Power series bounds

Question

Let $f(x)=\sum_{i} a_i x^i$ be real a power series which is absolutely convergent for all $x\in \mathbb{R}$, and in addition $f(x)$ is bounded, i.e. there is $M>0$ such that $|f(x)|<M$, for all $x\in \mathbb{R}$. Show that $g(x)= \sum_{i} a_i^2 x^i$ is also bounded. I suspect this is true, but don't have any proof. Does anyone know a proof of this?

Answer 1

The function $g(x)=\sum_{n=0}^\infty a_n^2 x^n$ cannot be bounded for all $x\in \mathbb{R}$ unless it is constant. If $x$ is positive and $a_m\neq 0$ for some $m>0$, then $$g(x)=\sum_{n=0}^\infty a_n^2 x^n =a_m^2 x^m +\sum_{m\neq n=0}^\infty a_n^2 x^n \geqslant a_m^2 x^m.$$ The inequality is because all terms in the sum are non-negative. Since $a_m\neq 0$, $a_m^2>0$, and $g(x)\geqslant a_m^2 x^m$ is unbounded.

More generally, a power series $\sum_{n=0}^\infty b_nx^n$, if all coefficients $b_n$ are non-negative and if the power series has infinite radius of convergence, it cannot be bounded unless $b_m=0$ for all $m>0$ (that is, unless it is constant).