# If $g: \mathbb{R}^n \to \mathbb{R}$ then there is a $h$ entire analytic function that extends $g$

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?

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

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!