Mixing
Proving $K= {x in mathbb R^n: exists t>0 (t^{-1}, text { for } x in A},$ given convex set $Asubset mathbb...
$begingroup$
I have a convex set $ A subset mathbb{R^n}$ and $K= {x in mathbb {R^n}mid exists t>0( t^{-1} text { for } x in A)}.$
I want to show, that $K$ is convex.
My idea:
$y in K Rightarrow exists t>0: t^{-1} y in A$ and $z in K Rightarrow exists t>0: t'^{-1} z in A$.
Thus $A$ convex, there is a $t''>0: t''^{-1}( lambda t^{-1} y + (1- lambda) t'^{-1} z) in A $ with $ lambda in [0,1]$.
How can I argue now, that this convex combination is in $K$?
convex-analysis
edited Jan 18 at 18:18
Andrés E. Caicedo
65.5k8159250
asked Jan 18 at 15:52
SvenMathSvenMath
377
$endgroup$
add a comment |
$begingroup$
I have a convex set $ A subset mathbb{R^n}$ and $K= {x in mathbb {R^n}mid exists t>0( t^{-1} text { for } x in A)}.$
I want to show, that $K$ is convex.
My idea:
$y in K Rightarrow exists t>0: t^{-1} y in A$ and $z in K Rightarrow exists t>0: t'^{-1} z in A$.
Thus $A$ convex, there is a $t''>0: t''^{-1}( lambda t^{-1} y + (1- lambda) t'^{-1} z) in A $ with $ lambda in [0,1]$.
How can I argue now, that this convex combination is in $K$?
convex-analysis
edited Jan 18 at 18:18
Andrés E. Caicedo
65.5k8159250
asked Jan 18 at 15:52
SvenMathSvenMath
377
$endgroup$
$begingroup$
What does ($t^{-1}$ for $xin A$)?
$endgroup$
– Mostafa Ayaz
Jan 18 at 16:10
$begingroup$
It should be $..exists t>0 : t^{-1} cdot x in A$
$endgroup$
– SvenMath
Jan 18 at 16:12
$begingroup$
@SvenMath so you should have $ty$ and $tz$ then?
$endgroup$
– LinAlg
Jan 18 at 16:16
add a comment |
$begingroup$
I have a convex set $ A subset mathbb{R^n}$ and $K= {x in mathbb {R^n}mid exists t>0( t^{-1} text { for } x in A)}.$
I want to show, that $K$ is convex.
My idea:
$y in K Rightarrow exists t>0: t^{-1} y in A$ and $z in K Rightarrow exists t>0: t'^{-1} z in A$.
Thus $A$ convex, there is a $t''>0: t''^{-1}( lambda t^{-1} y + (1- lambda) t'^{-1} z) in A $ with $ lambda in [0,1]$.
How can I argue now, that this convex combination is in $K$?
convex-analysis
edited Jan 18 at 18:18
Andrés E. Caicedo
65.5k8159250
asked Jan 18 at 15:52
SvenMathSvenMath
377
$endgroup$
I have a convex set $ A subset mathbb{R^n}$ and $K= {x in mathbb {R^n}mid exists t>0( t^{-1} text { for } x in A)}.$
I want to show, that $K$ is convex.
My idea:
$y in K Rightarrow exists t>0: t^{-1} y in A$ and $z in K Rightarrow exists t>0: t'^{-1} z in A$.
Thus $A$ convex, there is a $t''>0: t''^{-1}( lambda t^{-1} y + (1- lambda) t'^{-1} z) in A $ with $ lambda in [0,1]$.
How can I argue now, that this convex combination is in $K$?
convex-analysis
convex-analysis
edited Jan 18 at 18:18
Andrés E. Caicedo
65.5k8159250
asked Jan 18 at 15:52
SvenMathSvenMath
377
edited Jan 18 at 18:18
Andrés E. Caicedo
65.5k8159250
asked Jan 18 at 15:52
SvenMathSvenMath
377
edited Jan 18 at 18:18
Andrés E. Caicedo
65.5k8159250
edited Jan 18 at 18:18
Andrés E. Caicedo
65.5k8159250
edited Jan 18 at 18:18
Andrés E. Caicedo
65.5k8159250
65.5k8159250
asked Jan 18 at 15:52
SvenMathSvenMath
377
asked Jan 18 at 15:52
SvenMathSvenMath
377
asked Jan 18 at 15:52
SvenMathSvenMath
377
377
$begingroup$
What does ($t^{-1}$ for $xin A$)?
