TITLE: Functions $f$ such that $f(x+1)-f(x-1)=2f'(x)$.
QUESTION [6 upvotes]: What can one say about functions $f:\mathbb{R}\to\mathbb{R}$ satisfying the condition $f(x+1)-f(x-1)=2f'(x)$? Is is possible to find all such functions, or is this defining equation the best characterisation that one is likely to get?
I know that all polynomials with degree at most 2 satisfy the conditions. This is most easily seen by noting that constants, $f(x)=x$ and $f(x)=x^2$ satisfy the conditions and then noticing that if two functions $f$ and $g$ satisfy the condition then so does any linear combination of $f$ and $g$. 
One can also see that no polynomial with degree more than 2 will satisfy the condition by noting that $f(x)=x^3$ does not satisfy the condition and then noticing that if $f$ satisfies the condition then so does its derivative (Provided that $f'$ is also differentiable, which is obviously the case for polynomials). Thus the existence of any polynomial $f$ of degree more than 2 which satisfies the condition would imply that there exists such a cubic by repeatedly differentiating $f$, and the existence of such a cubic implies that $f(x)=x^3$ satisfies the condition because $f(x)=x^3$ is a linear combination of the obtained cubic and some quadratic polynomial.

REPLY [2 votes]: All such functions can be obtained in the following way.  Start with any smooth ($C^\infty$) function $f:[-1,1]\to\mathbb{R}$ such that $$f^{(n)}(1)-f^{(n)}(-1)=2f^{(n+1)}(0)$$ for all $n\geq 0$.  We can then extend $f$ smoothly to $[-1,2]$ by defining $f(x)=f(x-2)+2f'(x-1)$ for $x\in[1,2]$.  The equations above guarantee that all the higher derivatives of $f$ at $1$ from the left and right will agree, so $f$ is smooth at $1$.  But now, $f$ satisfies $f^{(n)}(x+1)-f^{(n)}(x-1)=2f^{(n+1)}(x)$ for all $n$ and all $x\in [0,1]$.  In particular, these equations for $x=1$ allow us to extend $f$ to $[2,3]$ in the same way.  We can similarly extend to $[3,4]$ and $[4,5]$ and so on, and also to $[-2,-1]$, $[-3,-2]$, and so on, and obtain a smooth function $f:\mathbb{R}\to\mathbb{R}$ satisfying $f(x+1)-f(x-1)=2f'(x)$ for all $x$.
Conversely, given any $f:\mathbb{R}\to\mathbb{R}$ such that $f(x+1)-f(x-1)=2f'(x)$ for all $x$, note that $f$ must be smooth: it is differentiable by assumption, but then its derivative can be written in terms of $f$ itself so its derivative is differentiable, and so on.  So restricting $f$ to $[-1,1]$ we get an initial function as above, and the values of $f$ on all of $\mathbb{R}$ can be recovered by the process above.
Let me finally remark that there are indeed many functions $f:[-1,1]\to\mathbb{R}$ satisfying the conditions above.  Indeed, those conditions only involve the derivatives of $f$ at $-1$, $0$, and $1$.  So we can first prescribe $f$'s behavior near those three points, and then fill in the values between them however we want, as long as we do so smoothly.  Moreover, given any two sequences $(a_n)$ and $(b_n)$ of real numbers, if we define $c_n=a_n+2b_{n+1}$, then it is possible to construct $f$ such that $f^{(n)}(-1)=a_n$, $f^{(n)}(0)=b_n$, and $f^{(n)}(1)=c_n$ (by Borel's lemma).