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Non-convergence counter example?

  1. Jul 27, 2010 #1

    strangerep

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    Non-convergence counter example??

    (This question occurred to me in the context of quantum field theory,
    but since it's purely mathematical, I'm asking it here...)

    Consider the universal vector space [itex]\Xi[/itex] of arbitrary-length
    sequences over C (the complex numbers). Denote

    [tex]
    \delta_k ~:=~ (0,0,\dots,0,1,0,0,\dots) ~~,
    ~~~~~ \mbox{where the 1 is in the k-th position.}
    [/tex]

    Every [itex]f\in\Xi[/itex] can be expressed as
    [tex]
    f ~:=~ \sum_{k=0}^\infty \, f_k \, \delta_k
    ~~~~~ \mbox{where all}~ f_k \in C.
    [/tex]

    For arbitrary [itex]f,g \in \Xi[/itex], we denote the usual positive definite
    (left antilinear) Hermitian formal inner product [itex](\cdot,\cdot)[/itex] by

    [tex]
    (f,g) ~:=~ \sum_{k=0}^\infty\, \bar{f_k} \, g_k ~~.
    [/tex]

    Such an inner product is ill-defined (divergent) in general, and one
    usually imposes restrictions on the allowed coefficients [itex]f_k[/itex] to
    ensure the inner product is well-defined. For example, one might demand
    that [itex]f_k=0[/itex] for all [itex]k[/itex] greater than some integer [itex]N_f[/itex] (different in
    general for each [itex]f[/itex]). Denote this space as V. Alternatively, one could
    demand that the [itex]\sum_k|f_k|^2[/itex] is finite, this space being denoted by H
    and is Cauchy-complete in the norm topology induced by the inner product.
    It's well known that H is self-dual, i.e., [itex]H^* = H[/itex].

    But my question concerns the universal vector space [itex]\Xi[/itex] above,
    in which there are no restrictions on the coefficients in the sequences,
    and this expression:

    [tex]
    P(g,f) ~:=~ {\frac{|(g,f)|^2}{(g,g) \, (f,f)}}
    ~~, ~~~~~~~~~(f,g \ne 0) ~.
    [/tex]

    Since the numerator and/or either factor in the denominator may be
    divergent, I'm interpreting this expression in the following sense:

    [tex]
    P(g,f)
    ~:=~ \lim_{N\to\infty} \, P_N(g,f) \,
    ~:=~ \lim_{N\to\infty} \, \frac{|(g,f)_N|^2}
    {\, (g,g)_N \; (f,f)_N} ~~,
    ~~~~~~~~~(f,g \ne 0) ~,
    [/tex]

    where the "N" subscripts mean that we have truncated the vectors
    to N terms (i.e., restricted to an N-dimensional subspace).

    By the usual Cauchy-Schwarz inequality, P(g,f) is bounded in [0,1],
    since every term of the sequence is so-bounded (i.e., for any value
    of N). This is true even if the numerator or denominator diverge
    separately as N increases arbitrarily (afaict)

    But here's my question: does the sequence necessarily converge?
    Or are there examples of f,g such that [itex]P_N(g,f)[/itex]
    wanders back and forth inside the range [0,1] forever with
    increasing N, never converging to a particular value therein?

    The random examples I've tried (where f and/or g have divergent
    norm, and/or (f,g) is divergent) all seem to converge to either
    0 or 1. I can't find any examples that oscillate back and
    forth forever in a non-convergent manner.

    Can anyone out there think of an example of f,g such that the sequence of
    [itex]P_N(g,f)[/itex] oscillates forever with increasing N? Or am I overlooking
    some standard result which says that such sequences do indeed always converge??

    Thanks in advance for any help/suggestions.
     
  2. jcsd
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