Is it possible to find a sequence of uniformly bounded, continuous and increasing functions $f _n:\mathbb{R} \rightarrow \mathbb{R}$ such that they converge pointwise to a continuous function $f$ but the convergence is not uniform?


  • $\begingroup$ $f_n(x) = \chi_{(-\infty,n]}$ is a discontinuous example. The idea should be the same with continuous example. (and by increasing, do you mean $f_n \leq f_{n+1}$ ?) $\endgroup$
    – Xiao
    Feb 3 '15 at 3:18
  • 1
    $\begingroup$ No, by increasing I mean $f_n(x)\leq f_n(y)$ whenever $x\leq y$ for all $n$. You are right, thanks! $\endgroup$
    – Vokram8
    Feb 3 '15 at 3:25
  • $\begingroup$ Then $f_n(x) = \chi_{[-n,\infty)}(x)$ is a discontinuous example. To get a continuous one, just "connect" the jump at around $-n$ $\endgroup$
    – Xiao
    Feb 3 '15 at 3:26

No, this does not hold. Consider e.g. $$f_n(x) := \begin{cases} 0, & x \leq 0, \\ \left( \frac{x}{n} \right)^n, & x \in (0,n), \\ 1, & x \geq n. \end{cases}$$

Note however that the claim does hold true if the sequence is increasing in the sense that $f_n(x) \leq f_{n+1}(x)$ for all $x \in \mathbb{R}$, $n \in \mathbb{N}$ (by Dini's theorem).


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