Let $(a_n) _{n\ge 0}$ be a sequence of real numbers such that the power series $\sum _{n=0}^{\infty}a_n x^n$ converges for every $x\in\mathbb{R}$, thereby defining a smooth function $f:\mathbb{R}\to\mathbb{R}$. Give a necessary and sufficient criterion on $(a_n)$ for $f$ to be bounded on $\mathbb{R}$.

A power series centred at the origin is a formal series $\sum_{n=0}^{\infty}a_n x^n$ with $a_n\in\mathbb{R}$. Its radius of convergence is $$ R = \frac{1}{\limsup_{n\to\infty}|a_n|^{1/n}} \in [0,+\infty]. $$

A function $f:\mathbb{C}\to\mathbb{C}$ is called entire if it is holomorphic on all of $\mathbb{C}$. Every power series with $R=+\infty$ defines a real-entire function, and by the identity theorem its complex extension is entire.

The radius of convergence of $\sum a_n z^n$ equals $$ R = \Bigl(\limsup_{n\to\infty}|a_n|^{1/n}\Bigr)^{-1}. $$

The condition $R=+\infty$ is equivalent to $a_n = O(r^n/n!)$ for every $r>0$, i.e., the coefficients decay faster than any geometric sequence. This is the Paley–Wiener type condition for entire functions of order $1$.

The order of an entire function $f$ is $$ \rho = \limsup_{r\to\infty}\frac{\log\log M(r)}{\log r}, \qquad M(r)=\max_{|z|=r}|f(z)|. $$ The type $\sigma$ (for $0<\rho<\infty$) is $$ \sigma = \limsup_{r\to\infty}\frac{\log M(r)}{r^{\rho}}. $$

Every bounded entire function $f:\mathbb{C}\to\mathbb{C}$ is constant.

Question 1 concerns real-valued functions $f:\mathbb{R}\to\mathbb{R}$. A function may be bounded on $\mathbb{R}$ while its complex extension is unbounded. For instance, $\cos z$ satisfies $|\cos z|\to\infty$ along the imaginary axis (since $\cos(iy)=\cosh y\to+\infty$). Liouville’s theorem therefore does not apply, and the problem is genuinely non-trivial.

Let $\mathcal{B}$ denote the set of all functions $f:\mathbb{R}\to\mathbb{R}$ that can be represented as a convergent power series $\sum_{n\ge 0}a_n x^n$ (with $R=+\infty$) and that are bounded on $\mathbb{R}$.

1. $\mathcal{B}$ is a linear subspace of $C^\infty(\mathbb{R})$: if $f,g\in\mathcal{B}$ and $\lambda\in\mathbb{R}$ then $f+\lambda g\in\mathcal{B}$.
2. $\mathcal{B}$ is closed under pointwise multiplication: if $f,g\in\mathcal{B}$ then $fg\in\mathcal{B}$.
3. $\mathcal{B}$ contains all functions of the form $c\cos(h(x))$, where $c\in\mathbb{R}$ and $h:\mathbb{R}\to\mathbb{R}$ is any entire function.

Part (3) follows from $\cos(h(x)) = \operatorname{Re}(e^{ih(x)})$ together with $|\cos(h(x))|\le 1$. The class is strictly larger than ${c\cos(bx):c,b\in\mathbb{R}}$; for example, $\cos(x^3-x)\in\mathcal{B}$.

Suppose $f(x)=\sum_{n=0}^{\infty}a_n x^n$ is bounded on $\mathbb{R}$. Then either:

1. $a_0$ is the only non-zero coefficient (i.e., $f$ is the constant function $f\equiv a_0$), or
2. there are infinitely many indices $n$ with $a_n\neq 0$, and the signs of the non-zero $a_n$ change infinitely often.

The sign-change condition is necessary: if the non-zero coefficients are eventually of one sign, the dominant-term comparison shows $f(x)\to\pm\infty$ as $x\to+\infty$ or $x\to-\infty$.

Every non-constant polynomial is unbounded on $\mathbb{R}$.

Consider the geometric series $$ f(x) = \sum_{n=0}^{\infty}(-1)^n x^{2n} = \frac{1}{1+x^2}, \qquad |x|<1. $$ The coefficients alternate in sign, yet $R=1\neq+\infty$. One must first require $R=+\infty$ before the sign-change condition becomes meaningful.

For a subtler case with $R=+\infty$: take $a_n=(-1)^n/n!$, so $$ f(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^n = e^{-x}. $$ The signs alternate, yet $e^{-x}\to+\infty$ as $x\to-\infty$.

The $e^{-x}$ example reveals the key gap: sign alternation of the coefficients does not prevent the function from growing in one direction, because the series for $e^{-x}$ reconstructs exponential growth in the negative half-line. A complete criterion must capture cancellation in both directions.

Let $f$ be holomorphic in $|z|\le R$. Then for $0<r<R$, $$ M(r) \le \frac{2r}{R-r}\sup_{|z|=R}\operatorname{Re}f(z) + \frac{R+r}{R-r},|f(0)|. $$

Borel–Carathéodory shows that the real part of a complex-valued entire function controls its modulus. For a real-valued function on $\mathbb{R}$ the analogous control is more delicate, since we only observe the function on a line, not on a disk.

Every entire function of finite order $\rho$ can be written as $$ f(z) = z^m e^{g(z)}\prod_{n=1}^{\infty} E_p!\left(\frac{z}{z_n}\right), $$ where $m\ge 0$, $p=\lfloor\rho\rfloor$, $g$ is a polynomial of degree $\le\rho$, and the $E_p$ are Weierstrass elementary factors.

A bounded real entire function of infinite order (if one exists) would not be directly covered by the Hadamard factorisation. Understanding the zero set and the exponential factor in $e^{g(z)}$ may be key to classifying all $f\in\mathcal{B}$.

Does $\mathcal{B}$ consist precisely of functions of the form $c\cos(h(x))$ and their linear combinations and products, where $h:\mathbb{R}\to\mathbb{R}$ is entire and $c\in\mathbb{R}$?

By Proposition 1 (3), every $c\cos(h(x))$ belongs to $\mathcal{B}$, and $\mathcal{B}$ is an algebra, so all products and sums remain in $\mathcal{B}$. What is unknown is whether every element of $\mathcal{B}$ arises this way. Note that $\sin x = \cos(x-\pi/2) \in \mathcal{B}$, so sine is already covered.

Is there a decay condition on $|a_n|$, combined with the sign-change condition, that gives a sufficient criterion for $f\in\mathcal{B}$?

Can the Fourier theory for tempered distributions give a necessary and sufficient condition for $f\in\mathcal{B}$ in terms of the spectral support of $f$?

Characterise those linear differential operators $L$ with entire coefficients whose full solution space lies within $\mathcal{B}$.

If the partial sums $S_N(x)=\sum_{n=0}^{N}a_n x^n$ are uniformly bounded on growing intervals $[-R_N,R_N]$ (with $R_N\to\infty$), does it follow that $f\in\mathcal{B}$? Conversely, if $f\in\mathcal{B}$, how fast must $R_N$ grow relative to $N$ for the bound to hold?

• Analysis
• Entire Functions

• Function Theory
• Calculus
• Continuity