david macmanus

If you have access to Byron and Fuller's "Mathematics of Classical and
Quantum Mechanics" and would like to help me unravel some of their text
then please read on.
On page 222 (of volume 1) they have the following definition: Definition
5.7. A set of orthonormal functions is said to be *closed* if no nonzero
function is orthogonal to every function in the set.
My question is, is this defintion correct, or is it rather a *complete*
set of functions which is defined in this way?
They then go on to discuss the difference between closed and complete as
applied to set of orthogonal functions, but before I tackle this I need
to know if their defintion is right.
Many thanks,
David.


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robin chapman

Defnitions can't be correct or incorrect, but they may be
generally accepted or not generally accepted, alternatively,
standard or nonstandard.

I presume here that "function" effectively means
"element of a Hilbert (or simply an inner product) space".
It is certainly not standard to call such a set closed.
The word "closed" has a standard definition in general topology
and this is used in Hilbert space theory, as one often needs
to deal with closed subspaces. I presume these authors avoid this,
or otherwise have anothe nonstandard word for "closed"?


Hmmm. You just give the definition of "complete" as applied to orthogonal
sets. I wonder (well, not really) what they mean by "complete". On the other
hand complete is an overworked word: we also have complete metric spaces
(Hilbert spaces are an example of course :-)) and that's a completely (:-))
different concept!

--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"His mind has been corrupted by colours, sounds and shapes."
The League of Gentlemen

david macmanus

Thanks for your comments.
A definition of complete is given on page 220. They say, " It is in
terms of mean convergence that the completeness of an orthonormal set of
functions is defined.
Definition 5.6 Let g(x) be any function in Hilbert space and let
{f_i(x)} be an orthonormal set of functions in Hilbert space. If there
exist constant coefficients {a_i} such that the sequence of partial sums
g_n(x) ~ SUM (i=1 to n) a_i f_i(x) converges in the mean to g(x), then
the set of functions {f_i} is a complete orthonormal set. Equivalently,
if the mean square error can be made arbitrarily small then the set
{f_i} is a complete orthonormal set of functions."

This is how they define complete.

Thanks,
David.


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david c. ullrich

Well, they didn't phrase that very well - what they meant was
"Let {f_i(x)} be an orthonormal set of functions in Hilbert space.
If for any g there exist [etc] then {f_i} is complete."

Anyway, now I'm very curious what their discussion of the
"differences between" closed and complete amounts to,
because completeness and closedness (using their
definition) are equivalent. Does their discussion of the
differences end up by saying there are no differences?

************************

David C. Ullrich

david macmanus

not even a little bit?

what they mean by "complete".

I've posted their definition of complete in reply to David Ullrich's
reply.
Thanks for your comments - yes, I understand the use of the word
complete when describing spaces. In this particular context the word
refers to sets of functions, or vectors, I suppose, rather than spaces.
David.


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david macmanus

Hmmm...well, it is indeed the discussion of the difference between
closed and complete that I would like to understand - this occurs on
pages 222-223 of B+F. The simple answer to your question is "no", the
discussion does not end up saying there are no differences between the
terms. It would take me a while to type it into the PC but if you would
be willing to look at their discussion I'm more than willing type it in.
Just say the word (might be tomorrow now, though).
Thanks,
David.


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david macmanus

Here is their discussion - I hope I haven't made any typos!

Definition 5.6 Let g(x) be any function in Hilbert space and let
{f_i(x)} be an orthonormal set of functions in Hilbert space. If there
exist constant coefficients {a_i} such that the sequence of partial sums
g_n(x) ~ SUM (i=1 to n) a_i f_i(x) converges in the mean to g(x), then
the set of functions {f_i} is a complete orthonormal set. Equivalently,
if the mean square error can be made arbitrarily small then the set
{f_i} is a complete orthonormal set of functions.


Definition 5.7. A set of orthonormal functions is said to be closed if
no nonzero function is orthogonal to every function in the set.

Theorem 5.2. A set of orthonormal functions in Hilbert space is complete
if and only if it is closed.

Proof. We first prove that completeness of the set implies that the set
is closed. Assume that there is a nonzero function f(x) (and let it be
normalised), such that
(f_i, f) ~ c_i = INT (a to b) f_i(x)* f(x) dx = 0 for all i.
Then
Lim(n goes to infinity) INT (b to a) | f - SUM(i = 1 to n) c_i f_i |^2
dx =
INT (b to a) | f |^2 = 1 =/= 0,
(since f is normalised), so the set {f_i) is not complete. Thus
completeness of an orthonormal set of functions implies that there are
no functions that are orthonormal to every member of the set.
We now prove the converse: if the orthonormal set is closed, it is
complete.
If it is not complete, then the completeness relation:
(f, f) = SUM (i = 1 to infinity) | c_i |^2 = SUM (i = 1 to infinity) |
(f_i, f) |^2
is not satisfied.
Thus there exists some function f(x) such that
|| f ||^2 > SUM (n = 1 to infinity) | c_n |^2,
where c_n = (f_n, f). But since the above infinite series is convergent,
the sequence {g_m (x), where g_m (x) = SUM (n = 1 to m) c_n f_n (x) is a
Cauchy sequence in Hilbert space, and therefore, because of the
completeness of the SPACE, the g_m (x) must converge in the mean to a
limit in the space, call it g(x), such that c_n = (f_n, g). Therefore,
(f_n, g) - (f_n, f) so (f_n, f - g) = 0. Thus f(x) - g(x) is orthogonal
to f_n (x) for all n. We now show that the norm of f(x) - g(x) is not
equal to zero, so {f_n (x)} is not closed, contrary to our assumption.
It will then follow by contradiction that the set {f_n (x)} is complete
and the proof will be finished. Using the inequality
|| x - y || >= | || x || - || y || | we have
|| f - g || = || f - g_m - (g - g_m) || >= | || f - g_m || - || g - g_m
|| | for all m.
Now, as m goes to infinity, we know that || g - g_m || goes to zero,
whereas,
|| f - g_m ||^2 = || f - SUM (n =1 to m) c_n f_n ||^2 = || f || - SUM (n
= 1 to m) | c_n |^2 > 0 for all m by assumption.
Thus || f - g || > 0 and the proof is finished.
Note the crucial use that is made of the completeness of the Hilbert
space in proving it.


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david c. ullrich

??? A minute ago you said that no, they didn't say there was no
difference between "closed" and complete. But this theorem says
exactly that there is no difference!


************************

David C. Ullrich

david macmanus

Okay, thanks. Sounds like I'm getting confused. I'll ponder this over
the weekend. But a couple of questions in the meantime. Having read
their discussion of the difference between complete and closed, are you
now happy with their definition of closed, and are you happy with their
discussion generally?
Thanks,
David.


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Perhaps the guy perused the posts rather than read them.

/BAH

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