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

  • Thread starter Cauchy1789
  • Start date
In summary, the conversation discusses the construction of an inner product in a vectorspace consisting of realvalued sequences where the sum of squared elements in the sequence is finite. The inner product is defined as the sum of the product of each element in the sequences, and it is shown that this satisfies the conditions for an inner product. The conversation also touches on the connection between the sum of squared elements and the inner product, and the possibility of finding this construction in other spaces such as Hilbert spaces.
  • #1
Cauchy1789
46
0

Homework Statement



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]

Homework 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].?

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:
Physics news on Phys.org
  • #2
Cauchy1789 said:

Homework Statement



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]

Homework 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].?
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?:

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
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.

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

Sincerely
Cauchy
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?
 
  • #3
HallsofIvy said:
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 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?


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.

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

x_j and y_j.

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]\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?


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:

1. What is Cauchy-Schwarz inequality?

The Cauchy-Schwarz inequality is a mathematical concept that states the absolute value of the dot product of two vectors is always less than or equal to the product of their magnitudes. In other words, it states that the angle between two vectors cannot exceed 90 degrees.

2. How is Cauchy-Schwarz inequality applied to a sum?

When applying Cauchy-Schwarz inequality to a sum, it is used to find an upper bound for the sum of products of two sequences. This can be used to prove various theorems and inequalities in mathematics.

3. Can Cauchy-Schwarz inequality be applied to any type of sum?

Yes, Cauchy-Schwarz inequality can be applied to any sum where the terms can be represented as vectors. This includes real numbers, complex numbers, and even abstract vectors in vector spaces.

4. How do I know if I have correctly applied Cauchy-Schwarz inequality to a sum?

To ensure that you have correctly applied Cauchy-Schwarz inequality to a sum, you should check that the inequality holds true for all the terms in the sum. If the inequality is satisfied, then you have correctly applied it.

5. Are there any limitations to applying Cauchy-Schwarz inequality to a sum?

One limitation to applying Cauchy-Schwarz inequality to a sum is that it can only be used to find an upper bound for the sum of products. It cannot be used to find a lower bound. Additionally, the terms in the sum must be represented as vectors for the inequality to hold true.

Similar threads

  • General Math
Replies
8
Views
1K
  • Topology and Analysis
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Topology and Analysis
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
907
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
7
Views
1K
Replies
3
Views
726
Replies
2
Views
875
Back
Top