Mixing
Polar representation of convex sets
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Consider a set $G$ of the form
$$G={ru:uinmathbb S^{d-1}, 0leq rleq phi(u)},$$
Where $mathbb S^{d-1}$ is the unit sphere in $mathbb R^d$ and $phi:mathbb S^{d-1}to [0,infty)$ is a given function. Is it true that if $G$ is convex, then $phi$ needs to be Lipschitz ?
Thank you !
convex-analysis polar-coordinates
asked Aug 29 '17 at 21:00
TrivialPursuit
12
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show 2 more comments
up vote
0
down vote
favorite
Consider a set $G$ of the form
$$G={ru:uinmathbb S^{d-1}, 0leq rleq phi(u)},$$
Where $mathbb S^{d-1}$ is the unit sphere in $mathbb R^d$ and $phi:mathbb S^{d-1}to [0,infty)$ is a given function. Is it true that if $G$ is convex, then $phi$ needs to be Lipschitz ?
Thank you !
convex-analysis polar-coordinates
asked Aug 29 '17 at 21:00
TrivialPursuit
12
$k$-Lipschitz with which constant $k$ ? $k=1$ ?
– Jean Marie
Aug 29 '17 at 21:14
$k$-Lipschitz with some $k$.
– TrivialPursuit
Aug 29 '17 at 21:28
Idea: If $G$ is convex then locally at every point $phi$ is convex, which implies Lipschitz.
– Jose27
Aug 29 '17 at 21:30
Jose27, what does it mean to be convex for $phi$, when it is only defined on the sphere ?
– TrivialPursuit
Aug 29 '17 at 21:32
You should sarch the web with the term "support function" which gives you access to a rather vast amount of litterature, for example the very accessible article "Support Function Representation of Convex Bodies, Its Application in Geometric Computing, and Some Related Representations" by Pijush K. Ghosh and K. Vinod Kumar (Comp. Vision and Image Understanding Vol 72, no 3, 1998)
– Jean Marie
Aug 29 '17 at 21:37
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider a set $G$ of the form
$$G={ru:uinmathbb S^{d-1}, 0leq rleq phi(u)},$$
Where $mathbb S^{d-1}$ is the unit sphere in $mathbb R^d$ and $phi:mathbb S^{d-1}to [0,infty)$ is a given function. Is it true that if $G$ is convex, then $phi$ needs to be Lipschitz ?
Thank you !
convex-analysis polar-coordinates
asked Aug 29 '17 at 21:00
TrivialPursuit
12
Consider a set $G$ of the form
$$G={ru:uinmathbb S^{d-1}, 0leq rleq phi(u)},$$
Where $mathbb S^{d-1}$ is the unit sphere in $mathbb R^d$ and $phi:mathbb S^{d-1}to [0,infty)$ is a given function. Is it true that if $G$ is convex, then $phi$ needs to be Lipschitz ?
Thank you !
convex-analysis polar-coordinates
convex-analysis polar-coordinates
asked Aug 29 '17 at 21:00
TrivialPursuit
12
asked Aug 29 '17 at 21:00
TrivialPursuit
12
asked Aug 29 '17 at 21:00
TrivialPursuit
12
asked Aug 29 '17 at 21:00
TrivialPursuit
12
asked Aug 29 '17 at 21:00
TrivialPursuit
12
12
$k$-Lipschitz with which constant $k$ ? $k=1$ ?
– Jean Marie
Aug 29 '17 at 21:14
$k$-Lipschitz with some $k$.
– TrivialPursuit
Aug 29 '17 at 21:28
Idea: If $G$ is convex then locally at every point $phi$ is convex, which implies Lipschitz.
– Jose27
Aug 29 '17 at 21:30
Jose27, what does it mean to be convex for $phi$, when it is only defined on the sphere ?
– TrivialPursuit
Aug 29 '17 at 21:32
You should sarch the web with the term "support function" which gives you access to a rather vast amount of litterature, for example the very accessible article "Support Function Representation of Convex Bodies, Its Application in Geometric Computing, and Some Related Representations" by Pijush K. Ghosh and K. Vinod Kumar (Comp. Vision and Image Understanding Vol 72, no 3, 1998)
– Jean Marie
Aug 29 '17 at 21:37
|
show 2 more comments
$k$-Lipschitz with which constant $k$ ? $k=1$ ?
– Jean Marie
Aug 29 '17 at 21:14
$k$-Lipschitz with some $k$.
– TrivialPursuit
Aug 29 '17 at 21:28
Idea: If $G$ is convex then locally at every point $phi$ is convex, which implies Lipschitz.
– Jose27
Aug 29 '17 at 21:30
Jose27, what does it mean to be convex for $phi$, when it is only defined on the sphere ?
