# I Scalar Product Problem

1. Jul 1, 2017

### Gear300

Alright, so we ran into a peculiarity in answering this question.

Let R be the set of all functions f defined on the interval [0,1] such that -

(1) f(t) is nonzero at no more than countably many points t1, t2, . . .
(2) Σi = 1 to ∞ f2(ti) < ∞ .

Define addition of elements and multiplication of elements by scalars in the ordinary way, i.e., (f + g)(t) = f(t) + g(t), (αf)(t) = αf(t). If f and g are two elements of R, nonzero only at the points t1, t2, . . . and t'1, t'2, . . . respectively, define the scalar product of f and g as

(3) (f,g) = Σi,j = 1 to ∞ f(ti)g(t'j) .

Prove that this scalar product makes R into a Euclidean space.

By the looks of it, (3) is not referring to a sum across all pairs (i,j), since that may induce (absolute) convergence issues for certain elements in the set, where it might not be possible for them to have a finite norm. So we figured that the sum (3) is such that i and j run in parallel across Z+, like it would be in l2.
The peculiarity we found next is in the ordering of the countable domain of points with non-zero image. It may not be well-ordered for some functions, and even if you were to assume only well-ordered domains, there can be several different countable orderings (given that the sums of functions are included). One possibility we considered is if f has smaller ordering than g, we can generalize the sum (3) so that it only goes up to the ordering of f (like we would if f had a finite domain and g had a countable domain). But even so, the list of plausible orderings goes a long way in [0,1], and then there is a problem with resolving one of the properties of the scalar product:

(iv) (f , g+h) = (f , g) + (f , h)

Typically, proving (iv) would involve showing that (f , g+h) remains absolutely convergent. But the problem is how to consider the domain of g+h in the left expression as opposed to the individual domains of g and h in the right expression. In any case, we're stuck.

2. Jul 1, 2017

### Staff: Mentor

Where does the question come from? The scalar product doesn't seem to make sense.
What I could imagine is $$(f,g) = \sum_{k=1}^{\infty} f(s_k) g(s_k)$$ where the sk are the union of ti and t'i. Or the intersection, doesn't make a difference here.

3. Jul 1, 2017

4. Jul 2, 2017

### Gear300

So for completion, we have managed to prune the problem statement to something doable. Altogether, it works when considering mfb's statement of the scalar product -

- which is the intuitive way of looking that things, since that is how the scalar product in C2[a,b] behaves. The general idea is that -

- is an instance of absolute convergence, where absolute convergence implies unconditional convergence. Then by "Sequences and Series: A Sourcebook", Pete L. Clark, Chapter 2 . Section 9 . pg 89 . Theorem 2.52:

For a : NR an ordinary sequence and A ∈ R, the following are equivalent:
i. The unordered sum Σn ∈ Z+ an is convergent, with sum A.
ii. The series Σn = 0 to ∞ an is unconditionally convergent, with sum A.

So given any countable (un)ordering of points of non-zero image, so long as there exists a reordering ω that is absolutely convergent, then everything should fit together.

Last edited: Jul 2, 2017