Prove that this is a norm, infimum norm The 2019 Stack Overflow Developer Survey Results Are InProof that $|cdot|_B$ defines a normproving that this function does not define a norm on $mathbb R^2$ since the convexityShow that M is closed convex and find the minimum normProve that this space is not BanachLinear operator normProving that a set infers a norm given certain conditionsProve that this is a normProve this function defined for a convex, bounded, open, symmetric set $V$ is a norm on $mathbbR^n$Prove that the following function is a normprove that infimum is not attained for this subset of C([0,1])
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Prove that this is a norm, infimum norm
The 2019 Stack Overflow Developer Survey Results Are InProof that $|cdot|_B$ defines a normproving that this function does not define a norm on $mathbb R^2$ since the convexityShow that M is closed convex and find the minimum normProve that this space is not BanachLinear operator normProving that a set infers a norm given certain conditionsProve that this is a normProve this function defined for a convex, bounded, open, symmetric set $V$ is a norm on $mathbbR^n$Prove that the following function is a normprove that infimum is not attained for this subset of C([0,1])
$begingroup$
I'm trying to prove that given a vector space $X$ of finite dimension, $U subset X$ an open subset, bounded, convex and symmetric with respect to $bar0 in X$ (that is if $x in U Rightarrow -x in U$). Prove that
$|x|_u = inf lambda > 0$ $
is a norm in X.
I prove the first two properties and only the triangular inequality remains. The problem has many parts, so I do not think that all hypotheses are necessary for this part.
I say "infimum norm" but I don't know it's name. If it does have a name, I would be grateful if you could tell me the name.
real-analysis functional-analysis
asked Mar 30 at 23:20
NewLichKingNewLichKing
85
$endgroup$
add a comment |
$begingroup$
I'm trying to prove that given a vector space $X$ of finite dimension, $U subset X$ an open subset, bounded, convex and symmetric with respect to $bar0 in X$ (that is if $x in U Rightarrow -x in U$). Prove that
$|x|_u = inf lambda > 0$ $
is a norm in X.
I prove the first two properties and only the triangular inequality remains. The problem has many parts, so I do not think that all hypotheses are necessary for this part.
I say "infimum norm" but I don't know it's name. If it does have a name, I would be grateful if you could tell me the name.
real-analysis functional-analysis
asked Mar 30 at 23:20
NewLichKingNewLichKing
85
$endgroup$
$begingroup$
I'm an not sure how to interpret the notation you use. Do you mean to write $$ |x|_u = inf , left lambda > 0 mid lambda^-1x in U right?$$
$endgroup$
– Brian
Mar 30 at 23:31
$begingroup$
This is called the Minkowski functional of $U$ and you can easily find a proof of triangle inequality on Wikipedia.
$endgroup$
– Kavi Rama Murthy
Mar 30 at 23:42
$begingroup$
The triangle inequality is where you use the hypothesis that $U$ is convex. I assume you have not used that for the other properties of norm.
$endgroup$
– GEdgar
Mar 30 at 23:56
add a comment |
$begingroup$
I'm trying to prove that given a vector space $X$ of finite dimension, $U subset X$ an open subset, bounded, convex and symmetric with respect to $bar0 in X$ (that is if $x in U Rightarrow -x in U$). Prove that
$|x|_u = inf lambda > 0$ $
is a norm in X.
I prove the first two properties and only the triangular inequality remains. The problem has many parts, so I do not think that all hypotheses are necessary for this part.
I say "infimum norm" but I don't know it's name. If it does have a name, I would be grateful if you could tell me the name.
real-analysis functional-analysis
asked Mar 30 at 23:20
NewLichKingNewLichKing
85
$endgroup$
I'm trying to prove that given a vector space $X$ of finite dimension, $U subset X$ an open subset, bounded, convex and symmetric with respect to $bar0 in X$ (that is if $x in U Rightarrow -x in U$). Prove that
$|x|_u = inf lambda > 0$ $
is a norm in X.
I prove the first two properties and only the triangular inequality remains. The problem has many parts, so I do not think that all hypotheses are necessary for this part.
