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Homework Help: Consequence of Schwarz Inequality

  1. Jan 23, 2008 #1
    [SOLVED] Consequence of Schwarz Inequality

    1. The problem statement, all variables and given/known data
    Use the Schwarz inequality to demonstrate the following inequality:

    [tex]\left(\sum_{j=1}^n |a_j + b_j|^2\right)^{1/2} \le \left(\sum_{j=1}^n |a_j|^2\right)^{1/2} + \left(\sum_{j=1}^n |b_j|^2\right)^{1/2}[/tex]

    2. Relevant equations
    The Schwarz inequality:

    [tex]\left|\sum_{j=1}^n a_j \bar{b}_j\right|^2 \le \left(\sum_{j=1}^n |a_j|^2\right) \left(\sum_{j=1}^n |b_j|^2\right)[/tex]

    3. The attempt at a solution
    Expanding the term on the LHS, squaring both sides and simplifying the inequality of the problem produces:

    [tex]\sum_{j=1}^n \Re(a_j\bar{b}_j) \le \left(\sum_{j=1}^n |a_j|^2\right) \left(\sum_{j=1}^n |b_j|^2\right)[/tex]

    At this point, I use the Schwarz inequality: If I can show that

    [tex]\sum_{j=1}^n \Re(a_j\bar{b}_j) \le \left|\sum_{j=1}^n a_j \bar{b}_j\right|^2[/tex]

    then the problem is solved. [itex]a_j\bar{b}_j = \Re(a_j\bar{b}_j) + i \Im(a_j\bar{b}_j)[/itex]. Let [itex]\alpha[/itex] be

    [tex]\sum_{j=1}^n \Re(a_j\bar{b}_j)[/tex]

    and let [itex]\beta[/itex] be

    [tex]\sum_{j=1}^n \Im(a_j\bar{b}_j)[/tex]

    Then [itex]\alpha^2 \le |\alpha + i\beta|^2[/itex]. I would like to think that [itex]\alpha \le \alpha^2[/itex], but this only works if [itex]\alpha \ge 1[/itex] which is not necessarily true. What can I do?
  2. jcsd
  3. Jan 23, 2008 #2
    Try taking a look at < A + B,A + B>, where <,> is the inner product. Also, note that |A|^2 = <A,A>. Your problem is equivalent to |A+B|<=|A|+|B|.
    Last edited: Jan 23, 2008
  4. Jan 23, 2008 #3
    I'm not familiar with inner products. The inequality of the problem does have the form of the triangle inequality doesn't it.
  5. Jan 23, 2008 #4

    [tex]\left|\sum_{j=1}^n a_j \bar{b}_j\right|^2 \le \left(\sum_{j=1}^n |a_j|^2\right) \left(\sum_{j=1}^n |b_j|^2\right)[/tex]

    Yes, the quantity on the left is of the form <A,B>^2 and the one on the right is (|A|*|B|)^2

    Sorry to hear you're not familiar with inner products, because they make this much swifter due to the bilinear properties.

    I'll try something else, but I will be a couple of hours. Good luck til then.

    Oh, and by the way, [itex]\sum_{j=1}^n a_j \bar{b}_j [/itex] = <A,B> for vector n-tuples A and B with entries from the complex numbers. Perhaps that, in combination with the previous info may give some help.
    Last edited: Jan 23, 2008
  6. Jan 23, 2008 #5
    I was actually talking about the inequality of the problem, not the Schwarz inequality.

    Sounds interesting. However, at this point, it seems to be a notational convenience more than anything. If the inequality holds because of some property of inner products, then I would need to understand said property and I think that amounts to understanding why the inequality holds.
  7. Jan 23, 2008 #6
    I'm getting at this: look at what happens if you expand out
    [tex] \sum_{j=1}^n (a_j + b_j ) (\bar{a}_j+ \bar{b}_j) [/tex]

    Your inequality should fall out of that expansion. You don't need to understand the general properties of inner products to demonstrate it in a specific case, as long as you make the correct replacements that the given Schwarz inequality provides. See if that helps you at all.
    Last edited: Jan 23, 2008
  8. Jan 24, 2008 #7


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    I don't think you need dot products for this: You are given the Schwartz inequality: [itex]|a+ b|\le |a|+ |b|[/itex] (|a| is the general "norm"- for finite sequences of real numbers, it is shorthand for the sum you have.)

    Let x= b, y= a-b. Then the Schwartz inequality in x and y, [itex]|x+y|\le |x|+ |y|[/itex] becomes [itex]|b+ a- b|= |a|\le |b|+ |a-b|[/itex]. Subtract |b| from both sides and you have your inequality.
  9. Jan 24, 2008 #8
    I just found a mistake in my algebra. Darn! This is wrong:

    [tex]\sum_{j=1}^n \Re(a_j\bar{b}_j) \le \left(\sum_{j=1}^n |a_j|^2\right) \left(\sum_{j=1}^n |b_j|^2\right)[/tex]

    It should be

    [tex]\sum_{j=1}^n \Re(a_j\bar{b}_j) \le \left(\sum_{j=1}^n |a_j|^2\right)^{1/2} \left(\sum_{j=1}^n |b_j|^2\right)^{1/2}[/tex]

    or upon squaring

    [tex]\left( \sum_{j=1}^n \Re(a_j\bar{b}_j) \right)^2 \le \left(\sum_{j=1}^n |a_j|^2\right) \left(\sum_{j=1}^n |b_j|^2\right)[/tex]

    The LHS of the above is certainly less than the LHS of the Schwarz inequality which is less than the RHS of the above (by the Schwarz inequality).

    Mathdope and HallsofIvy: thanks for the help.
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