I Solve Scalar Product Problem in Set R of Functions [0,1]

[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) Σ_{i = 1 to ∞} f²(t_i) < ∞ .

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) = Σ_{i,j = 1 to ∞} f(t_i)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.

 

[mfb]

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 s_k are the union of t_i and t'_i. Or the intersection, doesn't make a difference here.

 

[Gear300]

I was confused as well. Here is a picture.

https://goo.gl/photos/oDSL9zRYn54y265n8

 

[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 -

What I could imagine is $$(f,g) = \sum_{k=1}^{\infty} f(s_k) g(s_k)$$ where the s_k are the union of t_i and t'_i. Or the intersection, doesn't make a difference here.

- 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 -

(2) Σ_{i = 1 to ∞} f²(t_i) < ∞ .

- is an instance of absolute convergence, where absolute convergence implies unconditional convergence. Then by http://math.uga.edu/%7Epete/3100supp.pdf, Chapter 2 . Section 9 . pg 89 . Theorem 2.52:

For a_⋅ : N → R an ordinary sequence and A ∈ R, the following are equivalent:
i. The unordered sum Σ_{n ∈ Z⁺} a_n is convergent, with sum A.
ii. The series Σ_{n = 0 to ∞} a_n 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.