I say "infimum norm" but I don't know it's name. If it does have a name, I would be grateful if you could tell me the name.
real-analysis functional-analysis
real-analysis functional-analysis
asked Mar 30 at 23:20
NewLichKingNewLichKing
85
asked Mar 30 at 23:20
NewLichKingNewLichKing
85
edited Mar 31 at 1:11
NewLichKing
asked Mar 30 at 23:20
NewLichKingNewLichKing
85
asked Mar 30 at 23:20
NewLichKingNewLichKing
85
asked Mar 30 at 23:20
NewLichKingNewLichKing
85
85
$begingroup$
I'm an not sure how to interpret the notation you use. Do you mean to write $$ |x|_u = inf , left lambda > 0 mid lambda^-1x in U right?$$
$endgroup$
– Brian
Mar 30 at 23:31
$begingroup$
This is called the Minkowski functional of $U$ and you can easily find a proof of triangle inequality on Wikipedia.
$endgroup$
– Kavi Rama Murthy
Mar 30 at 23:42
$begingroup$
The triangle inequality is where you use the hypothesis that $U$ is convex. I assume you have not used that for the other properties of norm.
$endgroup$
– GEdgar
Mar 30 at 23:56
add a comment |
$begingroup$
I'm an not sure how to interpret the notation you use. Do you mean to write $$ |x|_u = inf , left lambda > 0 mid lambda^-1x in U right?$$
$endgroup$
– Brian
Mar 30 at 23:31
$begingroup$
This is called the Minkowski functional of $U$ and you can easily find a proof of triangle inequality on Wikipedia.
$endgroup$
– Kavi Rama Murthy
Mar 30 at 23:42
$begingroup$
The triangle inequality is where you use the hypothesis that $U$ is convex. I assume you have not used that for the other properties of norm.
$endgroup$
– GEdgar
Mar 30 at 23:56
$begingroup$
I'm an not sure how to interpret the notation you use. Do you mean to write $$ |x|_u = inf , left lambda > 0 mid lambda^-1x in U right?$$
$endgroup$
– Brian
Mar 30 at 23:31
$begingroup$
I'm an not sure how to interpret the notation you use. Do you mean to write $$ |x|_u = inf , left lambda > 0 mid lambda^-1x in U right?$$
$endgroup$
– Brian
Mar 30 at 23:31
$begingroup$
This is called the Minkowski functional of $U$ and you can easily find a proof of triangle inequality on Wikipedia.
$endgroup$
– Kavi Rama Murthy
Mar 30 at 23:42
$begingroup$
This is called the Minkowski functional of $U$ and you can easily find a proof of triangle inequality on Wikipedia.
$endgroup$
– Kavi Rama Murthy
Mar 30 at 23:42
$begingroup$
The triangle inequality is where you use the hypothesis that $U$ is convex. I assume you have not used that for the other properties of norm.
$endgroup$
– GEdgar
Mar 30 at 23:56
$begingroup$
The triangle inequality is where you use the hypothesis that $U$ is convex. I assume you have not used that for the other properties of norm.
$endgroup$
– GEdgar
Mar 30 at 23:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$frac 1 lambda +mu (x+y) =frac lambda lambda +mu frac 1 lambda x+ frac mu lambda +mu frac 1 mu y in U$ whenever $ frac 1 lambda x in U$ and $ frac 1 mu y in U$. Can you complete the proof of triangle inequality, given this information?
edited Mar 30 at 23:50
Stallmp
21219
answered Mar 30 at 23:49
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
$begingroup$
I see, $ |x+y|_u leq lambda + mu $, particularly $ |x+y|_u leq ( |x|_u + epsilon / 2) + ( |y|_u + epsilon / 2) $ for all $epsilon > 0$, then we get the inequality. A lot of thanks!
$endgroup$
– NewLichKing
Mar 31 at 1:05
add a comment |
Your Answer
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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$begingroup$
$frac 1 lambda +mu (x+y) =frac lambda lambda +mu frac 1 lambda x+ frac mu lambda +mu frac 1 mu y in U$ whenever $ frac 1 lambda x in U$ and $ frac 1 mu y in U$. Can you complete the proof of triangle inequality, given this information?
edited Mar 30 at 23:50
Stallmp
21219
answered Mar 30 at 23:49
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
$begingroup$
I see, $ |x+y|_u leq lambda + mu $, particularly $ |x+y|_u leq ( |x|_u + epsilon / 2) + ( |y|_u + epsilon / 2) $ for all $epsilon > 0$, then we get the inequality. A lot of thanks!
