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Proof of convergence (intro topology)

  1. Jun 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that if x = (x1, x2,...) and y = (y1, y2,...) are members of l^2, then

    [itex]\sum^{\infty}_{i=1} |x_{i}y_{i}|[/itex]


    2. Relevant equations

    My book defines l^2 to be:

    [itex]{ x=(x_{1}, x_{2}, ... ) \in ℝ^{\omega} : \sum^{\infty}_{i=1} (x_{i})^{2} converges }[/itex]

    (should be set brackets around that, don't know why they don't show up.)

    3. The attempt at a solution


    -I have no idea if Cauchy-Schwarts inequality actually holds in an infinite dimensional space, and can't find any information regarding whether it does or not.

    -I can't think of any way to relate my last line to the expression I am showing convergence for.

    -I have no idea if just because two series are convergent, then their product is convergent. I assumed it was anyway.



    Since x and y are both members of l^2,

    [itex] \sum^{\infty}_{i=1} (x_{i})^{2}[/itex]

    [itex]\sum^{\infty}_{i=1} (y_{i})^{2}[/itex]


    Then, the root of these series must converge too. The roots of these series are the norms of x and y.

    So, ||x|| and ||y|| converge.

    The Cauchy-Schwarts inequality states that

    [itex]|x \cdot y| \leq ||x|| ||y||[/itex]

    Since ||x|| ||y|| converges, so does |x (dot) y|

    SCRATCH that!

    I thought that was so because of the comparison test for convergent series, but the comparison test requires a comparison to be made between the terms of the series, not the series themselves...

    So, I'm at square one.

    Any help with this one?
  2. jcsd
  3. Jun 21, 2013 #2
    hmm I think trying first to prove the convergence of (x + y)^2 would be a good start..
  4. Jun 21, 2013 #3
    Could you shed some light on the three concerns I've outlined? I'm really shooting around in the dark here, I don't know how to show (x+y)^2 converges. I've never done a single proof regarding convergence before.
  5. Jun 22, 2013 #4
    No ideas?
  6. Jun 22, 2013 #5
    Use the inequality ##2ab \leq a^2 + b^2## for real numbers ##a## and ##b##.

    That said, Cauchy-Schwarz is valid for infinite dimensional space such as ##\ell^2##, but then you need to prove it.
  7. Jun 22, 2013 #6
    Hmm OK, I see where you're going with that. I'm away from all my books now, but is going to look like:

    Square each series, it converges (because they are absolutely convergent?) add the two convergent series, then due to your inequality I can use the comparison test and confirm that 2xy converges, then so does xy, and the series xy is exactly the dot product?
  8. Jun 22, 2013 #7
    Why do you use absolute convergence? You don't know that your series converge absolutely.
  9. Jun 22, 2013 #8
    I don't, but I wouldn't know how to proceed if they aren't. What I've read says that if a is convergent and b is convergent, I can only say that ab is convergent if a or b converges absolutely.
  10. Jun 22, 2013 #9
    You know ##\sum|x_n|^2## converges. Isn't that enough?
  11. Jun 22, 2013 #10
    Ohh, yes it is. I forgot that exponent was in the l^2 criteria.
  12. Jun 22, 2013 #11
    Okay, now I've learned a lot. From now one, when I'm attempting to prove convergence, I'll try to look for an algebraic identity that allows me to compare first.

    Since x and y are members of l^2,

    [itex]\sum^{\infty}_{i=1} (x_{i})^{2}[/itex]
    [itex]\sum^{\infty}_{i=1} (y_{i})^{2}[/itex]

    both converge.

    Since members of l^2 are points in R^omega, we know that xi and yi are real numbers, so

    [itex]\sum^{\infty}_{i=1} |x_{i}|^{2}[/itex]
    [itex]\sum^{\infty}_{i=1} |y_{i}|^{2}[/itex]

    both converge, since k^2 = |k|^2 for any real.

    The sum of two convergent series also converges.

    [itex]\sum^{\infty}_{i=1} |x_{i}|^{2} + |y_{i}|^{2}[/itex]

    Consider that

    [itex](a-b)^{2} \geq 0[/itex]
    [itex]\Rightarrow a^{2} - 2ab + b^{2} \geq 0[/itex]
    [itex]\Rightarrow a^{2} + b^{2} \geq 2ab[/itex].

    This inequality lets us say that

    [itex]|x_{i}|^{2} + |y_{i}|^{2} \geq 2|x_{i}||y_{i}|[/itex]

    Then, by the comparison convergence test,

    [itex]\sum^{\infty}_{i=1} 2|x_{i}||y_{i}|[/itex]


    Since a convergent series multiplied by a constant is still convergent, and |n||m| = |nm|, we can say that

    [itex]\sum^{\infty}_{i=1} |x_{i}y_{i}|[/itex]

    converges, which was the goal.
  13. Jun 22, 2013 #12
    That's it!
  14. Jun 22, 2013 #13
    Thanks again
  15. Jun 23, 2013 #14
    I've been trying to find more info on this l^2 set but googling doesn't work very well. Does it have another name?

    I'm wondering if it is defined this way because it makes many metrics and the operation above (which is not a metric, but is related to a basic operation) work.
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