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

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In summary, Peter is reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable". He is focused on Chapter 11: Metric Spaces and Normed Spaces and needs help understanding Corollary 11.1.5.
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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 ...

View attachment 7906
In the above proof by Garling we read the following:

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

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

Can someone please explain exactly why we have:$$\displaystyle \left( \sum_{ j = 1}^d ( \mid z_j \mid + \mid w_j \mid )^2 \right)^{ \frac{1}{2} } \le \left( \sum_{ j = 1 }^d \mid z_j \mid^2 \right)^{ \frac{1}{2} } + \left( \sum_{ j = 1 }^d \mid w_j \mid^2 \right)^{ \frac{1}{2} }= d(z,0) + d(w,0)$$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:View attachment 7907Hope the above scanned text helps readers understand the post ...

Peter

Peter said:
Question 1

Can someone please explain exactly why we have:

$$\displaystyle d( z + w , 0 ) = \left( \sum_{ j = 1 }^d \mid z_j + w_j \mid^2 \right)^{ \frac{1}{2} } \le \left( \sum_{ j = 1}^d ( \mid z_j \mid + \mid w_j \mid )^2 \right)^{ \frac{1}{2} }$$

How/why does this hold true?
The equality $$\displaystyle d( z + w , 0 ) = \left( \sum_{ j = 1 }^d | z_j + w_j |^2 \right)^{\!\! 1/2 }$$ is the definition of $d(z+w,0)$ (the $0$ in $d(z+w,0)$ is the zero in $\Bbb{C}^d$, so it is actually $(0,0,\ldots,0)$).

The inequality $$\displaystyle \left( \sum_{ j = 1 }^d | z_j + w_j |^2 \right)^{ \!\!1/2 } \leqslant \left( \sum_{ j = 1}^d (| z_j| + |w_j|)^2 \right)^{\!\!1/2}$$ follows from the fact that $| z_j + w_j | \leqslant | z_j| + |w_j|$, which is the triangle inequality for complex numbers.

Peter said:
Question 2

Can someone please explain exactly why we have:

$$\displaystyle \left( \sum_{ j = 1}^d ( \mid z_j \mid + \mid w_j \mid )^2 \right)^{ \frac{1}{2} } \le \left( \sum_{ j = 1 }^d \mid z_j \mid^2 \right)^{ \frac{1}{2} } + \left( \sum_{ j = 1 }^d \mid w_j \mid^2 \right)^{ \frac{1}{2} }= d(z,0) + d(w,0)$$

How/why does this hold true?
Since $|z_j|$ and $|w_j|$ are real, this inequality is just an application of the triangle inequality in $\Bbb{R}^d$, which is contained in Corollary 11.1.4.

Opalg said:
The equality $$\displaystyle d( z + w , 0 ) = \left( \sum_{ j = 1 }^d | z_j + w_j |^2 \right)^{\!\! 1/2 }$$ is the definition of $d(z+w,0)$ (the $0$ in $d(z+w,0)$ is the zero in $\Bbb{C}^d$, so it is actually $(0,0,\ldots,0)$).

The inequality $$\displaystyle \left( \sum_{ j = 1 }^d | z_j + w_j |^2 \right)^{ \!\!1/2 } \leqslant \left( \sum_{ j = 1}^d (| z_j| + |w_j|)^2 \right)^{\!\!1/2}$$ follows from the fact that $| z_j + w_j | \leqslant | z_j| + |w_j|$, which is the triangle inequality for complex numbers.Since $|z_j|$ and $|w_j|$ are real, this inequality is just an application of the triangle inequality in $\Bbb{R}^d$, which is contained in Corollary 11.1.4.
Thanks Opalg ...

... understand your points regarding Question 1 ...... still reflecting on what you have written regarding Question 2 ...

Thanks again for your help ...

Peter

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

The Standard Metric on C^d is a metric space that measures the distance between points in d-dimensional complex space. It is defined as the Euclidean metric, where the distance between two points is given by the square root of the sum of the squared differences in each coordinate.

## 2. Who is Garling and what is Corollary 11.1.5?

Garling refers to the mathematician Richard Garling, who is known for his work in metric spaces and complex analysis. Corollary 11.1.5 is a theorem in Garling's book "A Course in Mathematical Analysis" which discusses the behavior of the Standard Metric on C^d under certain conditions.

## 3. What does Corollary 11.1.5 state?

Corollary 11.1.5 states that if a sequence of points in C^d converges to a limit point under the Standard Metric, then the sequence converges to the same limit point under the Euclidean metric. In other words, the two metrics behave the same way for sequences that converge to a limit point.

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

The Standard Metric on C^d is different from other metrics in that it is specifically defined for complex numbers in d-dimensional space. Other metrics may use different formulas to calculate distance or may be defined for different types of spaces, such as real numbers or higher dimensions.

## 5. Why is the Standard Metric on C^d important in mathematics?

The Standard Metric on C^d is important in mathematics because it provides a useful tool for measuring distance and analyzing the behavior of sequences in complex space. It also has applications in various fields, such as physics, engineering, and computer science. Additionally, the Standard Metric is a fundamental concept in complex analysis, which is a branch of mathematics that has many important real-world applications.

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