$endgroup$
– NewLichKing
Mar 31 at 1:05
add a comment |
$begingroup$
$frac 1 lambda +mu (x+y) =frac lambda lambda +mu frac 1 lambda x+ frac mu lambda +mu frac 1 mu y in U$ whenever $ frac 1 lambda x in U$ and $ frac 1 mu y in U$. Can you complete the proof of triangle inequality, given this information?
edited Mar 30 at 23:50
Stallmp
21219
answered Mar 30 at 23:49
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
$begingroup$
I see, $ |x+y|_u leq lambda + mu $, particularly $ |x+y|_u leq ( |x|_u + epsilon / 2) + ( |y|_u + epsilon / 2) $ for all $epsilon > 0$, then we get the inequality. A lot of thanks!
$endgroup$
– NewLichKing
Mar 31 at 1:05
add a comment |
$begingroup$
$frac 1 lambda +mu (x+y) =frac lambda lambda +mu frac 1 lambda x+ frac mu lambda +mu frac 1 mu y in U$ whenever $ frac 1 lambda x in U$ and $ frac 1 mu y in U$. Can you complete the proof of triangle inequality, given this information?
edited Mar 30 at 23:50
Stallmp
21219
answered Mar 30 at 23:49
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
$frac 1 lambda +mu (x+y) =frac lambda lambda +mu frac 1 lambda x+ frac mu lambda +mu frac 1 mu y in U$ whenever $ frac 1 lambda x in U$ and $ frac 1 mu y in U$. Can you complete the proof of triangle inequality, given this information?
edited Mar 30 at 23:50
Stallmp
21219
answered Mar 30 at 23:49
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
edited Mar 30 at 23:50
Stallmp
21219
edited Mar 30 at 23:50
Stallmp
21219
edited Mar 30 at 23:50
Stallmp
21219
21219
answered Mar 30 at 23:49
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
answered Mar 30 at 23:49
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
answered Mar 30 at 23:49
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
74.1k53270
$begingroup$
I see, $ |x+y|_u leq lambda + mu $, particularly $ |x+y|_u leq ( |x|_u + epsilon / 2) + ( |y|_u + epsilon / 2) $ for all $epsilon > 0$, then we get the inequality. A lot of thanks!
$endgroup$
– NewLichKing
Mar 31 at 1:05
add a comment |
$begingroup$
I see, $ |x+y|_u leq lambda + mu $, particularly $ |x+y|_u leq ( |x|_u + epsilon / 2) + ( |y|_u + epsilon / 2) $ for all $epsilon > 0$, then we get the inequality. A lot of thanks!
$endgroup$
– NewLichKing
Mar 31 at 1:05
$begingroup$
I see, $ |x+y|_u leq lambda + mu $, particularly $ |x+y|_u leq ( |x|_u + epsilon / 2) + ( |y|_u + epsilon / 2) $ for all $epsilon > 0$, then we get the inequality. A lot of thanks!
$endgroup$
– NewLichKing
Mar 31 at 1:05
$begingroup$
I see, $ |x+y|_u leq lambda + mu $, particularly $ |x+y|_u leq ( |x|_u + epsilon / 2) + ( |y|_u + epsilon / 2) $ for all $epsilon > 0$, then we get the inequality. A lot of thanks!
$endgroup$
– NewLichKing
Mar 31 at 1:05
add a comment |
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$begingroup$
I'm an not sure how to interpret the notation you use. Do you mean to write $$ |x|_u = inf , left lambda > 0 mid lambda^-1x in U right?$$
$endgroup$
– Brian
Mar 30 at 23:31
$begingroup$
This is called the Minkowski functional of $U$ and you can easily find a proof of triangle inequality on Wikipedia.
$endgroup$
– Kavi Rama Murthy
Mar 30 at 23:42
$begingroup$
The triangle inequality is where you use the hypothesis that $U$ is convex. I assume you have not used that for the other properties of norm.
$endgroup$
– GEdgar
Mar 30 at 23:56