Measuring Fractal Dimension ?

Mark Dickinson dickinsm at
Wed Jun 17 07:52:29 EDT 2009

On Jun 17, 7:04 am, Paul Rubin <http://phr...@NOSPAM.invalid> wrote:
> I think a typical example of a curve that's continuous but not
> uniformly continuous is
>    f(t) = sin(1/t), defined when t > 0
> It is continuous at every t>0 but wiggles violently as you get closer
> to t=0.  You wouldn't be able to approximate it by sampling a finite
> number of points.  A sequence like
>    g_n(t) = sin((1+1/n)/ t)    for n=1,2,...
> obviously converges to f, but not uniformly.  On a closed interval,
> any continuous function is uniformly continuous.

Right, but pointwise convergence doesn't imply uniform
convergence even with continuous functions on a closed
bounded interval.  For an example, take the sequence
g_n (n >= 0), of continuous real-valued functions on
[0, 1] defined by:

g_n(t) = nt if 0 <= t <= 1/n else 1

Then for any 0 <= t <= 1, g_n(t) -> 0 as n -> infinity.
But the convergence isn't uniform:  max_t(g_n(t)-0) = 1
for all n.

Maybe James is thinking of the standard theorem
that says that if a sequence of continuous functions
on an interval converges uniformly then its limit
is continuous?


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