Definition of a Convex Cone
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In the definition of a convex cone, given that $x,y$ belong to the convex cone $C$,then $theta_1x+theta_2y$ must also belong to $C$, where $theta_1,theta_2 > 0$.
What I don't understand is why there isn't the additional constraint that $theta_1+theta_2=1$ to make sure the line that crosses both $x$ and $y$ is restricted to the segment in between them.
convex-analysis convex-optimization
edited Jul 3 '18 at 16:44
wjmccann
657118
asked Sep 15 '15 at 17:04
UndertherainbowUndertherainbow
314417
$endgroup$
|
show 4 more comments
$begingroup$
In the definition of a convex cone, given that $x,y$ belong to the convex cone $C$,then $theta_1x+theta_2y$ must also belong to $C$, where $theta_1,theta_2 > 0$.
What I don't understand is why there isn't the additional constraint that $theta_1+theta_2=1$ to make sure the line that crosses both $x$ and $y$ is restricted to the segment in between them.
convex-analysis convex-optimization
edited Jul 3 '18 at 16:44
wjmccann
657118
asked Sep 15 '15 at 17:04
UndertherainbowUndertherainbow
314417
$endgroup$
$begingroup$
The cone, by definition, contains rays, i.e. half-lines that extend out to the appropriate infinite extent. Adding the constraint that $theta_1 + theta_2 = 1$ would only give you a convex set, it wouldn't allow the extent of the cone.
$endgroup$
– postmortes
Sep 15 '15 at 17:07
1
$begingroup$
If $theta_1,theta_2 geq 0 $ this means in particular that all $theta_1,theta_2$ with $theta_1+theta_2 =1$ are also included
$endgroup$
– asterisk
Sep 15 '15 at 17:09
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@postmortes But how will you check for convexity if you don't? Wouldn't the set then have to be affine to satisfy that condition?
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– Undertherainbow
Sep 15 '15 at 17:11
$begingroup$
@postmortes I mean that if you don't restrict the line to the segment in between them, then the whole line would have to be in the set, so it would not just be convex, it would have to be an affine set to satisfy those conditions, right? I don't know.
$endgroup$
– Undertherainbow
Sep 15 '15 at 17:15
$begingroup$
It won't be an affine set unless the cone includes points either side of 0 (affine sets contain lines that go from postive infinity to negative infinity. @asterisk's comment points out that your cone will still be convex: pick any two points in it and all points on the line between them must lie in the cone, hence it is convex. Think about the positive orthant (i.e. ${xinmathbb{R}^n: x_igeq0}$) – that's a convex cone.
$endgroup$
– postmortes
Sep 15 '15 at 17:18
|
show 4 more comments
$begingroup$
In the definition of a convex cone, given that $x,y$ belong to the convex cone $C$,then $theta_1x+theta_2y$ must also belong to $C$, where $theta_1,theta_2 > 0$.
What I don't understand is why there isn't the additional constraint that $theta_1+theta_2=1$ to make sure the line that crosses both $x$ and $y$ is restricted to the segment in between them.
convex-analysis convex-optimization
edited Jul 3 '18 at 16:44
wjmccann
657118
asked Sep 15 '15 at 17:04
UndertherainbowUndertherainbow
314417
$endgroup$
In the definition of a convex cone, given that $x,y$ belong to the convex cone $C$,then $theta_1x+theta_2y$ must also belong to $C$, where $theta_1,theta_2 > 0$.
What I don't understand is why there isn't the additional constraint that $theta_1+theta_2=1$ to make sure the line that crosses both $x$ and $y$ is restricted to the segment in between them.
convex-analysis convex-optimization
convex-analysis convex-optimization
edited Jul 3 '18 at 16:44
wjmccann
657118
asked Sep 15 '15 at 17:04
UndertherainbowUndertherainbow
314417
edited Jul 3 '18 at 16:44
wjmccann
657118
asked Sep 15 '15 at 17:04
UndertherainbowUndertherainbow
314417
edited Jul 3 '18 at 16:44
wjmccann
657118
edited Jul 3 '18 at 16:44
wjmccann
657118
edited Jul 3 '18 at 16:44
wjmccann
657118
657118
asked Sep 15 '15 at 17:04
UndertherainbowUndertherainbow
314417
asked Sep 15 '15 at 17:04
UndertherainbowUndertherainbow
314417
asked Sep 15 '15 at 17:04
UndertherainbowUndertherainbow
314417
314417
$begingroup$
The cone, by definition, contains rays, i.e. half-lines that extend out to the appropriate infinite extent. Adding the constraint that $theta_1 + theta_2 = 1$ would only give you a convex set, it wouldn't allow the extent of the cone.
