Theorem: Let f(x) be a real valued function of one real variable such
that \sum_{n=1}^\infty f(a_n) converges whenever \sum_{n=1}^\infty a_n
converges. Then f(x) = cx, some constant c, on some neighbourhood of x =
0.
James Currie, Math
cur...@uwinnipeg.ca
>Theorem: Let f(x) be a real valued function of one real variable such
>that \sum_{n=1}^\infty f(a_n) converges whenever \sum_{n=1}^\infty a_n
>converges. Then f(x) = cx, some constant c, on some neighbourhood of x =
>0.
Presumably "whenever" means "if and only if" here? If f(x) is always 0
then sum f(a_n) certainly converges whenever sum a_n converges...
--
Gareth McCaughan Dept. of Pure Mathematics & Mathematical Statistics,
gj...@pmms.cam.ac.uk Cambridge University, England. [Research student]
This was proved by G. Waldenberg, American Mathematical Monthly 95 (1988)
542-544. Y. Benjamini's solution to Problem E3404, American Mathematical
Monthly 99 (1992) 466-467 contains an extension:
Let f: X -> Y be a mapping of normed spaces such that \sum_{n=1}^\infty f(a_n) converges whenever \sum_{n=1}^\infty a_n converges (both in the norm topology).
Then there is a neighbourhood of 0 on which f is equal to a bounded linear
operator.
I have a slight additional extension (unpublished):
Let f: X -> Y be a mapping of Banach spaces such that \sum_{n=1}^\infty f(a_n) converges weakly whenever \sum_{n=1}^\infty a_n converges (strongly). Then
there is a neighbourhood of 0 on which f is equal to a bounded linear
operator.
--
Robert Israel isr...@math.ubc.ca
Department of Mathematics (604) 822-3629
University of British Columbia fax 822-6074
Vancouver, BC, Canada V6T 1Y4
> In article <Pine.OSF.3.91.95120...@io.UWinnipeg.ca>,
> James Currie <cur...@io.UWinnipeg.ca> wrote:
>
> >Theorem: Let f(x) be a real valued function of one real variable such
> >that \sum_{n=1}^\infty f(a_n) converges whenever \sum_{n=1}^\infty a_n
> >converges. Then f(x) = cx, some constant c, on some neighbourhood of x =
> >0.
>
> Presumably "whenever" means "if and only if" here? If f(x) is always 0
> then sum f(a_n) certainly converges whenever sum a_n converges...
A couple of people mailed me to point out that this is just fine, with c=0.
My thought processes had gone something like "well, you can take f(x)=o(x)
and surely that will do. What's the simplest case? Ah, f=0."
However, I must apologise very completely, because even this is not
true (so my previous posting was wholly without foundation). For instance,
f(x)=x^3 won't do, because you could (for instance) let b_n=n^{-1/4}
and then have a_n go b1,-b1/2,-b1/2,b2,-b2/2,-b2/2,b3,-b3/2,-b3/2,...
whose sum obviously converges to 0, while the sum of f(a_n) is
(1-2^{3/4})(b1^3+b2^3+b3^3+...) and this obviously diverges.
So, I grovel. One of these years I'll manage to post something to
sci.math.research without silly mistakes in. Maybe.