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From: Kallikanzarid on 1 Dec 2009 03:33
Can you give me a constructive example of nowhere continous function
that satisfies a classical Holder condition? Thanks in advance.
From: Henry on 1 Dec 2009 04:18
On 1 Dec, 08:33, Kallikanzarid <lex.a...(a)gmail.com> wrote:
> Can you give me a constructive example of nowhere continous function
> that satisfies a classical Holder condition? Thanks in advance.

I assume you want an f(x) which for some C and a
|f(x)-f(y)| <= C * |x-y|^a for all x and y

So consider f(x)=1 for x rational and f(x)=0 for x irrational
which satisfies the condition for a=0 (and in this case C>=1).
Any other bounded nowhere continous function will do instead.
From: Kallikanzarid on 1 Dec 2009 04:38
On 1 дек, 16:18, Henry <s...(a)btinternet.com> wrote:
> On 1 Dec, 08:33, Kallikanzarid <lex.a...(a)gmail.com> wrote:
>
> > Can you give me a constructive example of nowhere continous function
> > that satisfies a classical Holder condition? Thanks in advance.
>
> I assume you want an f(x) which for some C and a
> |f(x)-f(y)| <= C * |x-y|^a for all x and y
>
> So consider f(x)=1 for x rational and f(x)=0 for x irrational
> which satisfies the condition for a=0 (and in this case C>=1).
> Any other bounded nowhere continous function will do instead.

Heh, it's too trivial :) Can you give an example for alpha in (0, 1)?
From: Henry on 1 Dec 2009 06:39
On Dec 1, 9:38 am, Kallikanzarid <lex.a...(a)gmail.com> wrote:
> On 1 дек, 16:18, Henry <s...(a)btinternet.com> wrote:
>
> > On 1 Dec, 08:33, Kallikanzarid <lex.a...(a)gmail.com> wrote:
>
> > > Can you give me a constructive example of nowhere continous function
> > > that satisfies a classical Holder condition? Thanks in advance.
>
> > I assume you want an f(x) which for some C and a
> > |f(x)-f(y)| <= C * |x-y|^a for all x and y
>
> > So consider f(x)=1 for x rational and f(x)=0 for x irrational
> > which satisfies the condition for a=0 (and in this case C>=1).
> > Any other bounded nowhere continous function will do instead.
>
> Heh, it's too trivial :) Can you give an example for alpha in (0, 1)?

What is your definition for continuous?
Perhaps: for each x and given e>0, there exists d>0
where for all y with |x-y| <= d we have |f(x)-f(y)| <= e

So let d=(e/C)^(1/a) and use the Holder condition to show continuity
From: Kallikanzarid on 1 Dec 2009 08:21
lol you're right :))))))))))
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