Mixing
Is $d(x,y)=frac {|x-y|} {sqrt {1+|x|^{2}}sqrt {1+|y|^{2}}}$ a metric on a normed linear space?
$begingroup$
Let $X$ be a normed linear space and $$d(x,y)=frac {|x-y|} {sqrt {1+|x|^{2}}sqrt {1+|y|^{2}}}$$ Is this a metric?
This question arose from the following post where the answer was given in the affirmative when $X$ is an inner product space. I am sorry that I have not made any progress so far. In the post below I asked the question in a comment and someone suggested that I should post this as a separate question.
How to show that the spherical metric satisfies the triangle inequality?
Some additional information: the triangle inequality for $d$ holds for an abstract norm on $X$ iff it holds in $C[0,1]$ with the supremum norm. This may or may not help in answering the question but it makes the question a bit more interesting because we can work with a specific norm. Justification for this claim: $C[0,1]$ is a universal space for the class of separable Banach spaces in the sense any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$. In particular any three dimensional subspace of $X$ is isometrically isomorphic to a subsapce of $C[0,1]$ and triangle inequality involves only a three diemnsional subspace.
normed-spaces
edited Jan 19 at 15:21
Blue
48.5k870154
asked Jan 16 at 10:30
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
add a comment |
$begingroup$
Let $X$ be a normed linear space and $$d(x,y)=frac {|x-y|} {sqrt {1+|x|^{2}}sqrt {1+|y|^{2}}}$$ Is this a metric?
This question arose from the following post where the answer was given in the affirmative when $X$ is an inner product space. I am sorry that I have not made any progress so far. In the post below I asked the question in a comment and someone suggested that I should post this as a separate question.
How to show that the spherical metric satisfies the triangle inequality?
Some additional information: the triangle inequality for $d$ holds for an abstract norm on $X$ iff it holds in $C[0,1]$ with the supremum norm. This may or may not help in answering the question but it makes the question a bit more interesting because we can work with a specific norm. Justification for this claim: $C[0,1]$ is a universal space for the class of separable Banach spaces in the sense any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$. In particular any three dimensional subspace of $X$ is isometrically isomorphic to a subsapce of $C[0,1]$ and triangle inequality involves only a three diemnsional subspace.
normed-spaces
edited Jan 19 at 15:21
Blue
48.5k870154
asked Jan 16 at 10:30
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
add a comment |
$begingroup$
Let $X$ be a normed linear space and $$d(x,y)=frac {|x-y|} {sqrt {1+|x|^{2}}sqrt {1+|y|^{2}}}$$ Is this a metric?
This question arose from the following post where the answer was given in the affirmative when $X$ is an inner product space. I am sorry that I have not made any progress so far. In the post below I asked the question in a comment and someone suggested that I should post this as a separate question.
How to show that the spherical metric satisfies the triangle inequality?
Some additional information: the triangle inequality for $d$ holds for an abstract norm on $X$ iff it holds in $C[0,1]$ with the supremum norm. This may or may not help in answering the question but it makes the question a bit more interesting because we can work with a specific norm. Justification for this claim: $C[0,1]$ is a universal space for the class of separable Banach spaces in the sense any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$. In particular any three dimensional subspace of $X$ is isometrically isomorphic to a subsapce of $C[0,1]$ and triangle inequality involves only a three diemnsional subspace.
normed-spaces
edited Jan 19 at 15:21
Blue
48.5k870154
asked Jan 16 at 10:30
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
$endgroup$
Let $X$ be a normed linear space and $$d(x,y)=frac {|x-y|} {sqrt {1+|x|^{2}}sqrt {1+|y|^{2}}}$$ Is this a metric?
This question arose from the following post where the answer was given in the affirmative when $X$ is an inner product space. I am sorry that I have not made any progress so far. In the post below I asked the question in a comment and someone suggested that I should post this as a separate question.
How to show that the spherical metric satisfies the triangle inequality?
