Consequence of Schwarz Inequality

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In summary, the Consequence of Schwarz Inequality is that if you can show that\sum_{j=1}^n \Re(a_j\bar{b}_j) \le \left|\sum_{j=1}^n a_j \bar{b}_j\right|^2, then the problem is solved.
  • #1
e(ho0n3
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[SOLVED] Consequence of Schwarz Inequality

Homework Statement


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]


Homework 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]


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?
 
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  • #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|.
 
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  • #3
I'm not familiar with inner products. The inequality of the problem does have the form of the triangle inequality doesn't it.
 
  • #4
e(ho0n3 said:
I'm not familiar with inner products. The inequality of the problem does have the form of the triangle inequality doesn't it.


[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.
 
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  • #5
Mathdope said:
[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
I was actually talking about the inequality of the problem, not the Schwarz inequality.

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

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.
 
  • #6
Mathdope said:
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.

e(ho0n3 said:
I was actually talking about the inequality of the problem, not the Schwarz inequality.
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]


e(ho0n3 said:
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.
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.
 
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  • #7
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.
 
  • #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.
 

1. What is the Schwarz Inequality?

The Schwarz Inequality, also known as the Cauchy-Schwarz Inequality, is a mathematical theorem that establishes a relationship between the inner product of two vectors and their lengths. It states that the absolute value of the inner product of two vectors is always less than or equal to the product of their lengths.

2. What are the consequences of the Schwarz Inequality?

The consequences of the Schwarz Inequality are numerous and have applications in various fields such as physics, engineering, and economics. Some of the main consequences include the Triangle Inequality, the Hölder's Inequality, and the Bessel's Inequality.

3. How is the Schwarz Inequality used in real-world problems?

The Schwarz Inequality is used in many real-world problems, particularly in optimization and constraint satisfaction. For example, it can be used to find the shortest distance between a point and a line, or to determine the minimum or maximum value of a function subject to certain constraints.

4. Can the Schwarz Inequality be extended to more than two vectors?

Yes, the Schwarz Inequality can be extended to any number of vectors. This is known as the Minkowski Inequality and it states that the absolute value of the inner product of several vectors is less than or equal to the product of their lengths multiplied by the product of the lengths' corresponding coefficients.

5. What is the relationship between the Schwarz Inequality and the dot product?

The Schwarz Inequality is closely related to the dot product, as the dot product is a specific case of the inner product. The dot product is used to calculate the inner product of two vectors, and the Schwarz Inequality can be used to determine the upper bound of the dot product's absolute value.

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