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Re: Measuring Fractal Dimension ?

 
 
Paul Rubin
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      06-17-2009
Jaime Fernandez del Rio <(E-Mail Removed)> writes:
> I am pretty sure that a continuous sequence of
> curves that converges to a continuous curve, will do so uniformly.


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.
 
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Charles Yeomans
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      06-17-2009

On Jun 17, 2009, at 2:04 AM, Paul Rubin wrote:

> Jaime Fernandez del Rio <(E-Mail Removed)> writes:
>> I am pretty sure that a continuous sequence of
>> curves that converges to a continuous curve, will do so uniformly.

>
> 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.


Isn't (-∞, ∞) closed?

Charles Yeomans

 
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Mark Dickinson
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      06-17-2009
On Jun 17, 7:04*am, Paul Rubin <http://(E-Mail Removed)> 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?

Mark
 
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Mark Dickinson
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      06-17-2009
On Jun 17, 12:52*pm, Mark Dickinson <(E-Mail Removed)> wrote:
> g_n(t) = nt if 0 <= t <= 1/n else 1


Whoops. Wrong definition. That should be:

g_n(t) = nt if 0 <= t <= 1/n else
n(2/n-t) if 1/n <= t <= 2/n else 0

Then my claim that g_n(t) -> 0 for all t might
actually make sense...
 
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Jaime Fernandez del Rio
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      06-17-2009
On Wed, Jun 17, 2009 at 1:52 PM, Mark Dickinson<(E-Mail Removed)> wrote:
> 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?


Jaime was simply plain wrong... The example that always comes to mind
when figuring out uniform convergence (or lack of it), is the step
function , i.e. f(x)= 0 if x in [0,1), x(x)=1 if x >= 1, being
approximated by the sequence f_n(x) = x**n if x in [0,1), f_n(x) = 1
if x>=1, where uniform convergence is broken mostly due to the
limiting function not being continuous.

I simply was too quick with my extrapolations, and have realized I
have a looooot of work to do for my "real and functional analysis"
exam coming in three weeks...

Jaime

P.S. The snowflake curve, on the other hand, is uniformly continuous, right?

--
(\__/)
( O.o)
( > <) Este es Conejo. Copia a Conejo en tu firma y aydale en sus
planes de dominacin mundial.
 
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Mark Dickinson
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      06-17-2009
On Jun 17, 1:26*pm, Jaime Fernandez del Rio <(E-Mail Removed)>
wrote:
> On Wed, Jun 17, 2009 at 1:52 PM, Mark Dickinson<(E-Mail Removed)> wrote:
> > 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?


s/James/Jaime. Apologies.

> P.S. The snowflake curve, on the other hand, is uniformly continuous, right?


Yes, at least in the sense that it can be parametrized
by a uniformly continuous function from [0, 1] to the
Euclidean plane. I'm not sure that it makes a priori
sense to describe the curve itself (thought of simply
as a subset of the plane) as uniformly continuous.

Mark
 
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pdpi
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      06-17-2009
On Jun 17, 1:26*pm, Jaime Fernandez del Rio <(E-Mail Removed)>
wrote:
> P.S. The snowflake curve, on the other hand, is uniformly continuous, right?



The definition of uniform continuity is that, for any epsilon > 0,
there is a delta > 0 such that, for any x and y, if x-y < delta, f(x)-f
(y) < epsilon. Given that Koch's curve is shaped as recursion over the
transformation from ___ to _/\_, it's immediately obvious that, for a
delta of at most the length of ____, epsilon will be at most the
height of /. It follows that, inversely, for any arbitrary epsilon,
you find the smallest / that's still taller than epsilon, and delta is
bound by the respective ____. (hooray for ascii demonstrations)

Curiously enough, it's the recursive/self-similar nature of the Koch
curve so easy to prove as uniformly continuous.
 
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Mark Dickinson
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      06-17-2009
On Jun 17, 2:18*pm, pdpi <(E-Mail Removed)> wrote:
> On Jun 17, 1:26*pm, Jaime Fernandez del Rio <(E-Mail Removed)>
> wrote:
>
> > P.S. The snowflake curve, on the other hand, is uniformly continuous, right?

>
> The definition of uniform continuity is that, for any epsilon > 0,
> there is a delta > 0 such that, for any x and y, if x-y < delta, f(x)-f
> (y) < epsilon. Given that Koch's curve is shaped as recursion over the
> transformation from ___ to _/\_, it's immediately obvious that, for a
> delta of at most the length of ____, epsilon will be at most the
> height of /. It follows that, inversely, for any arbitrary epsilon,
> you find the smallest / that's still taller than epsilon, and delta is
> bound by the respective ____. (hooray for ascii demonstrations)


I think I'm too stupid to follow this. It looks as though
you're treating (a portion of?) the Koch curve as the graph
of a function f from R -> R and claiming that f is uniformly
continuous. But the Koch curve isn't such a graph (it fails
the 'vertical line test', in the language of precalculus 101),
so I'm confused.

Here's an alternative proof:

Let K_0, K_1, K_2, ... be the successive generations of the Koch
curve, so that K_0 is the closed line segment from (0, 0) to
(1, 0), K_1 looks like _/\_, etc.

Parameterize each Kn by arc length, scaled so that the domain
of the parametrization is always [0, 1] and oriented so that
the parametrizing function fn has fn(0) = (0,0) and fn(1) = (1, 0).

Let d = ||f1 - f0||, a positive real constant whose exact value
I can't be bothered to calculate[*] (where ||f1 - f0|| means
the maximum over all x in [0, 1] of the distance from
f0(x) to f1(x)).

Then from the self-similarity we get ||f2 - f1|| = d/3,
||f3 - f2|| = d/9, ||f4 - f3|| = d/27, etc.

Hence, since sum_{i >= 0} d/(3^i) converges absolutely,
the sequence f0, f1, f2, ... converges *uniformly* to
a limiting function f : [0, 1] -> R^2 that parametrizes the
Koch curve. And since a uniform limit of uniformly continuous
function is uniformly continuous, it follows that f is
uniformly continuous.

Mark
[*] I'm guessing 1/sqrt(12).
 
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Paul Rubin
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      06-17-2009
Mark Dickinson <(E-Mail Removed)> writes:
> It looks as though you're treating (a portion of?) the Koch curve as
> the graph of a function f from R -> R and claiming that f is
> uniformly continuous. But the Koch curve isn't such a graph (it
> fails the 'vertical line test',


I think you treat it as a function f: R -> R**2 with the usual
distance metric on R**2.
 
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Mark Dickinson
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      06-17-2009
On Jun 17, 3:46*pm, Paul Rubin <http://(E-Mail Removed)> wrote:
> Mark Dickinson <(E-Mail Removed)> writes:
> > It looks as though you're treating (a portion of?) the Koch curve as
> > the graph of a function f from R -> R and claiming that f is
> > uniformly continuous. *But the Koch curve isn't such a graph (it
> > fails the 'vertical line test',

>
> I think you treat it as a function f: R -> R**2 with the usual
> distance metric on R**2.


Right. Or rather, you treat it as the image of such a function,
if you're being careful to distinguish the curve (a subset
of R^2) from its parametrization (a continuous function
R -> R**2). It's the parametrization that's uniformly
continuous, not the curve, and since any curve can be
parametrized in many different ways any proof of uniform
continuity should specify exactly which parametrization is
in use.

Mark
 
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