Proving $mid parallel x parallel - parallel y parallel mid$ $leqq$ $parallel x-y parallel$? [duplicate]
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Reverse Triangle Inequality Proof
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Let $V$ be a normed vector space and I need to prove the follwing inequality $mid parallel x parallel - parallel y parallel mid$ $leqq$ $parallel x-y parallel$ containing the norm and the absolute value of the real numbers. However things just get twisted and I cannot see how to prove it.... Could anyone please tell me how to prove it?
inequality vector-spaces norm
marked as duplicate by Jyrki Lahtonen, Martin R, Nosrati, Arnaud D., amWhy Aug 17 at 11:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
-2
down vote
favorite
This question already has an answer here:
Reverse Triangle Inequality Proof
3 answers
Let $V$ be a normed vector space and I need to prove the follwing inequality $mid parallel x parallel - parallel y parallel mid$ $leqq$ $parallel x-y parallel$ containing the norm and the absolute value of the real numbers. However things just get twisted and I cannot see how to prove it.... Could anyone please tell me how to prove it?
inequality vector-spaces norm
marked as duplicate by Jyrki Lahtonen, Martin R, Nosrati, Arnaud D., amWhy Aug 17 at 11:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
The question that presents itself : what can you infer from this inequality ?
â Neil hawking
Aug 17 at 6:54
Search for triangle inequality.
â Jyrki Lahtonen
Aug 17 at 7:00
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
This question already has an answer here:
Reverse Triangle Inequality Proof
3 answers
Let $V$ be a normed vector space and I need to prove the follwing inequality $mid parallel x parallel - parallel y parallel mid$ $leqq$ $parallel x-y parallel$ containing the norm and the absolute value of the real numbers. However things just get twisted and I cannot see how to prove it.... Could anyone please tell me how to prove it?
inequality vector-spaces norm
This question already has an answer here:
Reverse Triangle Inequality Proof
3 answers
Let $V$ be a normed vector space and I need to prove the follwing inequality $mid parallel x parallel - parallel y parallel mid$ $leqq$ $parallel x-y parallel$ containing the norm and the absolute value of the real numbers. However things just get twisted and I cannot see how to prove it.... Could anyone please tell me how to prove it?
This question already has an answer here:
Reverse Triangle Inequality Proof
3 answers
inequality vector-spaces norm
asked Aug 17 at 6:44
Keith
1,350720
1,350720
marked as duplicate by Jyrki Lahtonen, Martin R, Nosrati, Arnaud D., amWhy Aug 17 at 11:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jyrki Lahtonen, Martin R, Nosrati, Arnaud D., amWhy Aug 17 at 11:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
The question that presents itself : what can you infer from this inequality ?
â Neil hawking
Aug 17 at 6:54
Search for triangle inequality.
â Jyrki Lahtonen
Aug 17 at 7:00
add a comment |Â
The question that presents itself : what can you infer from this inequality ?
â Neil hawking
Aug 17 at 6:54
Search for triangle inequality.
â Jyrki Lahtonen
Aug 17 at 7:00
The question that presents itself : what can you infer from this inequality ?
â Neil hawking
Aug 17 at 6:54
The question that presents itself : what can you infer from this inequality ?
â Neil hawking
Aug 17 at 6:54
Search for triangle inequality.
â Jyrki Lahtonen
Aug 17 at 7:00
Search for triangle inequality.
â Jyrki Lahtonen
Aug 17 at 7:00
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
$||x||=||x-y+y|| le ||x-y||+||y||$, hence
$(1)$ $||x||-||y|| le ||x-y||$.
In a similar way we get
$(2)$ $||y||-||x|| le ||y-x||$.
Since $||y-x||=||x-y||$, $(1)$ and $(2)$ give the result.
Oh.............Thank you for your answer....
â Keith
Aug 17 at 6:52
Why the downvote ??????????????????????????????
â Fred
Aug 17 at 7:14
1
The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation.
â Jyrki Lahtonen
Aug 17 at 10:16
Yes, my sheriff !
â Fred
Aug 17 at 10:20
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$||x||=||x-y+y|| le ||x-y||+||y||$, hence
$(1)$ $||x||-||y|| le ||x-y||$.
In a similar way we get
$(2)$ $||y||-||x|| le ||y-x||$.
Since $||y-x||=||x-y||$, $(1)$ and $(2)$ give the result.
Oh.............Thank you for your answer....
â Keith
Aug 17 at 6:52
Why the downvote ??????????????????????????????
â Fred
Aug 17 at 7:14
1
The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation.
â Jyrki Lahtonen
Aug 17 at 10:16
Yes, my sheriff !
â Fred
Aug 17 at 10:20
add a comment |Â
up vote
1
down vote
accepted
$||x||=||x-y+y|| le ||x-y||+||y||$, hence
$(1)$ $||x||-||y|| le ||x-y||$.
In a similar way we get
$(2)$ $||y||-||x|| le ||y-x||$.
Since $||y-x||=||x-y||$, $(1)$ and $(2)$ give the result.
Oh.............Thank you for your answer....
â Keith
Aug 17 at 6:52
Why the downvote ??????????????????????????????
â Fred
Aug 17 at 7:14
1
The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation.
â Jyrki Lahtonen
Aug 17 at 10:16
Yes, my sheriff !
â Fred
Aug 17 at 10:20
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$||x||=||x-y+y|| le ||x-y||+||y||$, hence
$(1)$ $||x||-||y|| le ||x-y||$.
In a similar way we get
$(2)$ $||y||-||x|| le ||y-x||$.
Since $||y-x||=||x-y||$, $(1)$ and $(2)$ give the result.
$||x||=||x-y+y|| le ||x-y||+||y||$, hence
$(1)$ $||x||-||y|| le ||x-y||$.
In a similar way we get
$(2)$ $||y||-||x|| le ||y-x||$.
Since $||y-x||=||x-y||$, $(1)$ and $(2)$ give the result.
answered Aug 17 at 6:51
Fred
38.2k1238
38.2k1238
Oh.............Thank you for your answer....
â Keith
Aug 17 at 6:52
Why the downvote ??????????????????????????????
â Fred
Aug 17 at 7:14
1
The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation.
â Jyrki Lahtonen
Aug 17 at 10:16
Yes, my sheriff !
â Fred
Aug 17 at 10:20
add a comment |Â
Oh.............Thank you for your answer....
â Keith
Aug 17 at 6:52
Why the downvote ??????????????????????????????
â Fred
Aug 17 at 7:14
1
The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation.
â Jyrki Lahtonen
Aug 17 at 10:16
Yes, my sheriff !
â Fred
Aug 17 at 10:20
Oh.............Thank you for your answer....
â Keith
Aug 17 at 6:52
Oh.............Thank you for your answer....
â Keith
Aug 17 at 6:52
Why the downvote ??????????????????????????????
â Fred
Aug 17 at 7:14
Why the downvote ??????????????????????????????
â Fred
Aug 17 at 7:14
1
1
The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation.
â Jyrki Lahtonen
Aug 17 at 10:16
The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation.
â Jyrki Lahtonen
Aug 17 at 10:16
Yes, my sheriff !
â Fred
Aug 17 at 10:20
Yes, my sheriff !
â Fred
Aug 17 at 10:20
add a comment |Â
The question that presents itself : what can you infer from this inequality ?
â Neil hawking
Aug 17 at 6:54
Search for triangle inequality.
â Jyrki Lahtonen
Aug 17 at 7:00