A Gaussian Counterexample for the Uniform Convergence of Power Series

Published

10 November 2023

Suppose that

f (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}

has an infinite radius of convergence. Moreover, suppose that

lim_{x \to \pm \infty} f (x) = L < \infty

Do the sequence of functions

f_{n} (x) = \sum_{k = 0}^{n} a_{k} x^{k}

converge to $f (x)$ uniformly? No. Consider the counterexample:

e^{- x^{2}} = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} x^{2 n}}{n!}

Consider

a_{n} = {\begin{cases} \frac{(- 1)^{k}}{k!} & if n = 2 k \\ 0 & if n = 2 k + 1 \end{cases}

Then,

lim sup | a_{n} |^{1 / n} = lim | a_{2 k} |^{1 / 2 k} = lim \frac{1}{k!^{1 / 2 k}} = 0

Thus, $\sum_{n = 0}^{\infty} \frac{(- 1)^{n} x^{2 n}}{n!}$ has infinite radius of convergence. Moreover,

lim_{x \to \pm \infty} e^{- x^{2}} = 0

However, if $f_{n} \to f$ uniformly, then

lim sup {| f_{n} (x) - f (x) | : x \in R} = 0

But any partial sum will be unbounded on $R$ .

\begin{aligned} sup {| f_{n} (x) - f (x) | : x \in R} \\ = sup {| \sum_{k = 0}^{n} \frac{(- 1)^{k} x^{2 k}}{k!} - e^{- x^{2}} | : x \in R} \\ = + \infty x \to \infty \end{aligned}

Thus, $f_{n} (x)$ does not converge uniformly to $f (x)$ . It turns out that any power series with infinite radius of convergence that converges uniformly must be a polynomial.