I Divergent Sums of Linearly Independent Elements

1. Jun 24, 2017

Gear300

Suppose we had an infinite series -

z = i = 1 to ∞ ( α1(i)x1 + α2(i)x2 + . . . + αm(i)xm )

- rewritten as the cumulative sequence -

z(n) = α1(n)x1 + α2(n)x2 + . . . + αm(n)xm

- where the xj 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.

2. Jun 24, 2017

Staff: Mentor

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

3. Jun 24, 2017

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?

4. Jun 24, 2017

Staff: Mentor

You can only reorder the infinite sum arbitrarily if it is absolute convergent. Which implies absolute convergence for every component.
Wait, what? I though xj doesn't depend on i.

5. Jun 24, 2017

Gear300

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

z = z1 + z2 + . . . + zn + . . .

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

z(n) = α1(n)x1 + α2(n)x2 + . . . + αm(n)xm

Given that the xj 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 xj are linearly independent).

6. Jun 24, 2017

Staff: Mentor

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

7. Jun 24, 2017

Gear300

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

8. Jun 24, 2017

WWGD

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.

9. Jun 24, 2017

Staff: Mentor

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

10. Jun 24, 2017

Gear300

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

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

11. Jun 24, 2017

Staff: Mentor

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

12. Jun 24, 2017

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?

13. Jun 24, 2017

Staff: Mentor

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

14. Jun 25, 2017

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.

Last edited: Jun 25, 2017
15. Jun 26, 2017