Re: Differentiability of the Cantor function

From: José Carlos Santos (jcsantos_at_fc.up.pt)
Date: 03/29/05 

Date: Tue, 29 Mar 2005 17:21:01 +0100

On 29-03-2005 17:07, David C. Ullrich wrote:

>>It is well known that the Cantor function f is differentiable at any
>>point x of the complement of the Cantor set and that then f'(x) = 0.
>
> Saying this is "well known" almost sounds like there's something
> to be proved...
>

Oops, right. :-(

>>What I would like to know is this: exactly at which points is the
>>Cantor function not differentiable? Is it the Cantor set or some
>>smaller subset?
>
>
> I think it's clear that the function is differentiable at no
> point of the Cantor set.
>
> Exercise: If f is differentiable at x, x_n -> x, y_n -> x,
> x_n <= x, y_n >= x and x_n < y_n then
>
> (f(y_n)-f(x_n))/(y_n-x_n) -> f'(x).
>
> Supposing that, if f is the Cantor function and x is in
> the Cantor set then for every n we have x in I_n = [x_n, y_n]
> where y_n - x_n = 1/3^n and f(y_n) - f(x_n) = 1/2^n, hence
> f is not differentiable at x.

Thanks. It's quite easy, indeed. It's interesting to notice that "my"
statement (I mean, the assertion the set of non-differentiability of f
is exactly the Cantor set) is not true if f is replaced by the similar
function obtained starting from a fat Cantor set, because Lebesgue's
differentiation theorem tells that such a function is differentiable
almost everywhere.

Best regards,

Jose Carlos Santos

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