– TrivialPursuit
Aug 29 '17 at 21:32
You should sarch the web with the term "support function" which gives you access to a rather vast amount of litterature, for example the very accessible article "Support Function Representation of Convex Bodies, Its Application in Geometric Computing, and Some Related Representations" by Pijush K. Ghosh and K. Vinod Kumar (Comp. Vision and Image Understanding Vol 72, no 3, 1998)
– Jean Marie
Aug 29 '17 at 21:37
$k$-Lipschitz with which constant $k$ ? $k=1$ ?
– Jean Marie
Aug 29 '17 at 21:14
$k$-Lipschitz with which constant $k$ ? $k=1$ ?
– Jean Marie
Aug 29 '17 at 21:14
$k$-Lipschitz with some $k$.
– TrivialPursuit
Aug 29 '17 at 21:28
$k$-Lipschitz with some $k$.
– TrivialPursuit
Aug 29 '17 at 21:28
Idea: If $G$ is convex then locally at every point $phi$ is convex, which implies Lipschitz.
– Jose27
Aug 29 '17 at 21:30
Idea: If $G$ is convex then locally at every point $phi$ is convex, which implies Lipschitz.
– Jose27
Aug 29 '17 at 21:30
Jose27, what does it mean to be convex for $phi$, when it is only defined on the sphere ?
– TrivialPursuit
Aug 29 '17 at 21:32
Jose27, what does it mean to be convex for $phi$, when it is only defined on the sphere ?
– TrivialPursuit
Aug 29 '17 at 21:32
You should sarch the web with the term "support function" which gives you access to a rather vast amount of litterature, for example the very accessible article "Support Function Representation of Convex Bodies, Its Application in Geometric Computing, and Some Related Representations" by Pijush K. Ghosh and K. Vinod Kumar (Comp. Vision and Image Understanding Vol 72, no 3, 1998)
– Jean Marie
Aug 29 '17 at 21:37
You should sarch the web with the term "support function" which gives you access to a rather vast amount of litterature, for example the very accessible article "Support Function Representation of Convex Bodies, Its Application in Geometric Computing, and Some Related Representations" by Pijush K. Ghosh and K. Vinod Kumar (Comp. Vision and Image Understanding Vol 72, no 3, 1998)
– Jean Marie
Aug 29 '17 at 21:37
|
show 2 more comments
1 Answer
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In general no, take $G=[-1, 1]$ and $d=2$ then function $phi$ is neither continuous. If $0$ isn't an interior point you can consider on $mathbb R^2$ the closed square $G$ with vertices $(0, 1)$, $(2, 1)$, $(2, -1)$ and $(0, -1)$ and $phi$ isn't continuous on $mathbb S^1$.
You need to suppose also that $G$ is bounded and $0$ is an interior point of $G$, so exist $M, r>0$ such that $B_rsubseteq Gsubseteq B_M$ so the function $v:mathbb{R}^drightarrow [0, +infty[$ such that $v(0)=0$ and for every $xneq 0$
$$
v(x)=minleft{t>0:frac{x}{t}in Gright}
$$
that's well defined due to $B_rsubseteq G$.
Observe that $v(x)geq frac{lVert xrVert}{M}$ for every $x$ and $v(kx)=kv(x)$ for every $kgeq 0$. We prove now that $v$ is convex, let $x, yinmathbb{R}^dsetminus{0}$ such that $v(x)=a, v(y)=b$ and $lambdain [0, 1]$.
$$
frac{lambda x +(1-lambda)y}{lambda a +(1-lambda)b}=frac{lambda x}{lambda a +(1-lambda)b}+frac{(1-lambda)y}{lambda a +(1-lambda)b}=frac{lambda a}{lambda a +(1-lambda)b}frac{x}{a}+frac{(1-lambda)b}{lambda a +(1-lambda)b}frac{y}{b}in G
$$
because $frac xa, frac ybin G$ by definition, so
$$
vleft[lambda x +(1-lambda)yright]leqlambda a +(1-lambda)b=lambda v(x) +(1-lambda)v(y)
$$
Then exists $L>0$ such that for every $x, yinmathbb S^{d-1}$
$$
lvert v(x)-v(y)rvertleq LlVert x-yrVert
$$
Observe that if $xinmathbb S^{n-1}$ then $phi(x)=frac{1}{v(x)}$ so we can conclude with
$$
lvert phi(x)-phi(y)rvert=frac{lvert v(x)-v(y)rvert}{v(x)v(y)}leq LM^2lVert x-yrVert
$$
answered yesterday
P De Donato
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
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up vote
2
down vote
In general no, take $G=[-1, 1]$ and $d=2$ then function $phi$ is neither continuous. If $0$ isn't an interior point you can consider on $mathbb R^2$ the closed square $G$ with vertices $(0, 1)$, $(2, 1)$, $(2, -1)$ and $(0, -1)$ and $phi$ isn't continuous on $mathbb S^1$.
You need to suppose also that $G$ is bounded and $0$ is an interior point of $G$, so exist $M, r>0$ such that $B_rsubseteq Gsubseteq B_M$ so the function $v:mathbb{R}^drightarrow [0, +infty[$ such that $v(0)=0$ and for every $xneq 0$
$$
v(x)=minleft{t>0:frac{x}{t}in Gright}
$$
that's well defined due to $B_rsubseteq G$.
Observe that $v(x)geq frac{lVert xrVert}{M}$ for every $x$ and $v(kx)=kv(x)$ for every $kgeq 0$. We prove now that $v$ is convex, let $x, yinmathbb{R}^dsetminus{0}$ such that $v(x)=a, v(y)=b$ and $lambdain [0, 1]$.
$$
frac{lambda x +(1-lambda)y}{lambda a +(1-lambda)b}=frac{lambda x}{lambda a +(1-lambda)b}+frac{(1-lambda)y}{lambda a +(1-lambda)b}=frac{lambda a}{lambda a +(1-lambda)b}frac{x}{a}+frac{(1-lambda)b}{lambda a +(1-lambda)b}frac{y}{b}in G
$$
because $frac xa, frac ybin G$ by definition, so
$$
vleft[lambda x +(1-lambda)yright]leqlambda a +(1-lambda)b=lambda v(x) +(1-lambda)v(y)
$$
Then exists $L>0$ such that for every $x, yinmathbb S^{d-1}$
$$
lvert v(x)-v(y)rvertleq LlVert x-yrVert
$$
Observe that if $xinmathbb S^{n-1}$ then $phi(x)=frac{1}{v(x)}$ so we can conclude with
$$
lvert phi(x)-phi(y)rvert=frac{lvert v(x)-v(y)rvert}{v(x)v(y)}leq LM^2lVert x-yrVert
$$
answered yesterday
P De Donato
864
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
2
down vote
In general no, take $G=[-1, 1]$ and $d=2$ then function $phi$ is neither continuous. If $0$ isn't an interior point you can consider on $mathbb R^2$ the closed square $G$ with vertices $(0, 1)$, $(2, 1)$, $(2, -1)$ and $(0, -1)$ and $phi$ isn't continuous on $mathbb S^1$.
You need to suppose also that $G$ is bounded and $0$ is an interior point of $G$, so exist $M, r>0$ such that $B_rsubseteq Gsubseteq B_M$ so the function $v:mathbb{R}^drightarrow [0, +infty[$ such that $v(0)=0$ and for every $xneq 0$
$$
v(x)=minleft{t>0:frac{x}{t}in Gright}
$$
that's well defined due to $B_rsubseteq G$.
Observe that $v(x)geq frac{lVert xrVert}{M}$ for every $x$ and $v(kx)=kv(x)$ for every $kgeq 0$. We prove now that $v$ is convex, let $x, yinmathbb{R}^dsetminus{0}$ such that $v(x)=a, v(y)=b$ and $lambdain [0, 1]$.
$$
frac{lambda x +(1-lambda)y}{lambda a +(1-lambda)b}=frac{lambda x}{lambda a +(1-lambda)b}+frac{(1-lambda)y}{lambda a +(1-lambda)b}=frac{lambda a}{lambda a +(1-lambda)b}frac{x}{a}+frac{(1-lambda)b}{lambda a +(1-lambda)b}frac{y}{b}in G
$$
because $frac xa, frac ybin G$ by definition, so
$$
vleft[lambda x +(1-lambda)yright]leqlambda a +(1-lambda)b=lambda v(x) +(1-lambda)v(y)
$$
Then exists $L>0$ such that for every $x, yinmathbb S^{d-1}$
$$
lvert v(x)-v(y)rvertleq LlVert x-yrVert
$$
Observe that if $xinmathbb S^{n-1}$ then $phi(x)=frac{1}{v(x)}$ so we can conclude with
$$
lvert phi(x)-phi(y)rvert=frac{lvert v(x)-v(y)rvert}{v(x)v(y)}leq LM^2lVert x-yrVert
$$
answered yesterday
P De Donato
864
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
2
down vote
up vote
2
down vote
In general no, take $G=[-1, 1]$ and $d=2$ then function $phi$ is neither continuous. If $0$ isn't an interior point you can consider on $mathbb R^2$ the closed square $G$ with vertices $(0, 1)$, $(2, 1)$, $(2, -1)$ and $(0, -1)$ and $phi$ isn't continuous on $mathbb S^1$.
You need to suppose also that $G$ is bounded and $0$ is an interior point of $G$, so exist $M, r>0$ such that $B_rsubseteq Gsubseteq B_M$ so the function $v:mathbb{R}^drightarrow [0, +infty[$ such that $v(0)=0$ and for every $xneq 0$
$$
v(x)=minleft{t>0:frac{x}{t}in Gright}
$$
that's well defined due to $B_rsubseteq G$.
