TITLE: Existence of an holomorphic function
QUESTION [25 upvotes]: Is there a simple way to prove this fact : 

For all holomorphic functions $f : \mathbb C \to \mathbb C$, there is an holomorphic function $\psi : \mathbb C \to \mathbb C$ such that $$\psi(z+1) = \psi(z) + f(z) $$

The solution I know use Galois covering spaces and summand of automorphy.
Thanks in advance.

REPLY [10 votes]: As I mentioned in a comment, there are two very simple ways of solving the functional equation $\psi(z+1)=\psi(z)+f(z)$, for $\psi$ in terms of $f$. They are,
\begin{align}
\psi(z)&=-f(z)-f(z+1)-f(z+2)-f(z+3)-\cdots,{\rm\ and,}\\
\psi(z)&=f(z-1)+f(z-2)+f(z-3)+\cdots.
\end{align}
These solutions will only apply if either of the respective sums converge and, if we want $\psi$ to be holomorphic, then we require locally uniform convergence. For example, taking $f(z)=e^{az}$ then, for $\Re(a)\not=0$, the solutions above give $\psi(z)=e^{az}/(e^a-1)$ (using the first method for $\Re(a)<0$ and the second for $\Re(a)>0$).
More generally, we can prove the following,

Lemma: Let $f\colon\mathbb{C}\to\mathbb{C}$ be entire. Then, there exists entire $g,h$ with $f=g+h$ and such that, for each $K>0$, $z^2g(z)$ and $z^2h(-z)$ are uniformly bounded on the horizontal strip $-K\le\Im(z)\le K$ with $\Re(z)>0$.

I'll prove this in a moment. First, however, we can see that it gives an easy solution to the functional equation by applying the first method above to $g$ and the second to $h$,
\begin{align}
\psi(z)=&-g(z)-g(z+1)-g(z+2)-\cdots\\
&+h(z-1)+h(z-2)+h(z-3)+\cdots.
\end{align}

I'll now prove the lemma stated above, using a result on approximating functions by entire functions due to Arakelian. Let $E$ be the set of $z\in\mathbb{C}$ such that $2\lvert\Im(z)\rvert+1\le\lvert\Re(z)\rvert$. This is an unbounded closed subset of $\mathbb{C}$ and consists of two connected components, one for $\Re(z)\le-1$ and one for $\Re(z)\ge1$. Furthermore, for each $K>0$, the strip $-K\le\Im(z)\le K$ is contained in the union of $E$ and a bounded set. So, it is enough that the functions $g,h$ constructed in the lemma are such that $z^2g(z)$ and $z^2h(-z)$ are bounded on the component of $E$ with $\Re(z)\ge1$.
We can define a continuous function $F\colon E\to\mathbb{C}$ by setting $F(z)=e^{z^2}f(z)$ for $\Re(z)\le-1$ and $F(z)=0$ otherwise. This is also analytic on the interior of $E$.
Now, $E$ is an Arakelian set. That is, it is a closed subset of $\mathbb{C}$ whose complement does not contain any bounded connected components and such that, for each closed disk $D$, the union of the closed bounded components of $\mathbb{C}\setminus(E\cup D)$ is bounded (actually, it is empty, so trivially bounded). Then, Arakelian's approximation theorem states that, for any $\epsilon > 0$, there is an entire function $G$ with $\lvert F-G\rvert\le\epsilon$ on $E$. If we set $g(z)=e^{-z^2}G(z)$ and $h(z)=f(z)-g(z)$ then we see that $\lvert e^{z^2}g(z)\rvert$ and $\rvert e^{z^2} h(-z)\rvert$ are bounded by $\epsilon$ on the component of $E$ with $\Re(z)\ge1$.
Finally, on $E$,
$$
\Re(z^2)=\Re(z)^2-\Im(z)^2\ge\frac35\left(\Re(z)^2+\Im(z)^2\right)=\frac35\lvert z\rvert^2.
$$
So,
$$
\lvert e^{z^2}\rvert=\exp(\Re(z^2))\ge\exp(3\lvert z\rvert^2/5)\ge\frac35\lvert z\rvert^2,
$$
and $z^2g(z)$, $z^2h(-z)$ are bounded on the component of $E$ with $\Re(z)\ge1$ as required.