)|\le Ce^{|z|^p}$, $p<1$. – fedja Jun 15 '14 at 20:09
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    @fedja how do u know that the Newton series for such function converges? – Anixx Jun 15 '14 at 20:41
  • @fedja if you have a link to this property as you stated, this would be a great answer to this question, at least the first part: http://mathoverflow.net/questions/71206/discrete-analytic-functions – Anixx Jun 15 '14 at 20:47
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    I do not have a link, but I'll post the proof later today or tomorrow (unfortunately, it won't fit into the comment window, so I'll do it as an answer). – fedja Jun 16 '14 at 01:05
  • @fedja: what I had in mind is that for an everywhere absolutely convergent Newton series $f$, vanishing at all negative integers is equivalent, for the power series $F\in C^\infty([-1,1])$ with the same coefficients, having all derivatives vanishing at $-1$. It seems that the condition for absolute convergence for $f$ does not imply quasi-analyticity on $[-1,1]$ for the power series $F$, by Denjoy-Carleman condition. That's why I think there is such a Newton series. I did not try to fix the details after your comment with the simple counterexample, but maybe I could, and in case post. – Pietro Majer Jun 17 '14 at 11:08
  • @Pietro Majer Indeed, though the construction of a function like that is itself quite a non-trivial exercise. Thanks for the explanation :-) – fedja Jun 17 '14 at 18:51
  • @fedja What's up with the proof? Can we resonably hope to see it? – Anixx Jun 17 '14 at 18:57
  • @Anixx You can hope no longer: it is here already :-) – fedja Jun 17 '14 at 19:20
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    Regarding the function fedja asks for: Take $f(z) = \sqrt{z} \sin (2 \pi \sqrt{z})$. This is entire, since it is easy to check that $\sum_{n=0}^{\infty} (-1)^n (2 \pi)^{2n+1} z^{n+1}/(2n+1)!$ is everywhere convergent; we have the easy bound $|\sqrt{z} \sin(2 \pi \sqrt{z})| < \sqrt{|z|} e^{2 \pi \sqrt{|z|}}$; and $f$ clearly vanishes at squares of integers. – David E Speyer Jun 17 '14 at 20:01

2 Answers2

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Here goes, as promised.

Let $f$ be entire of order less than $1$, so $|f(z)|\le Ce^{|z|^p}$, $p<1$. Write the Newton polynomial $$ P(x)=\sum_{k=0}^n\Delta^kf(0) {x \choose k} $$ Note that $g(k)=f(k)-P(k)=0$ for $k=0,1,\dots,n$. On the other hand, we can crudely estimate $|g|$ in a disk of radius $R>2n$ by $Ce^{R^p}+\sum_{k=0}^n(2R)^k\frac 1{k!}|\Delta^k f(0)|$. Now, $\frac 1{k!}|\Delta^k f(0)|\le \max_{[0,R/2]}\frac{|f^{(k)}|}{k!}\le (2/R)^k Ce^{R^p}$ by Cauchy, so we finally get $$ |g|\le C 4^n e^{R^p} $$ in the disk of radius $R$ centered at the origin.

Now, for $|x|<n$, each corresponding Blaschke factor $\frac{R(x-k)}{R^2-kx}$ is at most $\frac{3n}R$ in absolute value, so $$ |g(x)|\le C\left(\frac{12}{R/n}\right)^n e^{R^p} $$ Choosing $R=n^{1/p}$, we get $|g(x)|\le \left(12en^{-\frac{1-p}p}\right)^n\to 0$ as $n\to\infty$.

fedja
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  • I have accepted this answer, although I would prefer it be posted in the linked question. Can you please tell, did you derive this theorem yourself or had you seen it before? – Anixx Jun 17 '14 at 20:22
  • Sorry I would not ask this, should I know that you're a renowned mathematician. Still your proof is explained in too difficult language at least for me to comprehend. – Anixx Jun 17 '14 at 21:17
  • @Anixx Neither, as usual. The correct statement is that I've seen enough stuff before to put it together in a few minutes from readily available blocks. As to "renowned mathematician", that is an obvious overstatement. "Someone who got a decent education as a student and still remembers about half of it" is a more accurate description. :-) – fedja Jun 17 '14 at 21:22
  • I doubt that is accurate for someone who has theorems in textbooks named after him. – Anixx Jun 17 '14 at 21:23
  • So it is good to know that this somehow touches your prevuous research. – Anixx Jun 17 '14 at 21:24
  • @Anixx I believe I have directly estimated the difference between the function and the $n$-th partial sum of its Newton series in the disk $n\mathbb D$, haven't I? – fedja Jun 20 '14 at 19:48
  • Oh sorry, I see – Anixx Jun 20 '14 at 20:22
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Another counter-example is extractable from Gerald Edgar's answer to this question, where he shows that $\sin (ax)$ is discrete analytic for $a \in (-\pi/3, \pi/3)$. So take $ f(x) = \sin ((\pi/4) x)$ (for example) and $g(x) = 9 x$ (for example). Then $f$ and $g$ are discrete analytic but the Newton series for $(f \circ g)(x) = \sin ((9 \pi/4) x)$ will converge to $\sin( (\pi/4) x)$, which equals $\sin ((9 \pi/4) x)$ for integer $x$, but not in general.

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David E Speyer
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