$endgroup$
– postmortes
Sep 15 '15 at 17:07
1
$begingroup$
If $theta_1,theta_2 geq 0 $ this means in particular that all $theta_1,theta_2$ with $theta_1+theta_2 =1$ are also included
$endgroup$
– asterisk
Sep 15 '15 at 17:09
$begingroup$
@postmortes But how will you check for convexity if you don't? Wouldn't the set then have to be affine to satisfy that condition?
$endgroup$
– Undertherainbow
Sep 15 '15 at 17:11
$begingroup$
@postmortes I mean that if you don't restrict the line to the segment in between them, then the whole line would have to be in the set, so it would not just be convex, it would have to be an affine set to satisfy those conditions, right? I don't know.
$endgroup$
– Undertherainbow
Sep 15 '15 at 17:15
$begingroup$
It won't be an affine set unless the cone includes points either side of 0 (affine sets contain lines that go from postive infinity to negative infinity. @asterisk's comment points out that your cone will still be convex: pick any two points in it and all points on the line between them must lie in the cone, hence it is convex. Think about the positive orthant (i.e. ${xinmathbb{R}^n: x_igeq0}$) – that's a convex cone.
$endgroup$
– postmortes
Sep 15 '15 at 17:18
|
show 4 more comments
$begingroup$
The cone, by definition, contains rays, i.e. half-lines that extend out to the appropriate infinite extent. Adding the constraint that $theta_1 + theta_2 = 1$ would only give you a convex set, it wouldn't allow the extent of the cone.
$endgroup$
– postmortes
Sep 15 '15 at 17:07
1
$begingroup$
If $theta_1,theta_2 geq 0 $ this means in particular that all $theta_1,theta_2$ with $theta_1+theta_2 =1$ are also included
$endgroup$
– asterisk
Sep 15 '15 at 17:09
$begingroup$
@postmortes But how will you check for convexity if you don't? Wouldn't the set then have to be affine to satisfy that condition?
$endgroup$
– Undertherainbow
Sep 15 '15 at 17:11
$begingroup$
@postmortes I mean that if you don't restrict the line to the segment in between them, then the whole line would have to be in the set, so it would not just be convex, it would have to be an affine set to satisfy those conditions, right? I don't know.
$endgroup$
– Undertherainbow
Sep 15 '15 at 17:15
$begingroup$
It won't be an affine set unless the cone includes points either side of 0 (affine sets contain lines that go from postive infinity to negative infinity. @asterisk's comment points out that your cone will still be convex: pick any two points in it and all points on the line between them must lie in the cone, hence it is convex. Think about the positive orthant (i.e. ${xinmathbb{R}^n: x_igeq0}$) – that's a convex cone.
$endgroup$
– postmortes
Sep 15 '15 at 17:18
$begingroup$
The cone, by definition, contains rays, i.e. half-lines that extend out to the appropriate infinite extent. Adding the constraint that $theta_1 + theta_2 = 1$ would only give you a convex set, it wouldn't allow the extent of the cone.
$endgroup$
– postmortes
Sep 15 '15 at 17:07
$begingroup$
The cone, by definition, contains rays, i.e. half-lines that extend out to the appropriate infinite extent. Adding the constraint that $theta_1 + theta_2 = 1$ would only give you a convex set, it wouldn't allow the extent of the cone.
$endgroup$
– postmortes
Sep 15 '15 at 17:07
1
1
$begingroup$
If $theta_1,theta_2 geq 0 $ this means in particular that all $theta_1,theta_2$ with $theta_1+theta_2 =1$ are also included
$endgroup$
– asterisk
Sep 15 '15 at 17:09
$begingroup$
If $theta_1,theta_2 geq 0 $ this means in particular that all $theta_1,theta_2$ with $theta_1+theta_2 =1$ are also included
$endgroup$
– asterisk
Sep 15 '15 at 17:09
$begingroup$
@postmortes But how will you check for convexity if you don't? Wouldn't the set then have to be affine to satisfy that condition?
