Entire function satisfying an iteration formula

I hope to figure out that what is the entire function $f$ that satisfies the following iteration formula $$f(z+1)-f(z)=Ce^{-z}$$ for some constant $C$.

Actually, I guess that $f$ has to be the form $f(z)=e^{-z+a}+be^{i2\pi{z}}+c$ where $a,b,c$ are complex numbers. Clearly, this kind of function satisfies the above formula. I hope to verify my guess, but I can't solve it completely.

Some partial results is obtained. Taking derivative gives $$f'(z+1)-f'(z)=-Ce^{-z}.$$ Then we have $$f(z+1)+f'(z+1)-f(z)-f'(z)=0,$$ that is, $g(z+1)=g(z)$ where $g(z)=f(z)+f'(z)$. If $f$ satisfies some growth condition so that $g$ meet the requirement of Carlson's theorem (https://en.wikipedia.org/wiki/Carlson%27s_theorem), then by Carlson's theorem $g$ has to be constant. Thus $f(z)+f'(z)=C$ which implies that $f''(z)+f'(z)=0$. Then $f'(z)=De^{-z}+A$. Combining these with the iteration formula, we can conclude that $f$ is of the form $e^{-z+a}+b$.

I wonder if it is true for the general case. Thank you for your attention.

Solution 1:

As Mike and amsmath has already pointed out in the comments, the space $L$ of all entire functions $g$ satisfying $g(z+1) = g(z)$ is of the form $h(e^{2\pi i z})$ where, $h:\mathbb{C}^\ast \to \mathbb{C}$ is analytic.

Set, $h(\zeta) = g\left(\dfrac{\log \zeta}{2\pi i }\right)$, then $h$ is analytic in $\mathbb{C}^\ast$ and consequently admits a Laurent Series expansion $\displaystyle h(\zeta) = \sum\limits_{k \in \mathbb{Z}} c_k\zeta^k$, converging uniformly and absolutely in annuli $A(0;r,R) = \{\zeta: 0 <r < |\zeta| < R\}$.

In particular, we may write $\displaystyle c_k = \frac{1}{2\pi i}\int_{|\zeta| = r_1} \frac{h(\zeta)}{\zeta^{k+1}}\,d\zeta$ and making the change of variable $\zeta = e^{2\pi i z}$ with $r_1 = |e^{2\pi iz_0}|$ we have, $\displaystyle c_k = \int_{z_0}^{z_0+1} g(z)e^{-2\pi i kz}\,dz$ (we might as well keep the contour a line joining $z_0$ and $z_0+1$, which is immaterial as $g$ is holomorphic). The series $\displaystyle g(z) = \sum\limits_{k \in \mathbb{Z}} c_ke^{2\pi i kz}$ converges absolutely and uniformly on every horizontal strip $\{|\Im z| < r\}$.

Now, to find a particular solution $f$ to the equation $f(z+1) - f(z) = w(z)$.

The generating function of the Bernoulli Polynomials $\displaystyle f(z,\zeta) = \frac{\zeta e^{\zeta z}}{e^{\zeta}-1}$ satisfies, $$f(z+1,\zeta) - f(z,\zeta) = \zeta e^{\zeta z} \tag{1}$$ (Taking $\zeta = -1$ finishes our case when $w(z) = Ce^{-z}$).

For, $|\zeta| < 2\pi$ it admits the series expansion $\displaystyle f(z,\zeta) = \sum\limits_{k \ge 0} B_k(z)\frac{\zeta^k}{k!}$, where, $B_k$ is the $k$-th Bernoulli Polynomail (degree $k$). Comparing both sides of eqn $(1)$, the coefficients must satisfy $$B_k(z+1) - B_k(z) = kz^{k-1}.$$

Going back to Cauchy integral formula we note that, $$B_k(z) = \frac{k!}{2\pi i}\int_{|\zeta| = r} \frac{\zeta e^{z\zeta}}{e^\zeta - 1}\frac{d\zeta}{\zeta^{k+1}}, \, \text{ for } r \in (0,2\pi) \tag{2}$$

When, $r \in ((2m-2)\pi, 2m\pi)$ if we denote the coefficients as $B_{k,m}(z)$, i.e., $$B_{k,m}(z) = \frac{k!}{2\pi i}\int_{|\zeta| = r_k} \frac{\zeta e^{z\zeta}}{e^\zeta - 1}\frac{d\zeta}{\zeta^{k+1}}, \, \text{ for } r_k \in ((2m-2)\pi, 2m\pi) \tag{3}$$

Then by Residue theorem we have, $$B_{k,m+1}(z) - B_{k,m}(z) = \frac{m!(e^{2\pi i mz} + (-1)^ke^{-2\pi i mz})}{(2\pi im)^k}$$

Therefore, $B_{k,m}$ also satiesfies, $$B_{k,m}(z+1) - B_{k,m}(z) = kz^{k-1}$$ for any $m \ge 1$.

If, $w(z)$ is an entire function with power series expansion $\displaystyle w(z) = \sum\limits_{k \ge 0} w_kz^k$ then atleast formally we expect, $$f(z) = \sum\limits_{k \ge 0} w_k\frac{B_{k+1,m_k}(z)}{k+1}$$ to satisfy $f(z+1) - f(z) = w(z)$, where, the sequence $\{m_k\}_{k \ge 0}$ we might use for ensuring convergence.

If we set $m_k = k+1$ and $r_k = 2\pi\left(m_k - \frac{1}{2}\right) = (2k+1)\pi$ in eqn $(3)$ we have, $$|B_{k+1,m_k}(z)| \le \frac{(k+1)!}{2\pi}\int_{|\zeta| = r_k}\left|\frac{\zeta^{-k-1} e^{z\zeta}}{e^\zeta - 1}\right|\,|d\zeta| \le \frac{C(k+1)!}{((2k+1)\pi)^{k+1}} \frac{e^{(2k+1)\pi|z|}}{\inf\limits_{|\zeta| = (2k+1)\pi} |e^\zeta - 1|}.$$

By Stirling's approximation we have $\displaystyle \frac{(k+1)!}{((2k+1)\pi)^{k+1}} \le Ce^{2k\pi}$ for some constant $C$. On the other hand $\inf\limits_{|\zeta| = (2k+1)\pi} |e^\zeta - 1|$ is uniformly bounded away from zero (bounded below by a constant $c$ independent of $k$) since, $|\zeta| = (2k+1)\pi$ remains uniformly bounded away from the zero set of $(e^\zeta - 1)$ as we vary $k$ (this in principle is a consequence of Lojasiewicz inequality).

Thus, $\displaystyle |B_{k+1,k+1}(z)| \le C'e^{2k\pi}e^{(2k+1)\pi|z|}$ has atmost exponential growth on compact sets, where as decay $|w_k|^{1/k} \to 0$ as $k \to \infty$ ensures normal convergence of the power series of $f$, i.e., $f$ is entire.

Thus, solutions to $f(z+1) - f(z) = w(z)$ for $w$ entire has the form, $$f(z) = \sum\limits_{k \ge 0} w_k\frac{B_{k+1,k+1}(z)}{k+1} + h(e^{2\pi i z}),$$ where, $h$ is analytic in $\mathbb{C}^\ast$.