$endgroup$
– Mostafa Ayaz
Jan 18 at 16:10
$begingroup$
It should be $..exists t>0 : t^{-1} cdot x in A$
$endgroup$
– SvenMath
Jan 18 at 16:12
$begingroup$
@SvenMath so you should have $ty$ and $tz$ then?
$endgroup$
– LinAlg
Jan 18 at 16:16
add a comment |
$begingroup$
What does ($t^{-1}$ for $xin A$)?
$endgroup$
– Mostafa Ayaz
Jan 18 at 16:10
$begingroup$
It should be $..exists t>0 : t^{-1} cdot x in A$
$endgroup$
– SvenMath
Jan 18 at 16:12
$begingroup$
@SvenMath so you should have $ty$ and $tz$ then?
$endgroup$
– LinAlg
Jan 18 at 16:16
$begingroup$
What does ($t^{-1}$ for $xin A$)?
$endgroup$
– Mostafa Ayaz
Jan 18 at 16:10
$begingroup$
What does ($t^{-1}$ for $xin A$)?
$endgroup$
– Mostafa Ayaz
Jan 18 at 16:10
$begingroup$
It should be $..exists t>0 : t^{-1} cdot x in A$
$endgroup$
– SvenMath
Jan 18 at 16:12
$begingroup$
It should be $..exists t>0 : t^{-1} cdot x in A$
$endgroup$
– SvenMath
Jan 18 at 16:12
$begingroup$
@SvenMath so you should have $ty$ and $tz$ then?
$endgroup$
– LinAlg
Jan 18 at 16:16
$begingroup$
@SvenMath so you should have $ty$ and $tz$ then?
$endgroup$
– LinAlg
Jan 18 at 16:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The best way to do things, is to note that if $a in A$, then every element of the form $at, t > 0$ belongs in $K$, since $(at)t^{-1} = a in A$. Conversely, if $l in K$, then for some $t > 0$ we have $lt^{-1} = b in A$, or that $l=tb$ for some $b in A$ and $t > 0$.
All this tells us that $K = {at : a in A, t > 0}$. If $A$ is convex, we claim this is convex.
With this description, indeed if $at in K$ and $bt' in K$, then if $lambda in [0,1]$ we want to show that $l = atlambda + bt'(1-lambda) in K$.
For this, note that $t lambda geq 0$ and $t'(1-lambda) geq 0$, with $k=tlambda + t'(1-lambda) > 0$, since one of $lambda,1-lambda$ is strictly positive.
Then, $$l = frac{l}{k}k = kleft(afrac{tlambda}{k} + b frac{t'(1-lambda)}{k}right)$$
is the product of $k>0$ and a convex combination of $a$ and $b$ which lies inside $A$ since it is convex. By definition of $K$, we get that $l in K$, or that $K$ is convex.
answered Jan 18 at 16:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
$begingroup$
Thank you for your proof. I need some time to unterstand what you have done.
$endgroup$
– SvenMath
Jan 18 at 16:37
$begingroup$
You are welcome. I will edit my answer if required.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:38
$begingroup$
No you musn't do that:)
$endgroup$
– SvenMath
Jan 18 at 16:39
$begingroup$
Ok , I will not do so.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:40
$begingroup$
Why is the last term a convex combination of a and b. Do you consider $ frac{ t lambda}{k} in [0,1] and frac{ t'(1- lambda)}{k} in [0,1] $
$endgroup$
– SvenMath
Jan 18 at 16:48
|
show 5 more comments
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1 Answer
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1 Answer
1
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oldest
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$begingroup$
The best way to do things, is to note that if $a in A$, then every element of the form $at, t > 0$ belongs in $K$, since $(at)t^{-1} = a in A$. Conversely, if $l in K$, then for some $t > 0$ we have $lt^{-1} = b in A$, or that $l=tb$ for some $b in A$ and $t > 0$.
All this tells us that $K = {at : a in A, t > 0}$. If $A$ is convex, we claim this is convex.
