Application of the Schwarz Inequality

In summary, I used the Schwarz Inequality and a geometric argument to show that:$$\sum|a_{j}+b_{j}|^{2})^{\frac{1}{2}}\leq(\sum|a_{j}|^{2})^{\frac{1}{2}}+(\sum|b_{j}|^{2})^{\frac{1}{2}}$$for all j from 1 to n.
  • #1
Tsunoyukami
215
11
Here's yet another assigned problem that I'm having difficulty with. I think I'm close to the end but am "nervous" (for lack of a better word) about whether or not I have used summation notation properly throughout the problem. Here it is:

"Use the Schwarz inequality to establish that

[itex]( \sum |a_{j} + b_{j}|^{2})^{1/2} \leq (\sum |a_{j}|^{2})^{1/2} + (\sum |b_{j}|^{2})^{1/2}[/itex]
" (From Complex Variables, 2nd Edition by Fisher, pg. 10; #24)

The Schwarz Inequality is:

[itex]| \sum a_{j}\overline{b_{j}}|^{2} \leq (\sum |a_{j}|^{2})(\sum |b_{j}|^{2})[/itex]


I will now describe my approach without writing out each step explicitly (I'm finding summation notation tedious to write for LaTeX and this is for a graded assignment so I'd rather not have the entire answer floating on the web). If anything I say is unclear or you require more details let me know and I can fill it out.

Also, please note that the sums in both the above inequalities are from j=1 to j=n. I was unsure how to code that properly in LaTeX.


I squared both side of the given inequality and expanded the left-hand side. After some algebra I obtained the following expression:

[itex](\sum Re(a_{j}\overline{b_{j}}))^{2} \leq (\sum |a_{j}|^{2})(\sum |b_{j}|^{2})[/itex]

The right-hand sides of this expression and the Schwarz Inequality are exactly equal. I argue that this inequality is true if it's left-hand side is less than or equal to the left-hand side of the Schwarz Inequality, that is:

[itex](\sum Re(a_{j}\overline{b_{j}}))^{2} \leq | \sum a_{j}\overline{b_{j}}|^{2} [/itex]


To me this is obvious when j = 1. Let [itex]w_{j} = a_{j}\overline{b_{j}}[/itex], and can be written in the form [itex]w = x + iy[/itex], then:

[itex]\sum Re(w_{j}) \leq | \sum w_{j}| [/itex]
[itex]Re(w_{j}) \leq | w_{j}| [/itex]
[itex]x \leq (x^{2} + y^{2})^{1/2} [/itex]
[itex]x^{2} \leq (x^{2} + y^{2}) [/itex]
[itex]0 \leq y^{2}[/itex]

So I am done. But how can I extend this to the case when [itex]j \neq 1[/itex]?


If anything I've described above seems incorrect please let me know and I can write out the steps I took explicitly to see where I went wrong. Thanks a lot in advance for any assistance! :)
 
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  • #2
Tsunoyukami said:
So I am done. But how can I extend this to the case when [itex]j \neq 1[/itex]?


If anything I've described above seems incorrect please let me know and I can write out the steps I took explicitly to see where I went wrong. Thanks a lot in advance for any assistance! :)

Consider this:

$$\sum_{j=\beta}^{n}\left[a_j\right] = \sum_{j=1}^{n}\left[a_j\right]-\sum_{j=1}^{\beta-1}\left[a_j\right]$$
 
  • #3
To be honest, I'm not sure how I should use that. I expect I would apply it to the right-hand side of my expression. Is it applicable even though the sum is inside the modulus. Is it possible to provide a hint of how to apply what you've recommended above? :s
 
  • #4
Tsunoyukami said:
To be honest, I'm not sure how I should use that. I expect I would apply it to the right-hand side of my expression. Is it applicable even though the sum is inside the modulus. Is it possible to provide a hint of how to apply what you've recommended above? :s
It's not so much the formula that's important. It's the concept that a summation can be written as a sum or difference of two summations.

If you have the solution for j=1, then you can prove something about a combination of sums with one of the sums being the sum with j=1. What can you prove that would make the j≠1 case follow from the j=1 case?
 
  • #5
Thank you for your suggestions, Mandelbroth, but I stumbled upon (what I believe) is a simple geometric argument for the proof of this statement. It is essentially what I posted in my first post:

[itex](\sum Re(a_{j}\overline{b_{j}}))^{2} \leq | \sum a_{j}\overline{b_{j}}|^{2} [/itex]
[itex]\sum Re(a_{j}\overline{b_{j}}) \leq | \sum a_{j}\overline{b_{j}}|[/itex]
[itex]\sum Re(z_{j}) \leq | \sum z_{j}|[/itex]

From here I argued that the left-hand side can be interpreted geometrically as the sum of the x-components of the products ##a_{j}\overline{b_{j}}## and the right-hand side can be interpreted geometrically as the modulus or distance (ie. in both the x- and y- directions). Then, by the triangle inequality this inequality is true.
 
  • #6
Tsunoyukami said:
Thank you for your suggestions, Mandelbroth, but I stumbled upon (what I believe) is a simple geometric argument for the proof of this statement. It is essentially what I posted in my first post:

[itex](\sum Re(a_{j}\overline{b_{j}}))^{2} \leq | \sum a_{j}\overline{b_{j}}|^{2} [/itex]
[itex]\sum Re(a_{j}\overline{b_{j}}) \leq | \sum a_{j}\overline{b_{j}}|[/itex]
[itex]\sum Re(z_{j}) \leq | \sum z_{j}|[/itex]

From here I argued that the left-hand side can be interpreted geometrically as the sum of the x-components of the products ##a_{j}\overline{b_{j}}## and the right-hand side can be interpreted geometrically as the modulus or distance (ie. in both the x- and y- directions). Then, by the triangle inequality this inequality is true.

