The Standard Metric on C^d .... Garling, Corollary 11.1.5 ....

In summary: } \le \left( \sum_{ j = 1}^d \mid z_j \mid^2 \right)^{...} + \left( \sum_{ j = 1}^d \mid w_j \mid^2 \right)^{...}##
  • #1
Math Amateur
Gold Member
MHB
3,998
48
I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable" ... ...

I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...

I need some help with Corollary 11.1.5 ...

Corollary 11.1.5 reads as follows:
Garling - Corollary 11.1.5 ... .png


In the above proof by Garling we read the following:

" ... ... : using the inequality of the previous corollary,##d( z + w , 0 ) = \left( \sum_{ j = 1 }^d \mid z_j + w_j \mid^2 \right)^{ \frac{1}{2} } \le \left( \sum_{ j = 1}^d ( \mid z_j \mid + \mid w_j \mid )^2 \right)^{ \frac{1}{2} } ####\le \left( \sum_{ j = 1 }^d \mid z_j \mid^2 \right)^{ \frac{1}{2} } + \left( \sum_{ j = 1 }^d \mid w_j \mid^2 \right)^{ \frac{1}{2} }= d(z,0) + d(w,0) ##
My questions are as follows:Question 1

Can someone please explain exactly why we have:## d( z + w , 0 ) = \left( \sum_{ j = 1 }^d \mid z_j + w_j \mid^2 \right)^{ \frac{1}{2} } \le \left( \sum_{ j = 1}^d ( \mid z_j \mid + \mid w_j \mid )^2 \right)^{ \frac{1}{2} } ##How/why does this hold true?
Question 2

Can someone please explain exactly why we have: ## \left( \sum_{ j = 1}^d ( \mid z_j \mid + \mid w_j \mid )^2 \right)^{ \frac{1}{2} } \le \left( \sum_{ j = 1 }^d \mid z_j \mid^2 \right)^{ \frac{1}{2} } + \left( \sum_{ j = 1 }^d \mid w_j \mid^2 \right)^{ \frac{1}{2} }= d(z,0) + d(w,0) ##How/why does this hold true?
Help will be appreciated ...

Peter========================================================================================Relevant to the above post is Proposition 11.2.3 and Corollary 11.1.4 ... so I am providing both ... as follows:
Garling - Proposition 11.1.3 and Corollary 11.1.4 ... .png

Hope the above scanned text helps readers understand the post ...

Peter
 

Attachments

  • Garling - Corollary 11.1.5 ... .png
    Garling - Corollary 11.1.5 ... .png
    35.9 KB · Views: 674
  • Garling - Proposition 11.1.3 and Corollary 11.1.4 ... .png
    Garling - Proposition 11.1.3 and Corollary 11.1.4 ... .png
    30.5 KB · Views: 545
Physics news on Phys.org
  • #2
What do you know about the relation between ##|a+b|## and ##|a|+|b|\,##?
And for the second question: square it and try to prove it by induction.
 
  • Like
Likes Math Amateur
  • #3
Math Amateur said:
Question 1

Can someone please explain exactly why we have:## d( z + w , 0 ) = \left( \sum_{ j = 1 }^d \mid z_j + w_j \mid^2 \right)^{ \frac{1}{2} } \le \left( \sum_{ j = 1}^d ( \mid z_j \mid + \mid w_j \mid )^2 \right)^{ \frac{1}{2} } ##How/why does this hold true?
It is sufficient to prove that, for any ##j##,
$$ \mid z_j + w_j \mid^2 \le ( \mid z_j \mid + \mid w_j \mid )^2 $$
which is equivalent to
$$ \mid z_j + w_j \mid \le \mid z_j \mid + \mid w_j \mid $$

We can prove that by cases, as is often best with absolute values. Because things are squared, there are only two cases that need considering, where the signs of ##z_j## and ##w_j## are the same, and where they are different. Each case is easy to prove.
 
  • Like
Likes Math Amateur
  • #4
The second one can be reduced to an instance of the Cauchy-Schwarz inequality by squaring both sides, then expanding the square inside the summation on the LHS, then cancelling terms that occur on both sides.
 
  • Like
Likes Math Amateur
  • #5
Thanks Andrew and fresh_42 ...

Appreciate your help as always ...

Hmm ... should have remembered that for ##z_1, z_2 \in \mathbb{C}## we have ##\mid z_1 + z_2 \mid \le \mid z_1 \mid + \mid z_2 \mid## ...

But ... still reflecting on your posts ...

Peter
 
Last edited:
  • #6
Math Amateur said:
Thanks Andrew and fresh_42 ...

Appreciate your help as always ...

Hmm ... should have remembers that for ##z_1, z_2 \in \mathbb{C}## we have ##\mid z_1 + z_2 \mid \le \mid z_1 \mid + \mid z_2 \mid## ...

But ... still reflecting on your posts ...

Peter
fresh_42, Andrew

Just a thought regarding Question 2 ... Garling suggests we use the previous Corollary (which is Corollary 11.1.4) ... but how do we use Corollary 11.1.4 to answer Question 2 ... ...

Peter
 
  • #7
Math Amateur said:
fresh_42, Andrew

Just a thought regarding Question 2 ... Garling suggests we use the previous Corollary (which is Corollary 11.1.4) ... but how do we use Corollary 11.1.4 to answer Question 2 ... ...

