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For which $Csubset Bbb R^n_++$, the set $log(C)$ is convex?

Ensuring strict concavity on log utility functionsRectangle in Euclidean n-spaceEquivalent definitions of piecewise affine convex function over a convex set of $mathbbR^n$.Convexity of sum and intersection of convex setsConvex sets in $mathbb R^n$: Do they have a particular form ? Does the gradient of a linear convex function $f$ exist on such a set?convex hull of union convex setIs every norm increasing?Every open set $ G subset mathbbR^n$ can be approximated by finite union of non-degenerate compact intervals contained in $G$?Let $A_1,A_2in GL(n,mathbb C)$ commutes, show that $log(A_1)$ commutes with $log(A_2)$ (for some pair)Why is the Fenchel conjugate of a piecewise linear function piecewise linear?

0

$begingroup$

Let $Bbb R^n_++=xinBbb R^nmid x_i>0, i=1,ldots,n$ and let $logcolon Bbb R^n_++to Bbb R^n$ be the component-wise logarithmic function, i.e.
$$log(x)=(log(x_1),ldots,log(x_n))qquad forall x in Bbb R^n_++$$ where $log$ is the usual logarithm in base $e$ (does it make a difference?). For a subset $SsubsetBbb R^n_++$, let $log(S)=log(x)mid xin S$.

For which $Csubset Bbb R^n_++$, the set $log(C)$ is convex in $Bbb R^n$?

Note

• Note that if $C$ is a line segment, i.e. $C=tx+(1-t)xmid 0leq t leq 1$, then $ln(C)$ can not be convex.




• Note that for every set $C=[a_1,b_1]times ldots times [a_n,b_n]$ with $0<a_i<b_i$, $log(C)=[log(a_1),log(b_1]times ldots times[log(a_n),log(b_n]$ is a convex set.




• If $C=Bbb R^n_++$, then $ln(C)=Bbb R^n$ which is convex.

real-analysis linear-algebra convex-analysis convex-geometry

share|cite|improve this question

edited Mar 30 at 15:15

Hello

asked Mar 29 at 21:02

HelloHello

319216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that if $C$ is not open, then the image of a straight line segment is not convex as it is of "dimension" 1 and not a line.
  $endgroup$
  – Hello
  Mar 29 at 21:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So let $S$ be all the points closer than $epsilon$ to some line segment $C$. Such a set is convex and open, and if $log(C)$ is not convex there exists an $epsilon$ small enough that $log(S)$ is not convex.
  $endgroup$
  – kimchi lover
  Mar 29 at 22:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see... So the question is probably better the other way around. Under which condition on $C$, the set $log(C)$ is convex. If $C$ is an interval, then we are fine :).
  $endgroup$
  – Hello
  Mar 29 at 22:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kimchilover I have edited my question
  $endgroup$
  – Hello
  Mar 29 at 22:38
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $Bbb R^n_++=xinBbb R^nmid x_i>0, i=1,ldots,n$ and let $logcolon Bbb R^n_++to Bbb R^n$ be the component-wise logarithmic function, i.e.
$$log(x)=(log(x_1),ldots,log(x_n))qquad forall x in Bbb R^n_++$$ where $log$ is the usual logarithm in base $e$ (does it make a difference?). For a subset $SsubsetBbb R^n_++$, let $log(S)=log(x)mid xin S$.

For which $Csubset Bbb R^n_++$, the set $log(C)$ is convex in $Bbb R^n$?

Note

• Note that if $C$ is a line segment, i.e. $C=tx+(1-t)xmid 0leq t leq 1$, then $ln(C)$ can not be convex.




• Note that for every set $C=[a_1,b_1]times ldots times [a_n,b_n]$ with $0<a_i<b_i$, $log(C)=[log(a_1),log(b_1]times ldots times[log(a_n),log(b_n]$ is a convex set.




• If $C=Bbb R^n_++$, then $ln(C)=Bbb R^n$ which is convex.

real-analysis linear-algebra convex-analysis convex-geometry

share|cite|improve this question

edited Mar 30 at 15:15

Hello

asked Mar 29 at 21:02

HelloHello

319216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that if $C$ is not open, then the image of a straight line segment is not convex as it is of "dimension" 1 and not a line.
  $endgroup$
  – Hello
  Mar 29 at 21:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So let $S$ be all the points closer than $epsilon$ to some line segment $C$. Such a set is convex and open, and if $log(C)$ is not convex there exists an $epsilon$ small enough that $log(S)$ is not convex.
  $endgroup$
  – kimchi lover
  Mar 29 at 22:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see... So the question is probably better the other way around. Under which condition on $C$, the set $log(C)$ is convex. If $C$ is an interval, then we are fine :).
  $endgroup$
  – Hello
  Mar 29 at 22:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kimchilover I have edited my question
  $endgroup$
  – Hello
  Mar 29 at 22:38
  

  
  
  
  

add a comment | 

$begingroup$

Let $Bbb R^n_++=xinBbb R^nmid x_i>0, i=1,ldots,n$ and let $logcolon Bbb R^n_++to Bbb R^n$ be the component-wise logarithmic function, i.e.
$$log(x)=(log(x_1),ldots,log(x_n))qquad forall x in Bbb R^n_++$$ where $log$ is the usual logarithm in base $e$ (does it make a difference?). For a subset $SsubsetBbb R^n_++$, let $log(S)=log(x)mid xin S$.