Some additional information: the triangle inequality for $d$ holds for an abstract norm on $X$ iff it holds in $C[0,1]$ with the supremum norm. This may or may not help in answering the question but it makes the question a bit more interesting because we can work with a specific norm. Justification for this claim: $C[0,1]$ is a universal space for the class of separable Banach spaces in the sense any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$. In particular any three dimensional subspace of $X$ is isometrically isomorphic to a subsapce of $C[0,1]$ and triangle inequality involves only a three diemnsional subspace.
normed-spaces
normed-spaces
edited Jan 19 at 15:21
Blue
48.5k870154
asked Jan 16 at 10:30
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
edited Jan 19 at 15:21
Blue
48.5k870154
asked Jan 16 at 10:30
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
edited Jan 19 at 15:21
Blue
48.5k870154
edited Jan 19 at 15:21
Blue
48.5k870154
edited Jan 19 at 15:21
Blue
48.5k870154
48.5k870154
asked Jan 16 at 10:30
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
asked Jan 16 at 10:30
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
asked Jan 16 at 10:30
Kavi Rama MurthyKavi Rama Murthy
61.7k42262
61.7k42262
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Here is a counter-example: Consider $X = Bbb R^2$ with the $1$-norm
$$
Vert (x_1, x_2) Vert = |x_1| + |x_2|
$$
and the points
$$
x = (1, 0) ,, quad y = (1, 1) ,, quad z = (0, 1) , .
$$
Then
$$
Vert x Vert = Vert z Vert = 1 ,, quad Vert y Vert = 2 , , \
Vert x - y Vert = Vert y - z Vert = 1 ,, quad Vert x - z Vert = 2 , ,
$$
so that
$$
d(x, z) = 1 , , quad d(x, y) = d(y, z) = frac{1}{ sqrt 2 sqrt 5}
$$
and the triangle inequality
$$
d(x, z) le d(x, y) + d(y, z) iff 1 le sqrt frac 25
$$
does not hold for these points.
More counter-examples can be constructed with the $p$-norm. Using the same points $x,y, z$ leads to the inequality
$$
2^{2/p} (1 + 2^{2/p}) le 8 , .
$$
which is not satisfied for sufficiently small $p$, e.g. for $p le frac 32$.
answered Jan 19 at 14:47
Martin RMartin R
29k33458
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– Kavi Rama Murthy
Jan 19 at 23:11
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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oldest
votes
$begingroup$
Here is a counter-example: Consider $X = Bbb R^2$ with the $1$-norm
$$
Vert (x_1, x_2) Vert = |x_1| + |x_2|
$$
and the points
$$
x = (1, 0) ,, quad y = (1, 1) ,, quad z = (0, 1) , .
$$
Then
$$
Vert x Vert = Vert z Vert = 1 ,, quad Vert y Vert = 2 , , \
Vert x - y Vert = Vert y - z Vert = 1 ,, quad Vert x - z Vert = 2 , ,
$$
so that
$$
d(x, z) = 1 , , quad d(x, y) = d(y, z) = frac{1}{ sqrt 2 sqrt 5}
$$
and the triangle inequality
$$
d(x, z) le d(x, y) + d(y, z) iff 1 le sqrt frac 25
$$
does not hold for these points.
More counter-examples can be constructed with the $p$-norm. Using the same points $x,y, z$ leads to the inequality
$$
2^{2/p} (1 + 2^{2/p}) le 8 , .
$$
which is not satisfied for sufficiently small $p$, e.g. for $p le frac 32$.
answered Jan 19 at 14:47
Martin RMartin R
29k33458
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– Kavi Rama Murthy
Jan 19 at 23:11
add a comment |
$begingroup$
Here is a counter-example: Consider $X = Bbb R^2$ with the $1$-norm
$$
Vert (x_1, x_2) Vert = |x_1| + |x_2|
$$
and the points
$$
x = (1, 0) ,, quad y = (1, 1) ,, quad z = (0, 1) , .
$$
Then
$$
Vert x Vert = Vert z Vert = 1 ,, quad Vert y Vert = 2 , , \
Vert x - y Vert = Vert y - z Vert = 1 ,, quad Vert x - z Vert = 2 , ,
$$
so that
$$
d(x, z) = 1 , , quad d(x, y) = d(y, z) = frac{1}{ sqrt 2 sqrt 5}
$$
and the triangle inequality
$$
d(x, z) le d(x, y) + d(y, z) iff 1 le sqrt frac 25
$$
does not hold for these points.
