Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Riemann's theorem for complex series

  1. Mar 24, 2010 #1
    Suppose the real numbers {a_k} are the terms of a conditionally convergent series. Let C({a_k}) denote the set of all sums of rearrangements of this series. A famous theorem of Riemann shows that, in this particular circumstance, C({a_k}) = R, the set of real numbers.

    Various generalizations of this result exist for Banach spaces, but I'm interested in the special case of complex series. In particular, if the complex numbers {a_k} are the terms of a conditionally convergent series, then C({a_k}) is an affine subspace of the complex plane.

    I was told this informally by a professor, but the result is completely fascinating to me, so I've been trying to find a proof, either online or by working one out myself. Well, so far I have failed in both endeavors!

    I was wondering if anyone had any insight into this problem, either a reference or an idea on how to proceed in proving it.

    One thing to note is that given an affine subspace of the plane, one can construct a series with terms {a_k} such that C({a_k}) is that particular subspace. For instance, if {alpha,beta} is a basis of the complex plane and {a_k} and {b_k} are *real* numbers that are the terms of two conditionally convergent series, then C({alpha * a_k + beta * a_k}) is the plane by Riemann's theorem as stated above for series with real terms.

    In short, it's evident that any affine subspace can be achieved in the form C({a_k})--the significant part is that C({a_k}) must always have this form! (Of course, this is provided that the associated series is conditionally convergent.)
  2. jcsd
  3. Mar 25, 2010 #2
    A convergent rearrangement of a_k determines that of b_k.The latter may not converge. Although your guess is right, we can't work with individual components.
    Here's a lemma which will prove the affinity of the subspace.
    Lemma : Let {c_n} be a conditionally convergent series ,with two rearrangements A & B converging to distinct complex numbers a,b. Then the series can be made to converge to ka+(1-k)b for any real number k.
    Here's why it works. Select a partial sum of A, which is close to a. The partial sums of B which include this partial sum eventually tend to b . Thus, we can 'move' along the direction a-->b ( back & forth by repeted selection) & generate the line.
    Suppose that complex number c not on this line is in C{c_n}. We can apply the lemma once more to infer that the range is ka+lb+mc for any real numbers k,l,m with
    k+l+m =1. This subspace is clearly the complex plane. (Essentially, if c c joined to every point on line a--b, the plane is spanned).
    The generalisation to real vector spaces follows similarly.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook