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

[Math Amateur]

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:

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

 

[Opalg]

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.

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.

 

[Math Amateur]

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