Let $g: \mathbb{R}^n \to \mathbb{R}$ be a continuous function, such that $g(x)>0$, for all $ x \in \mathbb{R}^n$. I want to prove that: there is an entire analytic function $h$ (this means that $h$ can be extended to $\mathbb{C}^n$ as a function everywhere holomorphic) such that, $$0 < h(x) < g(x),\; \forall \; x \in \mathbb{R}^n .$$

If $g$ has compact support, then, I know there exists a sequence $(f_k)_{k \in \mathbb{N}}$ of entires functions in $\mathbb{R}^n$ so that, for each, $k \in \mathbb{N}$, the function $f_k$ can be extended to the complex values as an entire function and $f_k \to g$ uniformly in $\mathbb{R}^n$ .

But, $g$ has compact support?

The only idea I had was involving the support of $ g $.

From there, how do I proceed? This is the way?

  • 4
    $\begingroup$ What do you mean by "$g$ has compact support"? Its support is all of $\mathbb R^n$. $\endgroup$ Jul 14, 2020 at 20:24
  • $\begingroup$ @RobertIsrael I wrote wrong. I will fix. $\endgroup$
    – Guilherme
    Jul 14, 2020 at 20:31

1 Answer 1


This is a special case of a famous theorem of Carleman ($m=1$) generalized by Scheinberg to arbitrary $m$, which for any continuos $f(x), x \in \mathbb R^n$ and error $\epsilon(x) >0, x \in \mathbb R^n$, gives an entire function $F(z)$ on $\mathbb C^n$ s.t its restriction to $\mathbb R^n$ satisfies $|F(x)-f(x)| < \epsilon(x)$, applied with $f(x)=g(x)/2, \epsilon(x) =g(x)/2$ and $h$ the resulting $F$.

One can give an easy direct proof in this particular case by constructing a non-zero entire function $G$ s.t $G(x) >1/g(x), x \in \mathbb R^n$ and then obviously $h=1/G$ will work

To construct $G$, we let:

$M_m=\max (1/g(x)), x=(x_q), ||x||_2 \le m+1, k_m \ge m, (\frac{m^2}{1+m})^{k_m} \ge M_m$ and let

$G_1(z)=M_0+\sum_{m \ge 1}(\frac{\sum_{q=1}^n z_q^2}{1+m})^{k_n}, z=(z_q) \in \mathbb C^n$.

Trivially $G_1$ is analytic on $\mathbb C^n$ since $k_m \to \infty$ and for $x \in \mathbb R^n, m \le ||x||_2 \le m+1$ the $m$ term is at least $M_m$ and the rest are positive so obviously $G_1(x)>1/g(x)$. But now $G(z)=e^{G_1(z)}$ is non-zero and clearly satisfies the same inequality since $e^y >y, y>0$ real and we are done!


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