More counter-examples can be constructed with the $p$-norm. Using the same points $x,y, z$ leads to the inequality
$$
2^{2/p} (1 + 2^{2/p}) le 8 , .
$$
which is not satisfied for sufficiently small $p$, e.g. for $p le frac 32$.
answered Jan 19 at 14:47
Martin RMartin R
29k33458
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– Kavi Rama Murthy
Jan 19 at 23:11
add a comment |
$begingroup$
Here is a counter-example: Consider $X = Bbb R^2$ with the $1$-norm
$$
Vert (x_1, x_2) Vert = |x_1| + |x_2|
$$
and the points
$$
x = (1, 0) ,, quad y = (1, 1) ,, quad z = (0, 1) , .
$$
Then
$$
Vert x Vert = Vert z Vert = 1 ,, quad Vert y Vert = 2 , , \
Vert x - y Vert = Vert y - z Vert = 1 ,, quad Vert x - z Vert = 2 , ,
$$
so that
$$
d(x, z) = 1 , , quad d(x, y) = d(y, z) = frac{1}{ sqrt 2 sqrt 5}
$$
and the triangle inequality
$$
d(x, z) le d(x, y) + d(y, z) iff 1 le sqrt frac 25
$$
does not hold for these points.
More counter-examples can be constructed with the $p$-norm. Using the same points $x,y, z$ leads to the inequality
$$
2^{2/p} (1 + 2^{2/p}) le 8 , .
$$
which is not satisfied for sufficiently small $p$, e.g. for $p le frac 32$.
answered Jan 19 at 14:47
Martin RMartin R
29k33458
$endgroup$
Here is a counter-example: Consider $X = Bbb R^2$ with the $1$-norm
$$
Vert (x_1, x_2) Vert = |x_1| + |x_2|
$$
and the points
$$
x = (1, 0) ,, quad y = (1, 1) ,, quad z = (0, 1) , .
$$
Then
$$
Vert x Vert = Vert z Vert = 1 ,, quad Vert y Vert = 2 , , \
Vert x - y Vert = Vert y - z Vert = 1 ,, quad Vert x - z Vert = 2 , ,
$$
so that
$$
d(x, z) = 1 , , quad d(x, y) = d(y, z) = frac{1}{ sqrt 2 sqrt 5}
$$
and the triangle inequality
$$
d(x, z) le d(x, y) + d(y, z) iff 1 le sqrt frac 25
$$
does not hold for these points.
More counter-examples can be constructed with the $p$-norm. Using the same points $x,y, z$ leads to the inequality
$$
2^{2/p} (1 + 2^{2/p}) le 8 , .
$$
which is not satisfied for sufficiently small $p$, e.g. for $p le frac 32$.
answered Jan 19 at 14:47
Martin RMartin R
29k33458
edited Jan 19 at 15:12
answered Jan 19 at 14:47
Martin RMartin R
29k33458
answered Jan 19 at 14:47
Martin RMartin R
29k33458
answered Jan 19 at 14:47
Martin RMartin R
29k33458
29k33458
$begingroup$
Thank you very much.
$endgroup$
– Kavi Rama Murthy
Jan 19 at 23:11
add a comment |
$begingroup$
Thank you very much.
$endgroup$
– Kavi Rama Murthy
Jan 19 at 23:11
$begingroup$
Thank you very much.
$endgroup$
– Kavi Rama Murthy
Jan 19 at 23:11
$begingroup$
Thank you very much.
$endgroup$
– Kavi Rama Murthy
Jan 19 at 23:11
add a comment |
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