From: ullrich@math.okstate.edu (David C. Ullrich)
Subject: Re: infinitely differentiable non analytic function not involving an   analytic function
Date: Mon, 08 Jan 2001 13:10:12 GMT
Newsgroups: sci.math

On Sun, 7 Jan 2001 20:00:22 GMT, Manu Militari <MM@never.ate.spam>
wrote:

>Do all the examples of real functions infinitely differentiable at a
>point but non analytic there involve an essential singularity of an
>analytic function?
>
>This is certianly a way to construct such (the canonical
>example of f(0) = 0, f(x) = exp(1/x^2), x != 0 is widely
>used, in fact it's the only example I recall ever seeing.)
>
>Is there a simple example that doesn't use such a "hack"?

Yes or no depending on your definition of "simple".
You can't expect to write down a simple _formula_
for the example, because the things we recognize as
closed-form formulas all seem to be in terms of
analytic functions. But it's very easy to construct
examples, (or to prove examples exist by the
Baire category theorem).

The following has the advantage that simple
modifications can be used to construct functions
with various desired properties. (That's more or
less a quote from Rudin, as is the construction,
more or less):

For h > 0 define

g_h(x) = 1/h if 0 < x < h
g_h(x) = 0 otherwise.

We will (well we won't here but we could) choose
a sequence h_j tending to 0 very very fast. Then
we set

f_n = g_h_1 * g_h_2 * ... * g_h_n,

where the * denotes convolution. The functions
f_n get smoother as n increases: f_1 is not
continuous, f_2 is continuous but not differentiable
(the graph of f_2 is a triangle), f_3 is differentiable
but not twice differentiable, etc. 

If we assume just that sum(h_j) < finite that's
enough to make f_n converge uniformly to f;
now f is continuous and has compact support
(but f is certainly non-zero since the integral
of f is 1). Assuming stronger conditions on
the h_j makes the derivatives f_n' converge
uniformly to something, hence f is differentible;
a somewhat stronger condition makes the
second derivatives converge, etc - if you
choose the right sequence h_j then f
is infinitely differentiable (but not analytic
since it has compact support but integral 1.)

See Rudin "Functional Analysis" for details -
not sure if it's at the start of the book or
at the start of the section on test functions.

>- Jim (jroth arrobas world point std point com)
>
==============================================================================

From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: infinitely differentiable non analytic function not involving an   analytic function
Date: 8 Jan 2001 10:21:49 -0500
Newsgroups: sci.math

In article <3a59b97d.2652222@nntp.sprynet.com>,
David C. Ullrich <ullrich@math.okstate.edu> wrote:
:On Sun, 7 Jan 2001 20:00:22 GMT, Manu Militari <MM@never.ate.spam>
:wrote:
:
:>Do all the examples of real functions infinitely differentiable
:>at a point but non analytic there involve an essential
:>singularity of an analytic function?
:>
:>This is certianly a way to construct such (the canonical
:>example of f(0) = 0, f(x) = exp(1/x^2), x != 0 is widely
:>used, in fact it's the only example I recall ever seeing.)
:>
:>Is there a simple example that doesn't use such a "hack"?
:
[D. Ullrich's response deleted]
(In my newserver, the original article floated out over the
weekend, so I have to addredd it indirectly.)

There is a function infinitely differentiable on R but nowhere
(real) analytic. It follows one of the recipes for a continuous
nowhere differentiable function:

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

At 0, one uses Cauchy-Hadamard formula for radius of
convergence; at multiples of 2*pi/2^k, observe that f becomes
(2*pi/2^k)-periodic after dropping first k terms,
and the points of non-analyticity are then dense.

Cheers, ZVK(Slavek).