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I proposed the following problem (from Bishop,
R. L. and Crittenden, R. J., Geometry of Manifolds).
Let {r_n} be an enumeration of the rational numbers,
and let g:R->R be the function
g(x) = exp(-1/x) if x>0
g(x) = 0 if x<=0
Prove that the function f:R->R defined by
f(x) = sum_{n=0 to inf} (2^(-n))*g(x - r_n) (I)
is nowhere analytic.
Unfortunately, nobody could give a proof. So I began
looking for a proof in the literature. I found a lot of
material about smooth nowhere analytic functions.
These are some very interesting sci.math threads:
http://mathforum.org/kb/message.jspa?
messageID=29381&tstart=0
http://sdp.mathforum.org/kb/message.jspa?
messageID=6615579&tstart=0
http://mathforum.org/kb/message.jspa?
messageID=197606&tstart=0
http://mathforum.org/kb/message.jspa?
messageID=155665&tstart=0
The two following essays by Dave L. Renfro
http://mathforum.org/kb/message.jspa?messageID=387148
http://mathforum.org/kb/message.jspa?
messageID=387149&tstart=0
are an invaluable source of material for those
interested in the subject. Dave gives a lot of
references, so I don't list them here.
Unfortunately, in all the works I referred to
there's no proof that a function like the f
in (I) is nowhere analytic. All that I found is
an editorial note in the Amer. Math. Month.,
70, 1963, p. 1109, which says that I. N. Baker
proved that the function
f(x) = sum_{n=0 to inf} (2^(-n))*g(x - r_n) (II)
where {r_n} is an enumeration of the rationals and
g(x) = exp(-1/(x^2)),
is nowhere analytic.
Anyhow, I'm not so sure that Baker did it. The fact
is that the so called "Principle of Condensation of
Singularities", introduced by Hankel, Cantor and
du Bois-Reymond, is not suited to produce an example
of a smooth nowhere analytic function (instead, it was
extensively used to produce many examples of other sorts
of "pathological" functions: see Hobson, The Theory of
Functions of a Real Variable, Ch. VI or Dini, Fondamenti
per la Teorica delle Funzioni di Variabili Reali). The
point is that the derivatives of functions like (I) or
(II) are not simple to study!
The only heuristic argument that induces me to believe
that a function like (I) is nowhere analytic is the
following. Replace the factors 2^(-n) in (I) with the
terms a_n of a generic absolute convergent series. Now,
suppose f isanalytic in some rational r_m. Then, for any
c=/=a_m, the function
h(x) = sum_{n=0 to inf} b_n*g(x - r_n),
where b_m = c and b_n = a_n for n=/=m, would not be
analytic in r_m, being g(x - r_m) not analytic there.
So, it seems quite strange that choosing the particular
sequence {(2^(-n)} you obtain an analytic function.
Anyhow, I tried to get a rigorous proof, but
unsuccessfully. Without any proof in the literature
that the principle of condensation of singularities can
provide examples of smooth nowhere analytic functions,
I must conclude that such examples should be considered
simply as conjectures.
What do you think about?
Thank you very much for your attention.
My Best Regards,
Maury Barbato