Elementary solutions to f(z+1)-f(z)=g(z) in entire functions

Question

Let g(z) be an entire function of a complex variable z. Does there exist an entire function f(z) such that f(z+1)-f(z)=g(z)? As I learned several years back, the answer to this is apparently 'yes', but I have not felt satisfied with the proof because it goes beyond my expertise.

I tried to find f using the power series expansion of g, for that works when g is a polynomial. But the results of partial inversions kept diverging. Representing g as an integral via Cauchy's formula, and doing inversion inside the integration led to similar problems. Perhaps I am overly optimistic, but a question this elementary should have equally elementary solution. Is there such a solution? If not, is there a reason to expect that no simple and elementary solution should exist?

Answer 1

It took me some time to find a solution that satisfies both requirements:

a) If should be based on the power series expansion

b) It should use no tools heavier than contour integration.

So, let $g(z)=\sum a_ k z^k$. We know that $a_ k$ decay faster than any geometric progression. We want analytic functions $F_ k(z)$ such that $F_ k(z+1)-F_ k(z)=z^k$ and $F_ k(z)$ grows not faster than a geometric progression as a function of $k$ in every disk. Then $f=\sum a_ k F_ k$ is what we want. So, just choose any odd multiples $r_ k\in(k,k+2\pi)$ of $\pi$ and put $$ F_ k(z)=\frac{k!}{2\pi i}\oint_ {|z|=r_ k}\frac {e^{z\zeta}}{e^\zeta-1}\frac{d\zeta}{\zeta^{k+1}} $$
The key is that $|e^\zeta-1|\ge \frac 12$ on the circle and $r_ k^k\ge k!$, so $|F_ k(z)|\le 2e^{|z|(k+2\pi)}$ on the plane.

Answer 2

An informal approach is to write the finite difference operator as exp(D)-1 where D is d/dx. We're then trying to evaluate 1/(exp(D)-1) f(x).

For example, consider the case f(x) = xcos(x).

Write cos(x)=(exp(ix)+exp(-ix))/(2i)

Quantum physicists make much use of the 'theorem': f(D) exp(ax) = exp(ax) f(a+D)

So 1/(exp(D)-1) (xexp(ix)) = exp(ix) 1/(exp(D+i)-1) x

1/(exp(D+i)-1) can be expanded as a power series in D. As we're applying it to x, only the constant and D^1 terms survive.

Reassembling and simplifying gives g(x) = ((1+x)(cos(x)-cos(1+x))+cos(1+x))/(4sin(1/2)^2)

Amazingly, Mathematica simplifies g(x+1)-g(x) to xcos(x) so it's probably correct. No doubt an analyst could make these kinds of shenanigans with D (which I believe go back to Heaviside) rigorous.

Answer 3

I suspect you want a constructive solution; if someone gives you g, you want to be able to say what f is.

Write Df(z) for f(z+1) - f(z). We want to solve Df(z) = g(z). Of course this only determines f up to a constant.

Let g(z) = (z)_k = z(z-1)(z-2)...(z-k+1). Then f(z) = (z)_(k+1)/(k+1) satisfies Df(z) = g(z). (The falling power (z)_k is to finite differences what z^k is to ordinary differential calculus.)

We can use this to "integrate" (i. e. given g, find f) the ordinary powers z^k by using the fact that ordinary powers can be written as linear combinations of falling powers. The coefficients of the basis change are Stirling numbers. So, for example,

g(z) = z³ = (z)₃ + 3(z)₂ + (z)₁

and so Df(z) = g(z) where f is

(z)₄/4 + (z)₃ + (z)₂/2 = z² (z-1)²/4. In general, given g(z) = z^k one gets

f(z) = ∑_k=1ⁿ S(n,k) (n)_k+1/(k+1)

where S(n,k) is a Stirling number of the second kind. But these polynomials are not that nice, and I'm not sure how one would use this to find, say, the solution to Df(z) = exp(z).

Answer 4

An answer to part of your question: it makes sense that a power series approach isn't so straightforward, since substitution of formal power series only makes formal sense when you're plugging in something with zero constant term, not something like 1+x. In particular, a sequence of polynomials p(x) converging to a power series f(x) in the Krull topology (i.e. eventually agreeing coefficients) doesn't imply that p(1+x) converges to f(1+x) in the Krull topology, even if f(1+x) happens to make sense due to f being the power series of an entire function.

Answer 5

First, you can find $f$ smooth with this property. Now you wish to replace $f$ by $f+h$, with $h$ periodic, so that $f+h$ is holomorphic. To do this, you need to be able to solve $\bar{d}\, h =$ some fixed smooth periodic function.

I think you can solve this by expanding the fixed periodic function into Fourier series and solving term-by-term. I suspect this is pretty close to (a translation of) the solution suggested in the link you gave.

Answer 6

I haven't thought through the necessary estimates, but it seems like you can use uniform approximation by polynomials on compact sets. For any disc of radius N, there is a polynomial g_N that differs from g on that disc by at most, say, 1/N or something, and a polynomial f_N that satisfies f_N(z+1)-f_N(z) = g_N(z). You can also demand bounds on how the derivatives of g_N differ from those of g. The hard part is putting bounds on how much f_N varies (on a disc of radius N-1) if you vary g_N by a small amount.

Anyway, if you get past that part, you can let N go to infinity.