- #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:

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:

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?

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:

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

Peter

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:

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:

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

Peter