Consequence of Schwarz Inequality

[e(ho0n3]

[SOLVED] Consequence of Schwarz Inequality

Homework Statement

Use the Schwarz inequality to demonstrate the following inequality:

\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}

Homework Equations

The Schwarz inequality:

\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)

The Attempt at a Solution

Expanding the term on the LHS, squaring both sides and simplifying the inequality of the problem produces:

\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)

At this point, I use the Schwarz inequality: If I 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. a_j\bar{b}_j = \Re(a_j\bar{b}_j) + i \Im(a_j\bar{b}_j). Let \alpha be

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

and let \beta be

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

Then \alpha^2 \le |\alpha + i\beta|^2. I would like to think that \alpha \le \alpha^2, but this only works if \alpha \ge 1 which is not necessarily true. What can I do?

 

[Mathdope]

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|.

 

[e(ho0n3]

I'm not familiar with inner products. The inequality of the problem does have the form of the triangle inequality doesn't it.

 

[Mathdope]

I'm not familiar with inner products. The inequality of the problem does have the form of the triangle inequality doesn't it.

\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)

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, \sum_{j=1}^n a_j \bar{b}_j = <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]

\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)

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.

 

[Mathdope]

Oh, and by the way, \sum_{j=1}^n a_j \bar{b}_j = <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.

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
\sum_{j=1}^n (a_j + b_j ) (\bar{a}_j+ \bar{b}_j)

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.

 

[HallsofIvy]

I don't think you need dot products for this: You are given the Schwartz inequality: |a+ b|\le |a|+ |b| (|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, |x+y|\le |x|+ |y| becomes |b+ a- b|= |a|\le |b|+ |a-b|. Subtract |b| from both sides and you have your inequality.

 

[e(ho0n3]

I just found a mistake in my algebra. Darn! This is wrong:

\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)

It should be

\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}

or upon squaring

\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)

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.