4
$\begingroup$

Consider $f:\mathbb R\to\mathbb R$, and let $P_n$ be a polynomial with degree $n\ge 1$. Now, if $$f(x)-f(x-c)=P_n(x)$$ for some fixed constant $c$, then what assumptions on $f$ could let us ensure that $f$ is also a polynomial?

This question comes up when I found out that many of the functional equations could be transformed into the form stated above, while all of them I met (well, in competitions) have polynomial solutions.

It should be easy to show that if $f$ is a polynomial, then its degree equals to $n+1$, which, if we could conclude that $f$ is a polynomial under some easier-to-prove conditions, then it would be a wonderful thing. Up to curiosity, I did the following 'primary' observations.

First, it could be shown that the claim that no condition needed is simply false. In fact, many counterexamples works. For example, if $g$ is a polynomial solution, then consider a periodic function $h$ with period $c$, $g+h$ shall also be a solution. This produced uncountably many counterexamples already since we could set $$h(x):=k\cos{(\frac{2\pi}{c} x)}$$ where $k$ is any nonzero real number.

Now, a simple approach, at least to get started, is to divide each sides by $c$, giving $$\frac{f(x)-f(x-c)}c=P_n(x)/c$$. This shows that at least the polynomial $$\frac1c\int P_n(x)\mathrm dx$$ could be used to approximate $f$ for some small $c$.

So, is there actually a set of functions $S(P_n, c)$ depending on a given real polynomial $P_n$ and a nonzero real number $c$, such that, $$ \begin{cases} S(P_n,c)\subseteq\mathbb R^\mathbb R\\ \mathcal P\subset S(P_n,c)\\ S(P_n,c)\cap A(P_n,c)\subset\mathcal P \end{cases} $$ where $\mathcal P$ is the set of all real polynomials with finite degree, and $A(P_n,c)$ is defined as $$A(P_n,c):=\{f\in\mathbb R^\mathbb R\mid\forall x\in\mathbb R, f(x)-f(x-c)=P_n(x)\}$$ And also, will the conclusion change if we replace all the $\mathbb R$ above to other domains?

Answers to any extent is appreciated. Thanks in advance.

$\endgroup$
  • $\begingroup$ $f(x)$ must have a finite number of roots on $\mathbb{R}$ must be $C^{\infty}(\mathbb{R})$ must exist an integer $k\ge 0$ such that for $n>k$ the $n-$th derivative is $0$ for any $x\in\mathbb{R}$ $\endgroup$ – Raffaele Jul 25 '17 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.