With this description, indeed if $at in K$ and $bt' in K$, then if $lambda in [0,1]$ we want to show that $l = atlambda + bt'(1-lambda) in K$.
For this, note that $t lambda geq 0$ and $t'(1-lambda) geq 0$, with $k=tlambda + t'(1-lambda) > 0$, since one of $lambda,1-lambda$ is strictly positive.
Then, $$l = frac{l}{k}k = kleft(afrac{tlambda}{k} + b frac{t'(1-lambda)}{k}right)$$
is the product of $k>0$ and a convex combination of $a$ and $b$ which lies inside $A$ since it is convex. By definition of $K$, we get that $l in K$, or that $K$ is convex.
answered Jan 18 at 16:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
$begingroup$
Thank you for your proof. I need some time to unterstand what you have done.
$endgroup$
– SvenMath
Jan 18 at 16:37
$begingroup$
You are welcome. I will edit my answer if required.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:38
$begingroup$
No you musn't do that:)
$endgroup$
– SvenMath
Jan 18 at 16:39
$begingroup$
Ok , I will not do so.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:40
$begingroup$
Why is the last term a convex combination of a and b. Do you consider $ frac{ t lambda}{k} in [0,1] and frac{ t'(1- lambda)}{k} in [0,1] $
$endgroup$
– SvenMath
Jan 18 at 16:48
|
show 5 more comments
$begingroup$
The best way to do things, is to note that if $a in A$, then every element of the form $at, t > 0$ belongs in $K$, since $(at)t^{-1} = a in A$. Conversely, if $l in K$, then for some $t > 0$ we have $lt^{-1} = b in A$, or that $l=tb$ for some $b in A$ and $t > 0$.
All this tells us that $K = {at : a in A, t > 0}$. If $A$ is convex, we claim this is convex.
With this description, indeed if $at in K$ and $bt' in K$, then if $lambda in [0,1]$ we want to show that $l = atlambda + bt'(1-lambda) in K$.
For this, note that $t lambda geq 0$ and $t'(1-lambda) geq 0$, with $k=tlambda + t'(1-lambda) > 0$, since one of $lambda,1-lambda$ is strictly positive.
Then, $$l = frac{l}{k}k = kleft(afrac{tlambda}{k} + b frac{t'(1-lambda)}{k}right)$$
is the product of $k>0$ and a convex combination of $a$ and $b$ which lies inside $A$ since it is convex. By definition of $K$, we get that $l in K$, or that $K$ is convex.
answered Jan 18 at 16:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
$begingroup$
Thank you for your proof. I need some time to unterstand what you have done.
$endgroup$
– SvenMath
Jan 18 at 16:37
$begingroup$
You are welcome. I will edit my answer if required.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:38
$begingroup$
No you musn't do that:)
$endgroup$
– SvenMath
Jan 18 at 16:39
$begingroup$
Ok , I will not do so.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:40
$begingroup$
Why is the last term a convex combination of a and b. Do you consider $ frac{ t lambda}{k} in [0,1] and frac{ t'(1- lambda)}{k} in [0,1] $
$endgroup$
– SvenMath
Jan 18 at 16:48
|
show 5 more comments
$begingroup$
The best way to do things, is to note that if $a in A$, then every element of the form $at, t > 0$ belongs in $K$, since $(at)t^{-1} = a in A$. Conversely, if $l in K$, then for some $t > 0$ we have $lt^{-1} = b in A$, or that $l=tb$ for some $b in A$ and $t > 0$.
All this tells us that $K = {at : a in A, t > 0}$. If $A$ is convex, we claim this is convex.
With this description, indeed if $at in K$ and $bt' in K$, then if $lambda in [0,1]$ we want to show that $l = atlambda + bt'(1-lambda) in K$.
For this, note that $t lambda geq 0$ and $t'(1-lambda) geq 0$, with $k=tlambda + t'(1-lambda) > 0$, since one of $lambda,1-lambda$ is strictly positive.
Then, $$l = frac{l}{k}k = kleft(afrac{tlambda}{k} + b frac{t'(1-lambda)}{k}right)$$
is the product of $k>0$ and a convex combination of $a$ and $b$ which lies inside $A$ since it is convex. By definition of $K$, we get that $l in K$, or that $K$ is convex.
answered Jan 18 at 16:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
The best way to do things, is to note that if $a in A$, then every element of the form $at, t > 0$ belongs in $K$, since $(at)t^{-1} = a in A$. Conversely, if $l in K$, then for some $t > 0$ we have $lt^{-1} = b in A$, or that $l=tb$ for some $b in A$ and $t > 0$.