For which $Csubset Bbb R^n_++$, the set $log(C)$ is convex in $Bbb R^n$?

Note

• Note that if $C$ is a line segment, i.e. $C=tx+(1-t)xmid 0leq t leq 1$, then $ln(C)$ can not be convex.




• Note that for every set $C=[a_1,b_1]times ldots times [a_n,b_n]$ with $0<a_i<b_i$, $log(C)=[log(a_1),log(b_1]times ldots times[log(a_n),log(b_n]$ is a convex set.




• If $C=Bbb R^n_++$, then $ln(C)=Bbb R^n$ which is convex.

real-analysis linear-algebra convex-analysis convex-geometry

share|cite|improve this question

edited Mar 30 at 15:15

Hello

asked Mar 29 at 21:02

HelloHello

319216

$endgroup$

share|cite|improve this question

edited Mar 30 at 15:15

Hello

asked Mar 29 at 21:02

HelloHello

319216

share|cite|improve this question

edited Mar 30 at 15:15

Hello

asked Mar 29 at 21:02

HelloHello

319216

asked Mar 29 at 21:02

HelloHello

319216

asked Mar 29 at 21:02

HelloHello

319216

1 Answer
1

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oldest

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0

$begingroup$

You have changed the title and point of the question, by dropping the restriction that $C$ is convex. As it stands now, the answer (tautologously, and I think uninterestingly) is that $C$ must be closed under component-wise weighted geometric means, in the sense that if $x=(x_i)$ and $y=(y_i)in C$ then so is $z=(z_i)$, where $z_i = x_i ^alpha y_i^beta$ for all $alpha,betain[0,1]$ with $alpha+beta=1$.

You might as well posit $L$ an arbitrary convex set, and then take $C$ to be the set of component-wise exponentials of elements of $C$, as in $x=(x_i)in C$ if and only if $x_i=exp(y_i)$ for $yin L$.

share|cite|improve this answer

answered Mar 31 at 15:44

kimchi loverkimchi lover

11.7k31229

$endgroup$

add a comment | 

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0

$begingroup$

You have changed the title and point of the question, by dropping the restriction that $C$ is convex. As it stands now, the answer (tautologously, and I think uninterestingly) is that $C$ must be closed under component-wise weighted geometric means, in the sense that if $x=(x_i)$ and $y=(y_i)in C$ then so is $z=(z_i)$, where $z_i = x_i ^alpha y_i^beta$ for all $alpha,betain[0,1]$ with $alpha+beta=1$.

You might as well posit $L$ an arbitrary convex set, and then take $C$ to be the set of component-wise exponentials of elements of $C$, as in $x=(x_i)in C$ if and only if $x_i=exp(y_i)$ for $yin L$.

share|cite|improve this answer

answered Mar 31 at 15:44

kimchi loverkimchi lover

11.7k31229

$endgroup$

add a comment | 

0

$begingroup$

You have changed the title and point of the question, by dropping the restriction that $C$ is convex. As it stands now, the answer (tautologously, and I think uninterestingly) is that $C$ must be closed under component-wise weighted geometric means, in the sense that if $x=(x_i)$ and $y=(y_i)in C$ then so is $z=(z_i)$, where $z_i = x_i ^alpha y_i^beta$ for all $alpha,betain[0,1]$ with $alpha+beta=1$.

You might as well posit $L$ an arbitrary convex set, and then take $C$ to be the set of component-wise exponentials of elements of $C$, as in $x=(x_i)in C$ if and only if $x_i=exp(y_i)$ for $yin L$.

share|cite|improve this answer

answered Mar 31 at 15:44

kimchi loverkimchi lover

11.7k31229

$endgroup$

add a comment | 

$begingroup$

You have changed the title and point of the question, by dropping the restriction that $C$ is convex. As it stands now, the answer (tautologously, and I think uninterestingly) is that $C$ must be closed under component-wise weighted geometric means, in the sense that if $x=(x_i)$ and $y=(y_i)in C$ then so is $z=(z_i)$, where $z_i = x_i ^alpha y_i^beta$ for all $alpha,betain[0,1]$ with $alpha+beta=1$.

You might as well posit $L$ an arbitrary convex set, and then take $C$ to be the set of component-wise exponentials of elements of $C$, as in $x=(x_i)in C$ if and only if $x_i=exp(y_i)$ for $yin L$.

share|cite|improve this answer

answered Mar 31 at 15:44

kimchi loverkimchi lover

11.7k31229

$endgroup$

share|cite|improve this answer

answered Mar 31 at 15:44

kimchi loverkimchi lover

11.7k31229

answered Mar 31 at 15:44

kimchi loverkimchi lover

11.7k31229

answered Mar 31 at 15:44

kimchi loverkimchi lover

11.7k31229