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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.

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Thanks in advance.

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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.

(*This is nothing but an expansion of my comment; it clarifies what can easily be done by a naïve power series method*)

The question is to prove that the *finite difference operator* $\Delta : f \mapsto f(z+1) – f(z)$ defines a surjection $\mathcal O(\mathbb C) \to \mathcal O(\mathbb C)$. ($\mathcal O(\mathbb C)$ is the algebra of *entire functions*, that is holomorphic functions on all of $\mathbb C$).

Because of standard complex analysis theorems, an entire function is the sum of a power series of infinite convergence radius. That is, series summation gives an isomorphism

$$ A_\infty = \left\{\sum_{n\geq 0} a_n z^n \in \mathbb C[[z]] \, \middle | \, \forall R > 0, a_n = O(R^{-n}) \right \} \to \mathcal O(\mathbb C).$$

So the finite difference operator also defines an operator $\Delta : A_\infty \to A_\infty$. Basically, it is given by Newton’s binomial formula: for $j \geq 0$, we have

$$\Delta(z^j) = (z+1)^j – z^j = \sum_{i=0}^{j-1} \binom j i z^i.$$

For a power series $\sum_{j \geq 0} a_j z^j \in \mathbb C[[z]]$, one can apply the last formula and linearity to get

$$\Delta\left( \sum_{j=0}^K a_j z^j \right) = \sum_{j=0}^K a_j \left( \sum_{i=0}^{j-1} \binom j i z^i \right) = \sum_{i=0}^{K-1} \left(\sum_{j=i+1}^K a_j \binom j i \right) z^i.$$ So, as, $K$ grows, all of the coefficients of this polynomial keep changing. Because $\binom j i$ is a degree-$i$ polynomial in $j$, the series $\sum_{j>i} a_j \binom j i$ converges and the operator $\Delta : A_\infty \to A_\infty$ is defined by the formula:

$$\begin{array}{ccc} A_\infty&\to& A_\infty\\ \sum_{n \geq 0} a_n z^n & \mapsto &\displaystyle \sum_{n\geq 0} \left( \sum_{k=n+1}^{+\infty} a_k \binom k n \right) z^n.\end{array}$$

So the original question is really equivalent to:

**Question:** is this operator $\Delta : A_\infty \to A_\infty$ surjective?

To address this question, it seems quite reasonable to look at polynomials of degree $N$ first, then to look at $N \to \infty$. Exactly like in the computation of $\Delta$, you need to have some kind of convergence to extend the definitions from the polynomial case to the power series case. In the previous computation, we needed the convergence of $\sum_{k \geq n} a_k \binom k n$ for all $n$ and the fact that $\sum_{n \geq 0} a_n z^n$ has an infinite convergence radius was more than enough for that.

What I’ll explain is that in order to get a map $I : A_\infty \to A_\infty$ such that $\Delta \circ I = \mathrm{id}_{A_\infty}$ (which is equivalent to proving that $\Delta$ is surjective) a similar approach works, but **only if we can assume that the decay of the $(a_n)$ is stronger than the $\forall R > 0, a_n = O(R^{-n})$ hypothesis.** That’s why this isn’t really an answer to the original question.

Obviously (either because of the definition or because of the big messy formula above) $\Delta$ induces a linear map $\mathbb C_{n}[z] \to \mathbb C_{n-1}[z]$. It’s quite easy to show that the only polynomials in $\ker \Delta$ are the constant ones, so a dimension computation shows that $\Delta : \mathbb C_{n}[z] \to \mathbb C_{n-1}[z]$ is onto, for all $n$. There is then a map $I : \mathbb C_{n-1}[z] \to \mathbb C_{n}[z]$ such that $\Delta \circ I = \mathrm{id}_{\mathbb C_{n-1}[z]}$ and this map is unique, up to adding constant polynomials.

In fact, one can be quite explicit about that: if $B_n$ is the $n$-th Bernoulli number, the polynomial

$$P_k = \frac{1}{k+1} \left( \sum_{n=0}^{k+1} \binom{k+1}{n} B_{k+1-n} z^n\right)$$ (which, up to the $1/(k+1)$ factor, is the $(k+1)$-th Bernoulli polynomial) satisfies the following relation

$$\Delta P_k = z^k,$$

so it seems like a good idea to define $I(z^k) = P_k$.

If $\sum_{k\geq 0} a_k z^k \in A_\infty$, we then have that

$$\begin{align*}I\left(\sum_{k=0}^K a_k z^k\right) &= \sum_{k=0}^K a_k \left(\frac{1}{k+1} \sum_{n=0}^{k+1} \binom{k+1}{n} B_{k+1-n} z^n\right)\\

& = \sum_{n=0}^{K+1} \left(\sum_{k=n-1}^K \frac{a_k}{k+1} \binom{k+1}n B_{k+1-n} \right) z^n.\end{align*}$$

Now, if the decay of $(a_k)$ was strong enough to guarantee the absolute convergence of

$\sum_{k\geq n-1} \frac{a_k}{k+1} \binom{k+1}n B_{k+1-n}$, the formula above (with $K$ replaced by $\infty$) would define a linear map $I : A_\infty \to A_\infty$ such that $\Delta \circ I = \mathrm{id}_{A_\infty}$. This would be the case if we had $|B_p| \leq C\cdot R^p$ for some constants $C$, $R$, for instance. The problem is that the real asymptotics for the Bernoulli numbers is

$$|B_{2n}| \sim 4\sqrt{\pi n}\left(\frac{n}{\pi e}\right)^{2n}$$ (for $n > 0$, $B_{2n+1} = 0$).

It is then well possible to find sequences $(a_n)$ such that $a_n = O(R^{-n})$ for all $R$, but which nevertheless satisfy $$\limsup_{k\to\infty} \left|\frac{a_k}{k+1} \binom{k+1}n B_{k+1-n}\right| = +\infty.$$ For example, $a_k = k^{-\alpha k}$ seems to work for $0 < \alpha < 1$.

So I think this method works if we restrict ourselves to classes of entire functions with strongly decaying Taylor coefficients (strongly enough for our $I$ to be defined), but not for any entire function.

Like I said in my comment, I’m surprised that the answer to the **Question** asked above could be “yes” with this natural method failing. If the OP hadn’t claimed the original result to be true, I would by now think the $\Delta$ operator not to be surjective on $A_\infty$ or $\mathcal O(\mathbb C)$.

On the other hand, if you try this method (using polynomials of higher and higher degree to approximate your power series solution) to construct elements of $\ker \Delta$, you would probably only find the constant functions, whereas the kernel of $\Delta$ in $\mathcal O(\mathbb C)$ [or $A_\infty$] is much larger than that (for instance it cointains $z \mapsto e^{2i\pi z}$). So perhaps there is a good reason for which this method was doomed to fail (some instability property?) and my surprise is really nothing but a consequence of my naïveté.

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