Consider the following definition:
Let $\Omega$ be an open set of $\mathbb{R}_x^n$, $x = (x_1, ..., x_n)$. A $\mathcal{C}^{\infty}$-function $\varphi(x)$ on $\Omega$ is said to be uniformly analytic on $\Omega$ if it has the following uniform bound: $$ \exists C>0, \forall \alpha \in \mathbb{N}^n, \sup_{x \in \Omega} {|\partial^{\alpha} \varphi(x)|} \leq C^{|\alpha|+1} |\alpha|!. $$
In the above definition, replace $\mathbb{R}_x^n$ by $\mathbb{C}_x^n$ and consider $\varphi(x)$ to be holomorphic on $\Omega$. Then we get the complex version of the definition of a uniformly analytic function.
My questions are as follows. Please note that they are more confirmatory as I just want to make sure that I've got things right.
- Is it true that any uniformly analytic function is analytic?
- Are all complex-analytic functions uniformly analytic? (Of course, I think not; but the situation might be different if $\Omega$ is compact?)
Any other insights you have are welcome. Thank you!
For (1), as you stated above, $\phi$ is uniformly analytic on the open set $\Omega \subset \mathbb{R}^n$ if there exists a constant $C > 0$ such that for any multi-index $\alpha \in \mathbb{N}^n$, we have the bound $\sup_{x \in \Omega} |\partial^{\alpha} \varphi(x) | \leq C^{|\alpha| + 1} |\alpha| !$. Thus, given $K \subset \Omega$ compact, we have that for any multi-index $\alpha$ and point $x \in K$ that $$ |\partial^{\alpha}\varphi (x) | \leq \sup_{x \in \Omega} |\partial^{\alpha} \varphi(x) | \leq C^{|\alpha| + 1} |\alpha| !, $$ i.e. $\varphi$ is (real) analytic in $\Omega$.
For (2), I don't believe that complex-analytic functions are necessarily uniformly complex-analytic: take an entire function such as $\varphi(z) = e^z$ defined on $\Omega = \mathbb{C}$, then for any $\alpha \in \mathbb{N}$, $\sup_{z \in \mathbb{C}} |\partial^{\alpha} e^z | = \infty$, so no constant $C > 0$ as above can exist. So at the very least we have to consider holomorphic functions defined on proper subsets of $\mathbb{C}$.