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From: Maury Barbato on 26 Apr 2010 01:46
Hello,
does there exist a nowhere continuous function f:R->R
(where R denotes the set of real numbers), such that

lim_{h->0} [f(x+h)-f(x-h)] = 0,

for every x in R?

Thank you very much for your attention.
My Best Regards,
Maury Barbato
From: kelsar777 on 26 Apr 2010 06:34
On 26 Kwi, 11:46, Maury Barbato <mauriziobarb...(a)aruba.it> wrote:
> Hello,
> does there exist a nowhere continuous function f:R->R
> (where R denotes the set of real numbers), such that
>
> lim_{h->0} [f(x+h)-f(x-h)] = 0,
>
> for every x in R?
>
> Thank you very much for your attention.
> My Best Regards,
> Maury Barbato

It's Dirichlet Function: d(x) := 1 for x rational and 0 for x
irrational.
From: Tim Little on 26 Apr 2010 06:42
On 2010-04-26, kelsar777 <kelsar777(a)gmail.com> wrote:
> On 26 Kwi, 11:46, Maury Barbato <mauriziobarb...(a)aruba.it> wrote:
>> lim_{h->0} [f(x+h)-f(x-h)] = 0 for every x in R?
>
> It's Dirichlet Function: d(x) := 1 for x rational and 0 for x
> irrational.

That only has the required property for rational x. For every
irrational x, we can choose arbitrarily small h such that x+h is
rational. Then x-h is irrational, and therefore f(x+h)-f(x-h) = 1.


- Tim
From: scattered on 26 Apr 2010 06:46
On Apr 26, 6:34 am, kelsar777 <kelsar...(a)gmail.com> wrote:
> On 26 Kwi, 11:46, Maury Barbato <mauriziobarb...(a)aruba.it> wrote:
>
> > Hello,
> > does there exist a nowhere continuous function f:R->R
> > (where R denotes the set of real numbers), such that
>
> > lim_{h->0} [f(x+h)-f(x-h)] = 0,
>
> > for every x in R?
>
> > Thank you very much for your attention.
> > My Best Regards,
> > Maury Barbato
>
> It's Dirichlet Function: d(x) := 1 for x rational and 0 for x
> irrational.

I don't see it: since x+h rational does not imply x-h rational, it is
not obvious that the Dirichlet function satisfies the specified limit
property.
From: William Elliot on 26 Apr 2010 06:50
On Mon, 26 Apr 2010, Maury Barbato wrote:

> Hello,
> does there exist a nowhere continuous function f:R->R
> (where R denotes the set of real numbers), such that
>
> lim_{h->0} [f(x+h)-f(x-h)] = 0,
>
> for every x in R?
>
f(x) = 0 if x rational
= 1 if x irrational ??

If x is rational, then f(x + h) = f(x - h).
If x is irrational, then f(x + h) = f(x - h), for all rational h.

What if x,h and x+h are irrational and x - h rational?

Let {sj} be a rational sequence with sj -> e, h = e - sj, x = e.

Then f(x + h) = f(2e - sj) = 1, f(x - h) = f(sj) = 0.
lim(h->0) (f(x + h) - f(x - h) /= 0. Shucks.
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