1
$\begingroup$

Good morning everybody. My question is as follows:

Let $K$ be a compact subset of $\mathbb R$ and let $v\in C^\infty(K)$.

Consider the finite difference operator $\Delta v(x)\doteq v(x+1)-v(x)$. It is known http://en.wikipedia.org/wiki/Indefinite_sum, that we can invert this operator $\Delta$ but the outcome is a family of functions which differ one from another by a periodic of modulo one function $p(x)=p(x+1)$.

It happens that in many cases e.g. if a Newton series exists that $p$ is actually a constant function.

Now the big issue here is to decide whether a Newton series exists which would be wonderful for my purposes. I am asking if this is ensured by the compactness of the set: to be more specific, we know that the Newton series holds for polynomials, and here maybe the result extends to smooth functions by some densities theorems like Stone Weierstrass or something along the same lines.

Is this result true? Can anybody provide me with a reference or a short proof if this is available?

I'm quite unused to these topics so any help is greatly appreciated.

Thanks again for the kindness and goodbye.

Improve this question
$\endgroup$
4
  • $\begingroup$ If $x\in M$, what is $x+1$? Are you perhaps on a geodesic, where $v(\gamma(t+1))-v(\gamma(t))$ makes sense? $\endgroup$
    – Joonas Ilmavirta
    Commented Aug 19, 2014 at 15:12
  • $\begingroup$ @JoonasIlmavirta I've edited the question. Let us assume that we are in a compact set in $\mathbb R$ and sorry for the confusion. Thank you $\endgroup$
    – user17697
    Commented Aug 19, 2014 at 15:16
  • 1
    $\begingroup$ What if $K=[0,\frac12]$? Then $\Delta v(x)$ is never defined, since either $x$ or $x+1$ is outside $K$. Try to formulate the question more carefully. $\endgroup$
    – Joonas Ilmavirta
    Commented Aug 19, 2014 at 15:19
  • $\begingroup$ @user17697 Maybe you want $\Delta$ as an operator $C^\infty(\mathbb{R})\to C^\infty(K)$ (by restriction) ? $\endgroup$
    – Pietro Majer
    Commented Sep 12, 2016 at 19:37

1 Answer 1

Reset to default
1
$\begingroup$

if a Newton series exists that p is actually a constant function.

Not exactly so. The solutions are always a set of functions different by a constant term. But among these solutions the one which is equal to its Newton series is unique.

This is due to the Carlson's theorem of 1914.

I have been pointed recently that convergence of Newton series is covered extensively in Gelfond, A. O. Calculus of finite differences. Translated from the Russian. International Monographs on Advanced Mathematics and Physics. Hindustan Publishing Corp., Delhi, 1971

The Russian original is available online:

http://inis.jinr.ru/sl/vol1/UH/_Ready/Mathematics/Gel%27fond%20A.%20Ischislenie%20konechnyh%20raznostej%20%281959%29%28ru%29%28L%29%28201s%29.pdf

Page 141 onwards.

Look also for some specific questions: Convergence of Newton series for sin ax

Improve this answer
$\endgroup$

You must log in to answer this question.