Observe that $v(x)geq frac{lVert xrVert}{M}$ for every $x$ and $v(kx)=kv(x)$ for every $kgeq 0$. We prove now that $v$ is convex, let $x, yinmathbb{R}^dsetminus{0}$ such that $v(x)=a, v(y)=b$ and $lambdain [0, 1]$.
$$
frac{lambda x +(1-lambda)y}{lambda a +(1-lambda)b}=frac{lambda x}{lambda a +(1-lambda)b}+frac{(1-lambda)y}{lambda a +(1-lambda)b}=frac{lambda a}{lambda a +(1-lambda)b}frac{x}{a}+frac{(1-lambda)b}{lambda a +(1-lambda)b}frac{y}{b}in G
$$
because $frac xa, frac ybin G$ by definition, so
$$
vleft[lambda x +(1-lambda)yright]leqlambda a +(1-lambda)b=lambda v(x) +(1-lambda)v(y)
$$
Then exists $L>0$ such that for every $x, yinmathbb S^{d-1}$
$$
lvert v(x)-v(y)rvertleq LlVert x-yrVert
$$
Observe that if $xinmathbb S^{n-1}$ then $phi(x)=frac{1}{v(x)}$ so we can conclude with
$$
lvert phi(x)-phi(y)rvert=frac{lvert v(x)-v(y)rvert}{v(x)v(y)}leq LM^2lVert x-yrVert
$$
answered yesterday
P De Donato
864
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In general no, take $G=[-1, 1]$ and $d=2$ then function $phi$ is neither continuous. If $0$ isn't an interior point you can consider on $mathbb R^2$ the closed square $G$ with vertices $(0, 1)$, $(2, 1)$, $(2, -1)$ and $(0, -1)$ and $phi$ isn't continuous on $mathbb S^1$.
You need to suppose also that $G$ is bounded and $0$ is an interior point of $G$, so exist $M, r>0$ such that $B_rsubseteq Gsubseteq B_M$ so the function $v:mathbb{R}^drightarrow [0, +infty[$ such that $v(0)=0$ and for every $xneq 0$
$$
v(x)=minleft{t>0:frac{x}{t}in Gright}
$$
that's well defined due to $B_rsubseteq G$.
Observe that $v(x)geq frac{lVert xrVert}{M}$ for every $x$ and $v(kx)=kv(x)$ for every $kgeq 0$. We prove now that $v$ is convex, let $x, yinmathbb{R}^dsetminus{0}$ such that $v(x)=a, v(y)=b$ and $lambdain [0, 1]$.
$$
frac{lambda x +(1-lambda)y}{lambda a +(1-lambda)b}=frac{lambda x}{lambda a +(1-lambda)b}+frac{(1-lambda)y}{lambda a +(1-lambda)b}=frac{lambda a}{lambda a +(1-lambda)b}frac{x}{a}+frac{(1-lambda)b}{lambda a +(1-lambda)b}frac{y}{b}in G
$$
because $frac xa, frac ybin G$ by definition, so
$$
vleft[lambda x +(1-lambda)yright]leqlambda a +(1-lambda)b=lambda v(x) +(1-lambda)v(y)
$$
Then exists $L>0$ such that for every $x, yinmathbb S^{d-1}$
$$
lvert v(x)-v(y)rvertleq LlVert x-yrVert
$$
Observe that if $xinmathbb S^{n-1}$ then $phi(x)=frac{1}{v(x)}$ so we can conclude with
$$
lvert phi(x)-phi(y)rvert=frac{lvert v(x)-v(y)rvert}{v(x)v(y)}leq LM^2lVert x-yrVert
$$
answered yesterday
P De Donato
864
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
answered yesterday
P De Donato
864
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
P De Donato
864
answered yesterday
P De Donato
864
864
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$k$-Lipschitz with which constant $k$ ? $k=1$ ?
– Jean Marie
Aug 29 '17 at 21:14
$k$-Lipschitz with some $k$.
– TrivialPursuit
Aug 29 '17 at 21:28
Idea: If $G$ is convex then locally at every point $phi$ is convex, which implies Lipschitz.
– Jose27
Aug 29 '17 at 21:30
Jose27, what does it mean to be convex for $phi$, when it is only defined on the sphere ?
– TrivialPursuit
Aug 29 '17 at 21:32
You should sarch the web with the term "support function" which gives you access to a rather vast amount of litterature, for example the very accessible article "Support Function Representation of Convex Bodies, Its Application in Geometric Computing, and Some Related Representations" by Pijush K. Ghosh and K. Vinod Kumar (Comp. Vision and Image Understanding Vol 72, no 3, 1998)
– Jean Marie
Aug 29 '17 at 21:37