Divergent Sums of Linearly Independent Elements

[Gear300]

Suppose we had an infinite series -

z = ∑_{i = 1 to ∞} ( α₁(i)x₁ + α₂(i)x₂ + . . . + α_m(i)x_m )

- rewritten as the cumulative sequence -

z(n) = α₁(n)x₁ + α₂(n)x₂ + . . . + α_m(n)x_m

- where the x_j are linearly independent and normalized (and serve as a finite basis across the sequence). If all the coefficients converged, then z(n) converges. If only one of the coefficients diverges, then z(n) diverges. How do we assess the case where more than one of the coefficients diverge (while noting that they are linearly independent)? Our space is a normed linear space.

 

[mfb]

If the whole system of x_j is linearly independent, you can analyze each component separately. The sum will converge if all components converge.

 

[Gear300]

Does that mean we can reorder the sum however we want, and that convergence is true iff all components converge? If so, would this change if the basis across the sum progressively became countable?

 

[mfb]

You can only reorder the infinite sum arbitrarily if it is absolute convergent. Which implies absolute convergence for every component.

would this change if the basis across the sum progressively became countable?

Wait, what? I though x_j doesn't depend on i.

 

[Gear300]

In general, what I was getting at was given a sum -

z = z₁ + z₂ + . . . + z_n + . . .

- where the basis for all of them is over the finite space spanned by { x₁ , x₂ , . . . , x_m }, then the sequence of progressive sums is -

z(n) = α₁(n)x₁ + α₂(n)x₂ + . . . + α_m(n)x_m

Given that the x_j are normalized, if the coefficients absolutely converge, then the sum can be reordered. But they do not necessarily have to absolutely converge for the sum z to converge. So I was unsure. Also, if the basis for the entire space is infinite, then the sequence of progressive sums can introduce new linearly independent elements. I was just wondering how to treat the case where several of the coefficients diverged in the sequence z(n) (given that the x_j are linearly independent).

 

[mfb]

As soon as one component diverges, the series diverges.
As soon as one component is not absolute convergent, reordering can change the limit.

 

[Gear300]

As soon as one component diverges, the series diverges.

Thanks. Is there a proof for this? I was considering the Hahn-Banach theorem, but I haven't made it work.

 

[WWGD]

Thanks. Is there a proof for this? I was considering the Hahn-Banach theorem, but I haven't made it work.

If I understood you, it comes down to $\infty +a = \infty$ ; a sum of numbers one of which is infinite ( meaning unbounded in norm, here ) cannot be finite.

 

[mfb]

3 lines of proof with epsilon/delta to show that convergence of the overall series requires convergence in every component.

 

[Gear300]

If I understood you, it comes down to $\infty +a = \infty$ ; a sum of numbers one of which is infinite ( meaning unbounded in norm, here ) cannot be finite.

Yeah. I thought it might be possible to extend a linear functional f(x₁) = ||x₁|| so that f(α₁(n)x₁ + α₂(n)x₂ + . . . + α_m(n)x_m) diverges while f(x) < ||x|| for all x ∈ R.

3 lines of proof with epsilon/delta to show that convergence of the overall series requires convergence in every component.

I figured it would turn out dubious if I didn't use some property of linear independence, so I never thought past that.

Thanks for the answers.

 

[mfb]

You need the linear independence. Otherwise the statement is false.

 

[Gear300]

Thanks again. I will work through this, but just as a quick question, does this remain true even for a sum across a countable set of linearly independent elements?

 

[mfb]

Then I would expect that the sum can converge even if the individual components do not, but it could depend on the x_j.
Which metric do you use?

 

[Gear300]

Sorry for the late response. I am not using any particular metric. I am just proving certain properties of Banach spaces (or normed linear spaces in general). Since these are metric spaces, I only have to worry about countable sums. But some of the smaller things, such as -

"Absolute convergence implies convergence, but the converse is not always true."

- have been slowing me down in my proofs. I was essentially just a little finicky about when we are allowed to reorder the sums and whatnot.

 

[Gear300]

I suppose for completion's sake, it would be a good to post this. It gives insight to a proof to the problem statement.

https://math.stackexchange.com/ques...ll-norms-are-equivalent-in-finite-dimensional

It essentially shows that all norms for finite-dimensional spaces are equivalent.