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From: Maury Barbato on 29 Oct 2009 11:26
Hello,
let a:N->Q a bijection, and let f:R->R be the
function

e^[-(1/x)] if x > 0
f(x) =
0 if x <= 0


Define g:R->R as follows

g(x) = sum_{n=0 to inf} [2^(-n)]*f(x - a(n))

g is well defined and it's easy to prove, using
the theorem on the derivative of a function series,
that g is C^(inf).
I think, anyhow, that g is nowhere analytic, but
I can't give a rigorous proof. Do you have some idea?

Thank you very much for your attention.
My Best Regards,
Maury Barbato
From: Timothy Murphy on 29 Oct 2009 15:40
Maury Barbato wrote:

> Hello,
> let a:N->Q a bijection, and let f:R->R be the
> function
>
> e^[-(1/x)] if x > 0
> f(x) =
> 0 if x <= 0
>
>
> Define g:R->R as follows
>
> g(x) = sum_{n=0 to inf} [2^(-n)]*f(x - a(n))
>
> g is well defined and it's easy to prove, using
> the theorem on the derivative of a function series,
> that g is C^(inf).
> I think, anyhow, that g is nowhere analytic, but
> I can't give a rigorous proof. Do you have some idea?

I don't see why your series is convergent, say for x = 0,
since you might have a(n) < 0 and very small
for arbitrarily large n.


--
Timothy Murphy
e-mail: gayleard /at/ eircom.net
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
From: Robert Israel on 29 Oct 2009 16:25
Timothy Murphy <gayleard(a)eircom.net> writes:

> Maury Barbato wrote:
>
> > Hello,
> > let a:N->Q a bijection, and let f:R->R be the
> > function
> >
> > e^[-(1/x)] if x > 0
> > f(x) =
> > 0 if x <= 0
> >
> >
> > Define g:R->R as follows
> >
> > g(x) = sum_{n=0 to inf} [2^(-n)]*f(x - a(n))
> >
> > g is well defined and it's easy to prove, using
> > the theorem on the derivative of a function series,
> > that g is C^(inf).
> > I think, anyhow, that g is nowhere analytic, but
> > I can't give a rigorous proof. Do you have some idea?
>
> I don't see why your series is convergent, say for x = 0,
> since you might have a(n) < 0 and very small
> for arbitrarily large n.

The problem is not when a(n) is small, but when it is large.
I think he means to use a bounded f, say replace
exp(-1/x) by exp(-1/x)/(1 + exp(-1/x)). Then the 2^(-n) ensures
that the series converges uniformly.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Zdislav V. Kovarik on 29 Oct 2009 15:55


On Thu, 29 Oct 2009, Maury Barbato wrote:

> Hello,
> let a:N->Q a bijection, and let f:R->R be the
> function
>
> e^[-(1/x)] if x > 0
> f(x) =
> 0 if x <= 0
>
>
> Define g:R->R as follows
>
> g(x) = sum_{n=0 to inf} [2^(-n)]*f(x - a(n))
>
> g is well defined and it's easy to prove, using
> the theorem on the derivative of a function series,
> that g is C^(inf).
> I think, anyhow, that g is nowhere analytic, but
> I can't give a rigorous proof. Do you have some idea?
>
> Thank you very much for your attention.
> My Best Regards,
> Maury Barbato

(There is always a risk that infinite summation may "iron out" the lack of
analyticity.)

If you settle for another example of a C^(inf) nowhere analytic function
with an easy proof, take

f(x) = sum(n=0 to infinity) cos(2^n * x)/n!

It is easy to calculate its derivatives at x=0, and when you apply
Cauchy-Hadamard formula for radius of convergence, it will come out as
zero.
1/R = lim sup (abs (f^(n)(0))/n!)

Now, f is (2*pi) periodic, so integer multiples of 2*pi are points of
non-analyticity.

If you drop finitely many initial terms of the series, the period shortens
by powers of 2, so you get lack of analyticity at all binary fractions of
2*pi and their integer multiples. That is a dense subset, and you are
done.

(Inspired by Weierstrass's example of an everywhere continuous, nowhere
differentiable function.)

Cheers, ZVK(Slavek).
From: Zdislav V. Kovarik on 29 Oct 2009 16:14
A correction of Cauchy-Hadamard formula below.

On Thu, 29 Oct 2009, Zdislav V. Kovarik wrote:

>
>
> On Thu, 29 Oct 2009, Maury Barbato wrote:
>
> > Hello,
> > let a:N->Q a bijection, and let f:R->R be the
> > function
> >
> > e^[-(1/x)] if x > 0
> > f(x) =
> > 0 if x <= 0
> >
> >
> > Define g:R->R as follows
> >
> > g(x) = sum_{n=0 to inf} [2^(-n)]*f(x - a(n))
> >
> > g is well defined and it's easy to prove, using
> > the theorem on the derivative of a function series,
> > that g is C^(inf).
> > I think, anyhow, that g is nowhere analytic, but
> > I can't give a rigorous proof. Do you have some idea?
> >
> > Thank you very much for your attention.
> > My Best Regards,
> > Maury Barbato
>
> (There is always a risk that infinite summation may "iron out" the lack of
> analyticity.)
>
> If you settle for another example of a C^(inf) nowhere analytic function
> with an easy proof, take
>
> f(x) = sum(n=0 to infinity) cos(2^n * x)/n!
>
> It is easy to calculate its derivatives at x=0, and when you apply
> Cauchy-Hadamard formula for radius of convergence, it will come out as
> zero.
> 1/R = lim sup (abs (f^(n)(0))/n!)
>
Correction: 1/R = lim sup (abs (f^(n)(0))/n!)^(1/n)

My apologies.

> Now, f is (2*pi) periodic, so integer multiples of 2*pi are points of
> non-analyticity.
>
> If you drop finitely many initial terms of the series, the period shortens
> by powers of 2, so you get lack of analyticity at all binary fractions of
> 2*pi and their integer multiples. That is a dense subset, and you are
> done.
>
> (Inspired by Weierstrass's example of an everywhere continuous, nowhere
> differentiable function.)
>
> Cheers, ZVK(Slavek).
>
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