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Applying Cauchy-Schwarz to a sum(Have I understood this correctly?)

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Given the vectorspace consisting of a realvalued sequences [tex]\{x_j\}[/tex] where [tex]\sum_{j=1}^{\infty} x_j^2 < \infty [/tex]. Show that M the vectorspace has an innerproduct given by

    [tex]\langle \{x_j\}, \{y_j\}\rangle = \sum_{j=1}^\infty x_j \cdot y_j [/tex]

    2. Relevant equations

    Since [tex]\{x_j\}[/tex] defines every possible vector component in M, then isn't that equal to that the square sum of every possible realvalued vectorcomponent of the M can be written as [tex]\sum_{j=1}^{\infty} x_j^2 \leq \infty (\mathrm{max} |x_j|)^2 = \infty, j = 1, \ldots, \infty[/tex].?

    3. The attempt at a solution

    All possible vector of either [tex]\{x_j\}[/tex]or [tex]\{y_j\}[/tex] are considered to be real valued, thus the definition of the inner product from Linear Algebra is true, hence by condition (1) of the innerproduct

    [tex]\langle \{x_j\} \cdot \{x_j\} \rangle = x_j ^2 [/tex] and thus the innerproduct of every component in either x or y can be written as

    [tex]\langle {x_j \} \cdot \{y_j \} \rangle = x_j \cdot y_j, j = 1, \ldots, \infty[/tex]

    Have I covered all what is required of me in showing the above?

    Sincerely
    Cauchy
     
    Last edited: Feb 15, 2009
  2. jcsd
  3. Feb 15, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Why would you say that? You are given that the sum is finite. Why would you then prove that sum is either finite or infinite?:

    I don't know what you mean by this. An inner product is an operator, it is neither true nor false. If you mean the conditions defining an inner product are satisfied, you should show them individually, not just assert it.

    I have no idea what you are trying to say here. It makes no sense to talk about "the innerproduct of every component". Components are numbers, not vectors and so have no innerproduct.

    Have you shown that, if [tex]\sum_{j=1}^{\infty} x_j^2 < \infty [/tex] and [tex]\sum_{j=1}^{\infty} y_j^2 < \infty [/tex], then [tex][tex]\sum_{j=1}^{\infty} x_j^2 < \infty[/tex]?

    Have you shown that [tex]\langle {x_i}, {y_i}\rangle= \sum_{j=1}^{\infty} x_j^2 < \infty[/tex] satisfies the conditions for an inner product?
     
  4. Feb 15, 2009 #3
    I was simply trying to understand why this sum of squared elements of the series {x_j}
    is put into the assigment? And why its less than infinity? I understand that maybe this square respresents that the inner product between elements in the above series.
    Isn't the series the same as [tex]\langle x_j, x_j \rangle = \sum_{j=1}^{\infty} x_j^2 = \| x \| ^2 \leq \infty \cdot (\mathrm{max}_{j=1}^{\infty}|x_j|) ^2 = \cdots ? \cdots < \infty[/tex]? Don't I need to understand what should be placed at the questionmark in order to understand how show that the Vectorspace M has an definable innerproduct the sequences x_j and y_j?
    Because connection between the sum of sequared real numbers of x_j and the inner product isn't that that my teacher is applying the first definition of inner product of Vectorspace? and then its my job to show that if the above definition hold then to show that definition part 2 and part 3 of the inner product also holds?


    What I was trying to say here is that the "component's" be elements of the sequences

    x_j and y_j.


    Because if that is true then [tex]\langle \{x_j\}, \{x_j\} \rangle = \sum_{j =1}^{\infty} x_j^2 = x_1 \cdot x_1 + x_2 \cdot x_2 + \cdots + x_\infty \cdot x_\infty < \infty[/tex]

    I will try to go to that :) But its possible to give me a hint on why this sum is less than infinity?

    Because a vectorspace M where the above is happening can that be found other place than in sequenced Hilbert spaces? But is that key understand why [tex]\sum x_j < \infty[/tex] where [tex]j = 1,\ldots, \infty[/tex]

    Best Regards
    Cauchy
     
    Last edited: Feb 15, 2009
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