Show that $|.|_{1/2}$ is not a norm. [duplicate]
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This question already has an answer here:
$l^p$ not norm, $p<1$
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Show that $|.|_{1/2}$ is not a norm. would anybody guide me that how can i prove or disprove?
thanks
optimization convex-analysis norm nonlinear-optimization regularization
asked Dec 29 '18 at 1:39
HumayooHumayoo
1013
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marked as duplicate by Holo, Eric Wofsey, Lee David Chung Lin, ancientmathematician, A. Pongrácz Dec 31 '18 at 8:17
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 |
$begingroup$
This question already has an answer here:
$l^p$ not norm, $p<1$
2 answers
Show that $|.|_{1/2}$ is not a norm. would anybody guide me that how can i prove or disprove?
thanks
optimization convex-analysis norm nonlinear-optimization regularization
asked Dec 29 '18 at 1:39
HumayooHumayoo
1013
$endgroup$
marked as duplicate by Holo, Eric Wofsey, Lee David Chung Lin, ancientmathematician, A. Pongrácz Dec 31 '18 at 8:17
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 |
$begingroup$
This question already has an answer here:
$l^p$ not norm, $p<1$
2 answers
Show that $|.|_{1/2}$ is not a norm. would anybody guide me that how can i prove or disprove?
thanks
optimization convex-analysis norm nonlinear-optimization regularization
asked Dec 29 '18 at 1:39
HumayooHumayoo
1013
$endgroup$
This question already has an answer here:
$l^p$ not norm, $p<1$
2 answers
Show that $|.|_{1/2}$ is not a norm. would anybody guide me that how can i prove or disprove?
thanks
This question already has an answer here:
$l^p$ not norm, $p<1$
2 answers
optimization convex-analysis norm nonlinear-optimization regularization
optimization convex-analysis norm nonlinear-optimization regularization
asked Dec 29 '18 at 1:39
HumayooHumayoo
1013
asked Dec 29 '18 at 1:39
HumayooHumayoo
1013
asked Dec 29 '18 at 1:39
HumayooHumayoo
1013
asked Dec 29 '18 at 1:39
HumayooHumayoo
1013
asked Dec 29 '18 at 1:39
HumayooHumayoo
1013
1013
marked as duplicate by Holo, Eric Wofsey, Lee David Chung Lin, ancientmathematician, A. Pongrácz Dec 31 '18 at 8:17
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 Holo, Eric Wofsey, Lee David Chung Lin, ancientmathematician, A. Pongrácz Dec 31 '18 at 8:17
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 |
add a comment |
1 Answer
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It doesn't satisfy the triangle inequality. All you need is a counter-example to show it isn't a norm. For example in $mathbb R^2$ consider $(1,0)$ and $(0,1)$.
answered Dec 29 '18 at 2:02
FortoxFortox
6718
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1
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Or $(1,4)$ and $(4,1).$ In either this case or yours we have an instance of $|a|+|b|<|a+b|$.
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– DanielWainfleet
Dec 29 '18 at 6:55
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It doesn't satisfy the triangle inequality. All you need is a counter-example to show it isn't a norm. For example in $mathbb R^2$ consider $(1,0)$ and $(0,1)$.
answered Dec 29 '18 at 2:02
FortoxFortox
6718
$endgroup$
1
$begingroup$
Or $(1,4)$ and $(4,1).$ In either this case or yours we have an instance of $|a|+|b|<|a+b|$.
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:55
add a comment |
$begingroup$
It doesn't satisfy the triangle inequality. All you need is a counter-example to show it isn't a norm. For example in $mathbb R^2$ consider $(1,0)$ and $(0,1)$.
answered Dec 29 '18 at 2:02
FortoxFortox
6718
$endgroup$
1
$begingroup$
Or $(1,4)$ and $(4,1).$ In either this case or yours we have an instance of $|a|+|b|<|a+b|$.
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:55
add a comment |
$begingroup$
It doesn't satisfy the triangle inequality. All you need is a counter-example to show it isn't a norm. For example in $mathbb R^2$ consider $(1,0)$ and $(0,1)$.
answered Dec 29 '18 at 2:02
FortoxFortox
6718
$endgroup$
It doesn't satisfy the triangle inequality. All you need is a counter-example to show it isn't a norm. For example in $mathbb R^2$ consider $(1,0)$ and $(0,1)$.
answered Dec 29 '18 at 2:02
FortoxFortox
6718
edited Dec 29 '18 at 3:53
answered Dec 29 '18 at 2:02
FortoxFortox
6718
answered Dec 29 '18 at 2:02
FortoxFortox
6718
answered Dec 29 '18 at 2:02
FortoxFortox
6718
6718
1
$begingroup$
Or $(1,4)$ and $(4,1).$ In either this case or yours we have an instance of $|a|+|b|<|a+b|$.
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:55
add a comment |
1
$begingroup$
Or $(1,4)$ and $(4,1).$ In either this case or yours we have an instance of $|a|+|b|<|a+b|$.
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:55
1
1
$begingroup$
Or $(1,4)$ and $(4,1).$ In either this case or yours we have an instance of $|a|+|b|<|a+b|$.
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:55
$begingroup$
Or $(1,4)$ and $(4,1).$ In either this case or yours we have an instance of $|a|+|b|<|a+b|$.
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:55
add a comment |
