Non-convergence counter example?

[strangerep]

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 $\Xi$ of arbitrary-length
sequences over C (the complex numbers). Denote

$$\delta_k ~:=~ (0,0,\dots,0,1,0,0,\dots) ~~,<br /> ~~~~~ \mbox{where the 1 is in the k-th position.}$$

Every $f\in\Xi$ can be expressed as
$$f ~:=~ \sum_{k=0}^\infty \, f_k \, \delta_k<br /> ~~~~~ \mbox{where all}~ f_k \in C.$$

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

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

Such an inner product is ill-defined (divergent) in general, and one
usually imposes restrictions on the allowed coefficients $f_k$ to
ensure the inner product is well-defined. For example, one might demand
that $f_k=0$ for all $k$ greater than some integer $N_f$ (different in
general for each $f$). Denote this space as V. Alternatively, one could
demand that the $\sum_k|f_k|^2$ 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., $H^* = H$.

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

$$P(g,f) ~:=~ {\frac{|(g,f)|^2}{(g,g) \, (f,f)}}<br /> ~~, ~~~~~~~~~(f,g \ne 0) ~.$$

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

$$P(g,f)<br /> ~:=~ \lim_{N\to\infty} \, P_N(g,f) \,<br /> ~:=~ \lim_{N\to\infty} \, \frac{|(g,f)_N|^2}<br /> {\, (g,g)_N \; (f,f)_N} ~~,<br /> ~~~~~~~~~(f,g \ne 0) ~,$$

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 $P_N(g,f)$
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
$P_N(g,f)$ 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.

 

[fresh_42]

Interesting question. I would expect that any behavior is possible in $\Xi$. There is a reason people consider bounded sequences instead. I would approach this subject via formal power series rather than analytical expressions. I guess your limits can diverge as well as oscillate.

 

[strangerep]

? Wow, an answer to a 9 yr old question that I thought (and still suspect) would never be resolved.

An additional complication is that, for the limit to be good, it must be independent of how it is approached. In this case, it means we must be able to re-order the sequence terms arbitrarily, such as (e.g.,) considering an equivalence class of all possible N-dimensional subspaces at each stage of the limit sequence.

 

[Infrared]

I don't see why the limit should exist. Say that $f$ and $g$ consist of only $0s$ and $1s$. If they start off alternating so that one is $1$ when the other is $0$, then $P_N=0$, and then for larger $N$ you can make them agree so that, you get $P_N$ arbitrarily close to $1$ by making this happen for long enough, and then for even larger $N$ make them go back to canceling so that $P_N$ can get arbitrarily close to $0$,...