$endgroup$
– Undertherainbow
Sep 15 '15 at 17:11
$begingroup$
@postmortes But how will you check for convexity if you don't? Wouldn't the set then have to be affine to satisfy that condition?
$endgroup$
– Undertherainbow
Sep 15 '15 at 17:11
$begingroup$
@postmortes I mean that if you don't restrict the line to the segment in between them, then the whole line would have to be in the set, so it would not just be convex, it would have to be an affine set to satisfy those conditions, right? I don't know.
$endgroup$
– Undertherainbow
Sep 15 '15 at 17:15
$begingroup$
@postmortes I mean that if you don't restrict the line to the segment in between them, then the whole line would have to be in the set, so it would not just be convex, it would have to be an affine set to satisfy those conditions, right? I don't know.
$endgroup$
– Undertherainbow
Sep 15 '15 at 17:15
$begingroup$
It won't be an affine set unless the cone includes points either side of 0 (affine sets contain lines that go from postive infinity to negative infinity. @asterisk's comment points out that your cone will still be convex: pick any two points in it and all points on the line between them must lie in the cone, hence it is convex. Think about the positive orthant (i.e. ${xinmathbb{R}^n: x_igeq0}$) – that's a convex cone.
$endgroup$
– postmortes
Sep 15 '15 at 17:18
$begingroup$
It won't be an affine set unless the cone includes points either side of 0 (affine sets contain lines that go from postive infinity to negative infinity. @asterisk's comment points out that your cone will still be convex: pick any two points in it and all points on the line between them must lie in the cone, hence it is convex. Think about the positive orthant (i.e. ${xinmathbb{R}^n: x_igeq0}$) – that's a convex cone.
$endgroup$
– postmortes
Sep 15 '15 at 17:18
|
show 4 more comments
1 Answer
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It is sufficient that $theta_1 x+ theta_2y in C, theta_1,theta_2ge 0$ implies $C$ is convex, which includes the case $theta x+(1-theta)y,theta in [0,1]$.
Conversely, is it necessary? Say that a cone is convex implies $theta_1 x+ theta_2y in C, theta_1,theta_2ge 0$. Convexity means $theta x+(1-theta)y in C$. For a cone, $xin C$ requires $lambda x in C, lambda ge 0$. We can then replace $x,y$ with $lambda x,lambda ge 0$ and $mu y,mu ge 0$ that both belong to $C$, like $theta lambda x+(1-theta)mu y in C$. Since $lambda,mu$ can be any non-negative real number, we can conclude that $theta_1 x +theta_2 y in C, theta_1, theta_2 ge 0$ is necessary, under the convexity condition.
answered Jan 4 at 6:40
Mingyuan ZhaoMingyuan Zhao
111
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add a comment |
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$begingroup$
It is sufficient that $theta_1 x+ theta_2y in C, theta_1,theta_2ge 0$ implies $C$ is convex, which includes the case $theta x+(1-theta)y,theta in [0,1]$.
Conversely, is it necessary? Say that a cone is convex implies $theta_1 x+ theta_2y in C, theta_1,theta_2ge 0$. Convexity means $theta x+(1-theta)y in C$. For a cone, $xin C$ requires $lambda x in C, lambda ge 0$. We can then replace $x,y$ with $lambda x,lambda ge 0$ and $mu y,mu ge 0$ that both belong to $C$, like $theta lambda x+(1-theta)mu y in C$. Since $lambda,mu$ can be any non-negative real number, we can conclude that $theta_1 x +theta_2 y in C, theta_1, theta_2 ge 0$ is necessary, under the convexity condition.
answered Jan 4 at 6:40
Mingyuan ZhaoMingyuan Zhao
111
$endgroup$
add a comment |
$begingroup$
It is sufficient that $theta_1 x+ theta_2y in C, theta_1,theta_2ge 0$ implies $C$ is convex, which includes the case $theta x+(1-theta)y,theta in [0,1]$.
