It is well known (or at least easy to see) that the complex exponential function $\exp:\mathbb{C}\to \mathbb{C}$ can be charaterized as the unique holomorphic function $f:\mathbb{C}\to \mathbb{C}$ such that $$ \lim_{h\to 0} \frac{f(h)-1}{h}=1 \quad \text{and} \quad f(z+w)=f(z)f(w) \quad \text{for all } z,w\in\mathbb{C}. $$ In fact we can make a stronger charaterization: $$ \lim_{h\to 0,\; h\in\mathbb{R}} \frac{f(h)-1}{h}=1 \quad \text{and} \quad f(z+w)=f(z)f(w) \quad \text{for all } z,w\in\mathbb{C}. $$
My question is: What are all the holomorphic functions $f:\mathbb{C}\to \mathbb{C}$ such that $$ \lim_{h\to 0,\; h\in\mathbb{R}} \frac{f(h)-1}{h}=1 \quad \text{and} \quad f(x+y)=f(x)f(y) \quad \text{for all } x,y\in\mathbb{R}? $$ Note the (perhaps big) difference: now the condition $f(x+y)=f(x)f(y)$ is given for all real $x$ and $y$. I know that the complex exponential function is an example of such $f$, but are there any others? If not, how to prove that the exponential is the only function satisfying this conditions?
Thank you in advance!
The answer is the same: If $f$ is holomorphic in $\Bbb C$ with $f(x+y)=f(x)f(y)$ for all $x, y\in \Bbb R$, then $f(z+w)=f(z)f(w)$ holds for all $z, w \in \Bbb C$, due to the identity theorem for holomorphic functions.
First fix $y\in \Bbb R$: Then $f(x+y)=f(x)f(y)$ for all $x \in \Bbb R$ implies that $f(z+y)=f(z)f(y)$ for all $z \in \Bbb C$.
Now fix $z \in \Bbb C$: Then $f(z+y)=f(z)f(y)$ for all $y \in \Bbb R$ implies that $f(z+w)=f(z)f(w)$ for all $z, w \in \Bbb C$