All this tells us that $K = {at : a in A, t > 0}$. If $A$ is convex, we claim this is convex.
With this description, indeed if $at in K$ and $bt' in K$, then if $lambda in [0,1]$ we want to show that $l = atlambda + bt'(1-lambda) in K$.
For this, note that $t lambda geq 0$ and $t'(1-lambda) geq 0$, with $k=tlambda + t'(1-lambda) > 0$, since one of $lambda,1-lambda$ is strictly positive.
Then, $$l = frac{l}{k}k = kleft(afrac{tlambda}{k} + b frac{t'(1-lambda)}{k}right)$$
is the product of $k>0$ and a convex combination of $a$ and $b$ which lies inside $A$ since it is convex. By definition of $K$, we get that $l in K$, or that $K$ is convex.
answered Jan 18 at 16:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
answered Jan 18 at 16:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
answered Jan 18 at 16:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
answered Jan 18 at 16:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
38.9k33477
$begingroup$
Thank you for your proof. I need some time to unterstand what you have done.
$endgroup$
– SvenMath
Jan 18 at 16:37
$begingroup$
You are welcome. I will edit my answer if required.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:38
$begingroup$
No you musn't do that:)
$endgroup$
– SvenMath
Jan 18 at 16:39
$begingroup$
Ok , I will not do so.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:40
$begingroup$
Why is the last term a convex combination of a and b. Do you consider $ frac{ t lambda}{k} in [0,1] and frac{ t'(1- lambda)}{k} in [0,1] $
$endgroup$
– SvenMath
Jan 18 at 16:48
|
show 5 more comments
$begingroup$
Thank you for your proof. I need some time to unterstand what you have done.
$endgroup$
– SvenMath
Jan 18 at 16:37
$begingroup$
You are welcome. I will edit my answer if required.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:38
$begingroup$
No you musn't do that:)
$endgroup$
– SvenMath
Jan 18 at 16:39
$begingroup$
Ok , I will not do so.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:40
$begingroup$
Why is the last term a convex combination of a and b. Do you consider $ frac{ t lambda}{k} in [0,1] and frac{ t'(1- lambda)}{k} in [0,1] $
$endgroup$
– SvenMath
Jan 18 at 16:48
$begingroup$
Thank you for your proof. I need some time to unterstand what you have done.
$endgroup$
– SvenMath
Jan 18 at 16:37
$begingroup$
Thank you for your proof. I need some time to unterstand what you have done.
$endgroup$
– SvenMath
Jan 18 at 16:37
$begingroup$
You are welcome. I will edit my answer if required.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:38
$begingroup$
You are welcome. I will edit my answer if required.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:38
$begingroup$
No you musn't do that:)
$endgroup$
– SvenMath
Jan 18 at 16:39
$begingroup$
No you musn't do that:)
$endgroup$
– SvenMath
Jan 18 at 16:39
$begingroup$
Ok , I will not do so.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:40
$begingroup$
Ok , I will not do so.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 16:40
$begingroup$
Why is the last term a convex combination of a and b. Do you consider $ frac{ t lambda}{k} in [0,1] and frac{ t'(1- lambda)}{k} in [0,1] $
$endgroup$
– SvenMath
Jan 18 at 16:48
$begingroup$
Why is the last term a convex combination of a and b. Do you consider $ frac{ t lambda}{k} in [0,1] and frac{ t'(1- lambda)}{k} in [0,1] $
$endgroup$
– SvenMath
Jan 18 at 16:48
|
show 5 more comments
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$begingroup$
What does ($t^{-1}$ for $xin A$)?
$endgroup$
– Mostafa Ayaz
Jan 18 at 16:10
$begingroup$
It should be $..exists t>0 : t^{-1} cdot x in A$
$endgroup$
– SvenMath
Jan 18 at 16:12
$begingroup$
@SvenMath so you should have $ty$ and $tz$ then?
$endgroup$
– LinAlg
Jan 18 at 16:16