That's not what you want to do, you want to prove the triangle inequality, not use it. Define a dot product ##a \cdot b=\sum a_{j}\overline{b_{j}}## and define a norm ##|a|=\sqrt{a \cdot a}##. Then Cauchy-Schwartz tells you ##|a \cdot b| \le |a| |b|##. Try to use that to prove ##|a+b| \le |a|+|b|##, and yes, square both sides. Start with ##|a+b|^2=(a+b) \cdot (a+b)##.
 
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  • #7
As you can probably tell I am rather confused by this problem. I'll begin again and show each step explicitly - let me know where I go astray, please!

The Schwarz Inequality is:

$$|\sum a_{j}\overline{b_{j}}|^{2}\leq(\sum|a_{j}|^{2})(\sum |b_{j}|^{2})$$

All sums are from j=1 to j=n.


$$(\sum|a_{j}+b_{j}|^{2})^{\frac{1}{2}}\leq(\sum|a_{j}|^{2})^{\frac{1}{2}}+(\sum|b_{j}|^{2})^{\frac{1}{2}}$$

$$\sum|a_{j}+b_{j}|^{2})\leq\sum|a_{j}|^{2}+2(\sum|a_{j}|)^{\frac{1}{2}}(\sum|b_{j}|)^{\frac{1}{2}}+\sum|b_{j}|^{2}$$

$$\sum(|a_{j}|^{2} + a_{j}\overline{b_{j}} + \overline{a_{j}}b_{j}+|b_{j}|^{2}\leq\sum|a_{j}|^{2} + 2 (\sum|a_{j}|)^{\frac{1}{2}} (\sum|b_{j}|)^{\frac{1}{2}}+\sum|b_{j}|^{2}$$

$$\sum(a_{j}\overline{b_{j}} + \overline{a_{j}}b_{j}) \leq 2(\sum|a_{j}|)^{\frac{1}{2}} (\sum|b_{j}|)^{\frac{1}{2}}$$

But here the left-hand side can be expressed equivalently as ##2\sum Re[a_{j}\overline{b_{j}}]##, so:

$$\sum(a_{j}\overline{b_{j}} + \overline{a_{j}}b_{j}) = 2\sum Re[a_{j}\overline{b_{j}}] \leq 2(\sum|a_{j}|)^{\frac{1}{2}} (\sum|b_{j}|)^{\frac{1}{2}}$$

$$\sum Re[a_{j}\overline{b_{j}}] \leq (\sum|a_{j}|)^{\frac{1}{2}} (\sum|b_{j}|)^{\frac{1}{2}}$$

$$(\sum Re[a_{j}\overline{b_{j}}])^{2} \leq (\sum|a_{j}|)(\sum|b_{j}|)$$

Here the right-hand side is equivalent to the right-hand side of the Schwarz Inequality (given above) and so I argue that this inequality is true iff:

$$(\sum Re[a_{j}\overline{b_{j}}])^{2} \leq |\sum a_{j}\overline{b_{j}}|^{2} \leq (\sum|a_{j}|)(\sum|b_{j}|)$$

So, in order to complete the problem I must show that:

$$(\sum Re[a_{j}\overline{b_{j}}])^{2} \leq |\sum a_{j}\overline{b_{j}}|^{2}$$

Let ##z_{j} = a_{j}\overline{b_{j}}##, then:

$$(\sum Re[z_{j}])^{2} \leq |\sum z_{j}|^{2}$$

This is true via the triangle inequality, as the left-hand side is the "x-component" of ##z_{j}## (ie. the magnitude of the real part) and the right-hand side is the modulus of ##z_{j}##. Is this geometric argument sufficient?

If yes, then I have (successfully) argued that ##(\sum Re[z_{j}])^{2} \leq |\sum z_{j}|^{2}## and therefore the initial inequality is true.
 
  • #8
This is really much easier to write if you use dot product notation. You don't need the 'Re' part of your argument, though it's certainly true. Just use CS on each of the two mixed sums separately ##|a \cdot b|=|b \cdot a| \le |a| |b|##.
 
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  • #9
Ahh, I see what you mean - that helps make it easier. I hadn't thought to consider using the dot product until your first response. Thanks a bunch!
 

1. What is the Schwarz Inequality?

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

2. What are the applications of the Schwarz Inequality?

The Schwarz Inequality has various applications in mathematics, physics, and engineering. It is used in optimization problems, functional analysis, and in proving other mathematical theorems. It also has applications in quantum mechanics, signal processing, and statistics.

3. How is the Schwarz Inequality used in optimization problems?

In optimization problems, the Schwarz Inequality is used to find the minimum or maximum values of a given function. It helps in finding the optimal solution by providing an upper bound for the function.

4. Can the Schwarz Inequality be applied to complex numbers?

Yes, the Schwarz Inequality can be applied to complex numbers. In this case, the dot product is replaced by the complex conjugate dot product, and the magnitudes are calculated using the complex modulus.

5. What is the significance of the Schwarz Inequality in quantum mechanics?

In quantum mechanics, the Schwarz Inequality plays a crucial role in proving the Heisenberg Uncertainty Principle. It also helps in understanding the relationship between the position and momentum of a particle and the limitations in measuring them simultaneously.

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