Let \(M:\mathbb R\to(0,\infty)\) be continuous.
Prove that there exists an entire function \(f\) such that \[ |f(x)|>M(x) \] for every \(x\in\mathbb R\).
Prove that there exists an entire function \(g\) such that \[ 0<|g(x)|<M(x) \] for every \(x\in\mathbb R\).
Prove that there is no entire function \(f\) satisfying \[ |f(z)|>|z| \] for every \(z\in\mathbb C\).
Solution
We first construct an entire real-valued-on-\(\mathbb R\) function that dominates an arbitrary prescribed continuous real profile. Exponentiation then gives (a), reciprocation gives (b), and Liouville’s theorem gives (c).
1. Let \(H:\mathbb R\to\mathbb R\) be continuous. There exists an entirefunction \(F\) with real coefficients such that\[ F(x)>H(x) \qquad(x\in\mathbb R). \]
For \(r\ge0\), set \[ B(r)=1+\max_{|x|\le r}\max\{H(x),0\}. \] Thus \(B(r)\) is finite, positive, and nondecreasing, and \[ B(r)>H(x) \qquad(|x|\le r). \]
Choose a positive constant \(c_0>B(1)\). For each integer \(n\ge1\), choose a positive integer \(N_n\) and set \[ c_n=\frac{B(n+1)}{n^{2N_n}}. \] Take \(N_1=1\). For \(n\ge2\), choose \(N_n\) so large that \[ c_n(n-1)^{2N_n} =B(n+1)\left(\frac{n-1}{n}\right)^{2N_n} <2^{-n}. \] This is possible because \((n-1)/n<1\).
Now define \[ F(z)=c_0+\sum_{n=1}^{\infty}c_nz^{2N_n}. \] For any fixed \(R>0\), if \(n>R+1\) and \(|z|\le R\), then \[ |c_nz^{2N_n}| \le c_n(n-1)^{2N_n} <2^{-n}. \] Hence the series converges uniformly on every compact disk, so \(F\) is entire. All coefficients are positive real numbers, so \(F(x)\) is real for real \(x\).
If \(|x|<1\), then \[ F(x)\ge c_0>B(1)>H(x). \] If \(|x|\ge1\), choose the integer \(n\ge1\) with \[ n\le |x|<n+1. \] Then the \(n\)-th term alone gives \[ F(x) \ge c_n|x|^{2N_n} \ge c_nn^{2N_n} =B(n+1) >H(x). \] Thus \(F\) dominates \(H\) on all of \(\mathbb R\).
2. There exists an entire function \(f\) satisfying\[ |f(x)|>M(x) \qquad(x\in\mathbb R). \]
Apply 1 to the continuous real function \[ H(x)=\log M(x). \] Let \(F\) be the resulting entire function. Since \(F(x)\) is real on \(\mathbb R\) and \(F(x)>\log M(x)\), the nowhere-zero entire function \[ f(z)=e^{F(z)} \] satisfies \[ |f(x)|=e^{F(x)}>M(x) \] for every real \(x\).
3. There exists an entire function \(g\) satisfying\[ 0<|g(x)|<M(x) \qquad(x\in\mathbb R). \]
Apply 2 to the positive continuous function \(1/M\). The construction in 2 gives a nowhere-zero entire function \(h=e^F\) such that \[ |h(x)|>\frac1{M(x)} \qquad(x\in\mathbb R). \] Therefore \[ g(z)=\frac1{h(z)}=e^{-F(z)} \] is entire and nowhere zero, and for real \(x\), \[ 0<|g(x)|<M(x). \]
4. No entire function can satisfy\[ |f(z)|>|z| \]for every \(z\in\mathbb C\).
Suppose such an \(f\) existed. The inequality implies that \(f(z)\neq0\) for every \(z\): this is immediate for \(z\neq0\), and at \(z=0\) the strict inequality gives \(|f(0)|>0\).
Hence \[ h(z)=\frac z{f(z)} \] is entire. Moreover, \[ |h(z)|<1 \] for every \(z\neq0\), while \(h(0)=0\). Thus \(h\) is a bounded entire function. By Liouville’s theorem it is constant, and since \(h(0)=0\) we must have \(h\equiv0\). But \(z/f(z)\neq0\) for every \(z\neq0\), a contradiction.