# Entire functions such that $\frac{f(z+1)-f(z-1)}{2}=f'(z)$

Consider the equation $$\frac{f(z+1)-f(z-1)}{2}=f'(z)\tag{*}.$$ Any polynomial of degree $\leq 2$ satisfies $(*)$ for all $z$. My question is:

If $f:\mathbb{C}\to\mathbb{C}$ is holomorphic and satisfies $(*)$ for all $z$, must $f$ be a polynomial of degree $\leq 2$?

I would also be interested in answers to weakened versions of the question, for instance where $f$ is only holomorphic on a strip $\{z:-c<\operatorname{Im}z<c\}$, or where $f$ is allowed to have isolated singularities.

Here are some things I know about this question so far. If we instead consider smooth functions $f:\mathbb{R}\to\mathbb{R}$, then there are lots more solutions to $(*)$ besides polynomials of degree $\leq 2$ (see Functions $f$ such that $f(x+1)-f(x-1)=2f'(x)$.).

On the other hand, if $f$ is a polynomial which satisfies $(*)$, it must have degree $\leq 2$. Indeed, if $f$ satisfies $(*)$, so does $f'$ (by differentiating the equation), so if a polynomial of degree $>2$ satisfied $(*)$, we could repeatedly differentiate to get a cubic. Subtracting the quadratic part (since $(*)$ is linear), we would conclude that $f(z)=z^3$ satisfies $(*)$, which is false.

You could attempt to build solutions to $(*)$ using Taylor series. Specifically, suppose $f(z)=\sum a_n z^n$ and let $g(z)=f'(z)-\frac{f(z+1)-f(z-1)}{2}$. To verify that $f$ satisifies $(*)$, it suffices to check that $g^{(n)}(0)=0$ for all $n\in\mathbb{N}$. This can be written as an infinite list of (infinitary) linear conditions on the $a_n$. For instance, the condition that $g(0)=0$ says that $$\sum_{n=1}^\infty a_{2n+1}=0$$ and the condition that $g'(0)=0$ says that $$\sum_{n=2}^\infty2na_{2n}=0.$$ In general, the equations for even derivatives involve only the $a_n$ for $n$ odd and the equations for odd derivatives involve only the $a_n$ for $n$ even, so you can consider odd $n$ and even $n$ separately. You could try to inductively construct the $a_n$ to make all these equations true one at a time. For instance, you might start by defining $a_3=1$ and $a_5=-1$, and then define $a_7$ and $a_9$ so that $g(0)=0$ remains true but $g''(0)=0$ becomes true. Then you could try to define $a_{11}$, $a_{13}$, and $a_{15}$ so that $g(0)=0$ and $g''(0)=0$ remain true and $g''''(0)=0$ becomes true. However, this has convergence issues: I don't know how to prove that such a construction will make the series $\sum_{n=1}^\infty a_{2n+1}$ actually converge (all the construction gives is that infinitely many of the partial sums are $0$), let alone that the $a_n$ shrink fast enough so that $\sum a_nz^n$ is entire.

• Not sure if it helps, but$$f^{(n)}(z)=\frac1{2^n}\sum_{k=0}^n\binom nk(-1)^kf(z+n-2k)$$ – Simply Beautiful Art Dec 10 '17 at 19:46
• We then happen to have$$f(z)=\sum_{n=0}^\infty\frac{z^n}{n!}f^{(n)}(0)=\sum_{n=0}^\infty\frac{(z/2)^n}{n!}\sum_{k=0}^n\binom nk(-1)^kf(n-2k)$$to which you can take some guess function $f_0$ and attempt to iterate$$f_{m+1}(z)=\sum_{n=0}^\infty\frac{(z/2)^n}{n!}\sum_{k=0}^n\binom nk(-1)^kf_m(n-2k)$$ – Simply Beautiful Art Dec 10 '17 at 20:08

If $f:\mathbb{C}\to\mathbb{C}$ is holomorphic and satisfies $(*)$ for all $z$, must $f$ be a polynomial of degree $\leq 2$?
No, there are more solutions. For $f(z) = \sin(az+b)$ we have $$f(z \pm 1) = \sin(az+b)\cos(a) \pm \cos(az+b) \sin(a)$$ and therefore $$\frac{f(z+1)-f(z-1)}{2} = \cos(az+b) \sin(a) \, .$$
It follows that for any $a \in \Bbb C$ satisfying $\sin(a) = a$ (and there are infinitely many such $a$) and arbitrary $b\in \Bbb C$ the function $$f(z) = \sin(az+b)$$ satisfies the equation $(*)$, and therefore also any linear combination $$f(z) = \sum_{j=1}^n c_j \sin(a_j z + b_j)$$ if all $a_j$ are fixed points of the sine function.
• Ah, that's very nice. I should have thought of trying exponential solutions. I'll wait a little while before accepting in case someone else comes up with a classification of all the solutions. (In particular, I would be very interested to see a general class of solutions that includes $z$ and $z^2$, similar to how yours includes constants as the case $a=0$.) – Eric Wofsey Dec 10 '17 at 21:35