Consider the sequence of functions \(f_n: (-1, 1) \to \mathbb{R}\), defined by
\[ f_n(x) = \sum_{k=1}^{n} k x^k. \]
Show that \((f_n)\) converges in the topology of compact convergence; conclude that the limit function is continuous. (This is a standard fact about power series.)
Show that \((f_n)\) does not converge in the uniform topology.
Solution
1. For \(x \in (-1, 1)\), \(f_n(x) \to f(x) = \dfrac{x}{(1-x)^2}\).
For \(\abs{x} < 1\), differentiating the geometric series \(\sum_{k \ge 0} x^k = \frac{1}{1-x}\) term by term gives \(\sum_{k \ge 1} k x^{k-1} = \frac{1}{(1-x)^2}\); multiply by \(x\).
2. (a) \(f_n \to f\) uniformly on every compact \(K \subseteq (-1, 1)\), and \(f\) is continuous.
Since \(K\) is compact and \(\abs{x} < 1\) on \(K\), \(r = \max_{x \in K} \abs{x} < 1\). For \(x \in K\), \[\abs{f(x) - f_n(x)} = \abs{\sum_{k=n+1}^\infty k x^k} \le \sum_{k=n+1}^\infty k r^k.\] The series \(\sum_k k r^k\) converges by the ratio test, since \(\frac{(k+1) r^{k+1}}{k r^k} \to r < 1\). So its tails tend to \(0\), and \(\sup_{x \in K} \abs{f(x) - f_n(x)} \to 0\). Each \(f_n\) is a polynomial, hence continuous, and \(f\) is continuous on each compact \(K\) as a uniform limit of continuous functions. Every point of \((-1, 1)\) has a compact neighborhood \([-r, r]\) in \((-1, 1)\), so \(f\) is continuous.
3. (b) \((f_n)\) does not converge in the uniform topology.
Convergence in the uniform topology implies pointwise convergence, so a uniform limit would equal \(f\). For fixed \(n\), as \(x \to 1^-\), \(f_n(x) \to n(n+1)/2\) while \(f(x) \to +\infty\). Hence \(\sup_{x \in (-1, 1)} \abs{f(x) - f_n(x)} = \infty\), and the uniform distance \(\bar{\rho}(f_n, f) = 1\) for every \(n\).
4. Q.E.D.
Steps 2 and 3 prove parts (a) and (b).