Peter
I only see that it gives you the equations for the ##d(0,a)## terms. The inequalities are the triangle inequality for question 1 and squaring the second should give you the one for question 2 by the help of Cauchy-Schwarz as @andrewkirk mentioned. It might be a little work to actually prove it, but it's a good exercise (you may assume Cauchy-Schwarz as given, as the Wiki link contains a proof):
$$
\left( \sum_{ j = 1}^n ( \mid z_j \mid + \mid w_j \mid )^2 \right)^{ \frac{1}{2} } \le \left( \sum_{ j = 1 }^n \mid z_j \mid^2 \right)^{ \frac{1}{2} } + \left( \sum_{ j = 1 }^n \mid w_j \mid^2 \right)^{ \frac{1}{2} }
$$
Try it for ##n=2,3## and you should see how it works.
 
  • Like
Likes Math Amateur
  • #8
Hi fresh_42, Andrew ...

... taking your advice regarding Question 2 ...

We have to show that ... :##\left( \sum_{ j = 1}^d ( \mid z_j \mid + \mid w_j \mid )^2 \right)^{ \frac{1}{2} } \le \left( \sum_{ j = 1 }^d \mid z_j \mid^2 \right)^{ \frac{1}{2} } + \left( \sum_{ j = 1 }^d \mid w_j \mid^2 \right)^{ \frac{1}{2} }## ... ... ... ... ... (1)Squaring both sides of (1) we get ...

##\sum_{ j = 1}^d ( \mid z_j \mid + \mid w_j \mid )^2 = \sum_{ j = 1}^d \mid z_j \mid^2 + \sum_{ j = 1}^d \mid w_j \mid^2 + 2 \left( \sum_{ j = 1 }^d \mid z_j \mid^2 \right)^{ \frac{1}{2} } \left( \sum_{ j = 1 }^d \mid w_j \mid^2 \right)^{ \frac{1}{2} }## ... ... ... ... ... (2)Now ... expanding the LHS of (2) we get

##\sum_{ j = 1}^d ( \mid z_j \mid + \mid w_j \mid )^2 = \sum_{ j = 1}^d \mid z_j \mid^2 + 2 \sum_{ j = 1}^d \mid z_j \mid \mid w_j \mid + \sum_{ j = 1}^d \mid w_j \mid^2## ... ... ... ... ... (3)Substituting (3) into (2) we get ...:

##\sum_{ j = 1}^d \mid z_j \mid^2 + 2 \sum_{ j = 1}^d \mid z_j \mid \mid w_j \mid + \sum_{ j = 1}^d \mid w_j \mid^2 \ \le \ \sum_{ j = 1}^d \mid z_j \mid^2 + \sum_{ j = 1}^d \mid w_j \mid^2 + 2 \left( \sum_{ j = 1 }^d \mid z_j \mid^2 \right)^{ \frac{1}{2} } \left( \sum_{ j = 1 }^d \mid w_j \mid^2 \right)^{ \frac{1}{2} }####\Longrightarrow \ \ \sum_{ j = 1}^d \mid z_j \mid \mid w_j \mid \ \le \ \left( \sum_{ j = 1 }^d \mid z_j \mid^2 \right)^{ \frac{1}{2} } \left( \sum_{ j = 1 }^d \mid w_j \mid^2 \right)^{ \frac{1}{2} }## ...

... BUT ... this is just Cauchy's Inequality! ... QED ...Is the above correct ... ?

Peter
 
  • Like
Likes WWGD and fresh_42
  • #9
Yes, that's correct. Well done.
 
  • #10
FWIW, note that question 2 really is triangle inequality as well.

Consider two vectors ##\mathbf a## and ##\mathbf x##, where ##\mathbf a## has the magnitude of the components of ##\mathbf z## and ##\mathbf x## has the magnitude of the components of ##\mathbf w##

(alternatively work directly with ##\mathbf z## and ##\mathbf w## and assume WLOG that components are real non-negative)

##
\left( \sum_{ j = 1}^d ( \mid z_j \mid + \mid w_j \mid )^2 \right)^{ \frac{1}{2} } =
\big \Vert \mathbf a + \mathbf x\big \Vert_2 \leq \big \Vert \mathbf a \big \Vert_2 +\big \Vert \mathbf x \big \Vert_2 = \left( \sum_{ j = 1 }^d \mid z_j \mid^2 \right)^{ \frac{1}{2} } + \left( \sum_{ j = 1 }^d \mid w_j \mid^2 \right)^{ \frac{1}{2} }##

by triangle inequality.
 
  • Like
Likes Math Amateur

FAQ: The Standard Metric on C^d .... Garling, Corollary 11.1.5 ....

1. What is the Standard Metric on C^d?

The Standard Metric on C^d is a mathematical concept used to measure distances between points in a complex vector space. It is defined using the Euclidean norm, which is the square root of the sum of squared components of a vector.

2. Who is Garling and what is their contribution to Corollary 11.1.5?

Garling is a mathematician who has made significant contributions to the field of complex analysis. In Corollary 11.1.5, Garling proves that the Standard Metric on C^d is complete, meaning that every Cauchy sequence in C^d converges to a point in C^d.

3. How is the Standard Metric on C^d different from other metrics?

The Standard Metric on C^d is a special case of the more general p-norm metric, which is commonly used in functional analysis. It differs from other metrics in that it is specifically tailored for use on complex vector spaces and has properties that make it particularly useful for certain mathematical applications.

4. What are some applications of the Standard Metric on C^d?

The Standard Metric on C^d has various applications in mathematics, including complex analysis, functional analysis, and differential geometry. It is also used in physics and engineering to study and model complex systems.

5. Is the Standard Metric on C^d used in real-world measurements?

While the Standard Metric on C^d is a fundamental concept in mathematics, it is not typically used in real-world measurements. Instead, it is used as a tool for theoretical calculations and proofs, rather than practical applications.

Back
Top