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.

  • $\begingroup$ Not sure if it helps, but$$f^{(n)}(z)=\frac1{2^n}\sum_{k=0}^n\binom nk(-1)^kf(z+n-2k)$$ $\endgroup$ – Simply Beautiful Art Dec 10 '17 at 19:46
  • $\begingroup$ 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$$f_{m+1}(z)=\sum_{n=0}^\infty\frac{(z/2)^n}{n!}\sum_{k=0}^n\binom nk(-1)^kf_m(n-2k)$$ $\endgroup$ – 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.

  • $\begingroup$ 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 $a=0$.) $\endgroup$ – Eric Wofsey Dec 10 '17 at 21:35
  • $\begingroup$ @EricWofsey: Sure! A complete characterization would be nice. $\endgroup$ – Martin R Dec 10 '17 at 22:30

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.