Conversely, is it necessary? Say that a cone is convex implies $theta_1 x+ theta_2y in C, theta_1,theta_2ge 0$. Convexity means $theta x+(1-theta)y in C$. For a cone, $xin C$ requires $lambda x in C, lambda ge 0$. We can then replace $x,y$ with $lambda x,lambda ge 0$ and $mu y,mu ge 0$ that both belong to $C$, like $theta lambda x+(1-theta)mu y in C$. Since $lambda,mu$ can be any non-negative real number, we can conclude that $theta_1 x +theta_2 y in C, theta_1, theta_2 ge 0$ is necessary, under the convexity condition.
answered Jan 4 at 6:40
Mingyuan ZhaoMingyuan Zhao
111
$endgroup$
add a comment |
$begingroup$
It is sufficient that $theta_1 x+ theta_2y in C, theta_1,theta_2ge 0$ implies $C$ is convex, which includes the case $theta x+(1-theta)y,theta in [0,1]$.
Conversely, is it necessary? Say that a cone is convex implies $theta_1 x+ theta_2y in C, theta_1,theta_2ge 0$. Convexity means $theta x+(1-theta)y in C$. For a cone, $xin C$ requires $lambda x in C, lambda ge 0$. We can then replace $x,y$ with $lambda x,lambda ge 0$ and $mu y,mu ge 0$ that both belong to $C$, like $theta lambda x+(1-theta)mu y in C$. Since $lambda,mu$ can be any non-negative real number, we can conclude that $theta_1 x +theta_2 y in C, theta_1, theta_2 ge 0$ is necessary, under the convexity condition.
answered Jan 4 at 6:40
Mingyuan ZhaoMingyuan Zhao
111
$endgroup$
It is sufficient that $theta_1 x+ theta_2y in C, theta_1,theta_2ge 0$ implies $C$ is convex, which includes the case $theta x+(1-theta)y,theta in [0,1]$.
Conversely, is it necessary? Say that a cone is convex implies $theta_1 x+ theta_2y in C, theta_1,theta_2ge 0$. Convexity means $theta x+(1-theta)y in C$. For a cone, $xin C$ requires $lambda x in C, lambda ge 0$. We can then replace $x,y$ with $lambda x,lambda ge 0$ and $mu y,mu ge 0$ that both belong to $C$, like $theta lambda x+(1-theta)mu y in C$. Since $lambda,mu$ can be any non-negative real number, we can conclude that $theta_1 x +theta_2 y in C, theta_1, theta_2 ge 0$ is necessary, under the convexity condition.
answered Jan 4 at 6:40
Mingyuan ZhaoMingyuan Zhao
111
answered Jan 4 at 6:40
Mingyuan ZhaoMingyuan Zhao
111
answered Jan 4 at 6:40
Mingyuan ZhaoMingyuan Zhao
111
answered Jan 4 at 6:40
Mingyuan ZhaoMingyuan Zhao
111
111
add a comment |
add a comment |
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$begingroup$
The cone, by definition, contains rays, i.e. half-lines that extend out to the appropriate infinite extent. Adding the constraint that $theta_1 + theta_2 = 1$ would only give you a convex set, it wouldn't allow the extent of the cone.
$endgroup$
– postmortes
Sep 15 '15 at 17:07
1
$begingroup$
If $theta_1,theta_2 geq 0 $ this means in particular that all $theta_1,theta_2$ with $theta_1+theta_2 =1$ are also included
$endgroup$
– asterisk
Sep 15 '15 at 17:09
$begingroup$
@postmortes But how will you check for convexity if you don't? Wouldn't the set then have to be affine to satisfy that condition?
$endgroup$
– Undertherainbow
Sep 15 '15 at 17:11
$begingroup$
@postmortes I mean that if you don't restrict the line to the segment in between them, then the whole line would have to be in the set, so it would not just be convex, it would have to be an affine set to satisfy those conditions, right? I don't know.
$endgroup$
– Undertherainbow
Sep 15 '15 at 17:15
$begingroup$
It won't be an affine set unless the cone includes points either side of 0 (affine sets contain lines that go from postive infinity to negative infinity. @asterisk's comment points out that your cone will still be convex: pick any two points in it and all points on the line between them must lie in the cone, hence it is convex. Think about the positive orthant (i.e. ${xinmathbb{R}^n: x_igeq0}$) – that's a convex cone.
$endgroup$
– postmortes
Sep 15 '15 at 17:18