A function that converges pointwise but not uniformly?

Question

Give an example of a sequence of functions $f_n \in C(X)$ (where $X$ is a compact metric space), and a function $f \in C(X)$ such that $f_n \rightarrow f$ pointwise, but not uniformly.

I am having trouble with the condition that $f_n$ and $f$ are in $C(X)$, so I assume they both need to be continuous?

Dini's theorem shows $f_n$ can't be monotone in that case. Hope this helps. — cronos2, Apr 22 '17 at 16:02
On $[0,1]$, let the graph of $f_n$ be the line segments connecting $(0,0)$ to $(1/n,1)$, $(1/n,1)$ to $(2/n,0)$, and $(2/n,0)$ to $(1,0)$ ($n>2$). — David Mitra, Apr 22 '17 at 16:11
I simply cannot believe there aren't alread several examples of this in the site. Have you looked? — Mariano Suárez-Álvarez, Apr 22 '17 at 16:17
@cronos2 that does help, thank you! So I assume that can apply to the moving hats example that David Mitra described which does in fact converge point wise to 0 but not uniformly — km24, Apr 22 '17 at 16:43

score 2 · Answer 1 · answered Aug 16 '18 at 04:40

If $f(x)=0$ for all $x\in[0,1],$ and if $f_n(x)=0$ for all $x\in\{0\}\cup[\frac1n,1],$ then we know enough to conclude that $f_n\to f$ pointwise. Can you somehow contrive to complete the definition of $f_n$ so that the sequence does not converge uniformly?

score 1 · Answer 2 · 2019-01-14T12:45:24.047

1

Consider $f_n$ = $\frac{nx}{1+n^2 x^2}$ where $x$ $\epsilon$ $[0,1]$.Here the limit function for the above sequence is $f$ = $0$ when for all $x$ $\epsilon$ $[0,1]$.

Clearly, $f_n$ = $\frac{nx}{1+n^2 x^2}$ is a sequence of continuous functions and $f$ is also continuous.

Now, $M_n$ $=$ $sup_{x \in [0,1]}|f_n(x) - f(x)|$ = $sup_{x \in [0,1]} \frac{nx}{1+n^2 x^2}$ $=$ $\frac{1}{2}$.

So by $M_n$ test, $f_n$ converges to f pointwise, not uniformly.

edited Jan 14 '19 at 12:45

answered Apr 22 '17 at 15:48

$f$ was supposed to be continuous – cronos2 Apr 22 '17 at 16:00
I hope this example works. – Apr 22 '17 at 17:25

A function that converges pointwise but not uniformly?

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