Let \(f:\mathbb R\to(0,\infty)\) be continuous. Prove that there exists a nonzero entire function \(g\) such that \[ 0\le g(x)<f(x) \] for every \(x\in\mathbb R\).
Solution
Set \[ q(x)=-\log f(x). \] This is a continuous real-valued function on \(\mathbb R\).
1. Choose a constant \(C>0\) such that\[ C>q(x) \]for every \(|x|\le2\).
The function \(q\) is continuous, hence bounded above on the compact interval \([-2,2]\). Choose \(C\) larger than that upper bound and also positive.
2. For every integer \(n\ge2\), set\[ Q_n=1+\max\left(0, \max_{n\le|x|\le n+1}q(x)\right). \]Choose a positive integer \(N_n\) so large that\[ Q_n n^{-N_n}\le2^{-n}, \]and define\[ c_n=Q_n n^{-2N_n}>0. \]
Each \(Q_n\) is finite and positive by continuity of \(q\) on the compact set \(\{x:n\le|x|\le n+1\}\). Since \(n\ge2\), the quantity \(n^{-N}\) tends to \(0\) as \(N\to\infty\), so such an integer \(N_n\) exists.
3. The series\[ H(z)=C+\sum_{n=2}^{\infty}c_nz^{2N_n} \]converges uniformly on compact subsets of \(\mathbb C\), and therefore definesan entire function.
Fix \(R>0\). For all sufficiently large \(n\), one has \(n>R^2\). Then for \(|z|\le R\), \[ \begin{aligned} c_n|z|^{2N_n} &\le Q_n\left(\frac{R}{n}\right)^{2N_n}\\ &=\bigl(Q_n n^{-N_n}\bigr) \left(\frac{R^2}{n}\right)^{N_n}\\ &\le2^{-n}. \end{aligned} \] Thus the tail is uniformly dominated on the disk \(|z|\le R\) by a geometric series. The Weierstrass \(M\)-test gives local uniform convergence, so \(H\) is entire.
4. For every real \(x\),\[ H(x)>q(x). \]
All coefficients \(c_n\) are positive and all exponents are even, so every summand \(c_nx^{2N_n}\) is nonnegative for real \(x\).
If \(|x|\le2\), then \[ H(x)\ge C>q(x) \] by 1.
If \(|x|>2\), choose the integer \(n\ge2\) such that \[ n\le|x|<n+1. \] Then \[ c_nx^{2N_n} \ge c_n n^{2N_n} =Q_n >q(x). \] Since every other term and \(C\) are nonnegative, this gives \(H(x)>q(x)\).
5. Define\[ g(z)=e^{-H(z)}. \]Then \(g\) is a nonzero entire function and, for every real \(x\),\[ 0<g(x)<f(x). \]
Because \(H\) is entire, so is \(e^{-H}\), and the exponential function never vanishes. For real \(x\), the even-power series defining \(H\) has real values, so \(g(x)=e^{-H(x)}>0\). By 4, \[ -H(x)< -q(x)=\log f(x). \] Exponentiating gives \[ 0<g(x)=e^{-H(x)}<f(x). \] This is stronger than the required \(0\le g(x)<f(x)\).