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A convex subset of a set has 'smaller' boundary than the set?
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Let $Asubseteq B$ be subsets of the real plane. Show that if $A$ is convex and $B$ is bounded, then the length of the border of $A$ $leq$ the length of the border of $B$.
real-analysis geometry
edited Feb 2 at 23:41
user26857
39.5k124284
asked Dec 20 '12 at 10:16
SgernestoSgernesto
173110
$endgroup$
add a comment |
$begingroup$
Let $Asubseteq B$ be subsets of the real plane. Show that if $A$ is convex and $B$ is bounded, then the length of the border of $A$ $leq$ the length of the border of $B$.
real-analysis geometry
edited Feb 2 at 23:41
user26857
39.5k124284
asked Dec 20 '12 at 10:16
SgernestoSgernesto
173110
$endgroup$
add a comment |
$begingroup$
Let $Asubseteq B$ be subsets of the real plane. Show that if $A$ is convex and $B$ is bounded, then the length of the border of $A$ $leq$ the length of the border of $B$.
real-analysis geometry
edited Feb 2 at 23:41
user26857
39.5k124284
asked Dec 20 '12 at 10:16
SgernestoSgernesto
173110
$endgroup$
Let $Asubseteq B$ be subsets of the real plane. Show that if $A$ is convex and $B$ is bounded, then the length of the border of $A$ $leq$ the length of the border of $B$.
real-analysis geometry
real-analysis geometry
edited Feb 2 at 23:41
user26857
39.5k124284
asked Dec 20 '12 at 10:16
SgernestoSgernesto
173110
edited Feb 2 at 23:41
user26857
39.5k124284
asked Dec 20 '12 at 10:16
SgernestoSgernesto
173110
edited Feb 2 at 23:41
user26857
39.5k124284
edited Feb 2 at 23:41
user26857
39.5k124284
edited Feb 2 at 23:41
user26857
39.5k124284
39.5k124284
asked Dec 20 '12 at 10:16
SgernestoSgernesto
173110
asked Dec 20 '12 at 10:16
SgernestoSgernesto
173110
asked Dec 20 '12 at 10:16
SgernestoSgernesto
173110
173110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Given any convex closed subset $A$ of a Euclidean space $mathbb R^n$, we can define the nearest-point projection $p_Acolonmathbb R^nto A$ by sending each point $xinmathbb R^n$ into $yin A$ that attains the minimum $min_{yin A}|x-y|$. Due to the convexity of $A$, $p_A$ is well-defined. This concept generalizes the orthogonal projection onto a linear subspace.
A notable property of orthogonal projections is shared by $p_A$: it is a contraction in the sense that $|p_A(x)-p_A(y)|le |x-y|$ for all $x,yin mathbb R^n$.
It remains to show that $p_A$ maps $partial B$ onto $partial A$. The claim then follows from the fact that contractions do not increase length (or area, etc).
The proofs of 2 and 3 are easier to carry out yourself than to watch/read someone else do it.
$endgroup$
add a comment |
$begingroup$
First, show that for a general half-space $H$, $Bcap H$ has smaller boundary than $B$.
If $A$ is sufficiently smooth, you can write $A$ as the intersection of $B$ with a countable family of half-spaces, say
$$A=Bcap(bigcap_{k=1}^infty H_k).$$
From this (and lower semicontinuity of the perimeter functional, see any text of sets of finite perimeter), it follows that $A$ has smaller boundary than $B$.
edited Feb 2 at 23:16
user26857
39.5k124284
answered Dec 20 '12 at 10:30
MartijnMartijn
301129
$endgroup$
$begingroup$
@Martjin. I know an elementary argument to show that the perimeter of a convex poligon 'inside' a simple closed curve is smaller than the length of the curve. I think this argument could be passed to the limit to show it for an arbitrary convex curve.
$endgroup$
– Sgernesto
Dec 20 '12 at 12:43
$begingroup$
In any case I think that an analogous statement is true in any dimension.
$endgroup$
– Sgernesto
Dec 20 '12 at 12:44
$begingroup$
@Sgernesto: probably, this is the same argument
$endgroup$
– Martijn
Dec 20 '12 at 15:32
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
Given any convex closed subset $A$ of a Euclidean space $mathbb R^n$, we can define the nearest-point projection $p_Acolonmathbb R^nto A$ by sending each point $xinmathbb R^n$ into $yin A$ that attains the minimum $min_{yin A}|x-y|$. Due to the convexity of $A$, $p_A$ is well-defined. This concept generalizes the orthogonal projection onto a linear subspace.
A notable property of orthogonal projections is shared by $p_A$: it is a contraction in the sense that $|p_A(x)-p_A(y)|le |x-y|$ for all $x,yin mathbb R^n$.
It remains to show that $p_A$ maps $partial B$ onto $partial A$. The claim then follows from the fact that contractions do not increase length (or area, etc).
The proofs of 2 and 3 are easier to carry out yourself than to watch/read someone else do it.
$endgroup$
add a comment |
$begingroup$
Given any convex closed subset $A$ of a Euclidean space $mathbb R^n$, we can define the nearest-point projection $p_Acolonmathbb R^nto A$ by sending each point $xinmathbb R^n$ into $yin A$ that attains the minimum $min_{yin A}|x-y|$. Due to the convexity of $A$, $p_A$ is well-defined. This concept generalizes the orthogonal projection onto a linear subspace.
A notable property of orthogonal projections is shared by $p_A$: it is a contraction in the sense that $|p_A(x)-p_A(y)|le |x-y|$ for all $x,yin mathbb R^n$.
It remains to show that $p_A$ maps $partial B$ onto $partial A$. The claim then follows from the fact that contractions do not increase length (or area, etc).
The proofs of 2 and 3 are easier to carry out yourself than to watch/read someone else do it.
$endgroup$
add a comment |
$begingroup$
Given any convex closed subset $A$ of a Euclidean space $mathbb R^n$, we can define the nearest-point projection $p_Acolonmathbb R^nto A$ by sending each point $xinmathbb R^n$ into $yin A$ that attains the minimum $min_{yin A}|x-y|$. Due to the convexity of $A$, $p_A$ is well-defined. This concept generalizes the orthogonal projection onto a linear subspace.
A notable property of orthogonal projections is shared by $p_A$: it is a contraction in the sense that $|p_A(x)-p_A(y)|le |x-y|$ for all $x,yin mathbb R^n$.
It remains to show that $p_A$ maps $partial B$ onto $partial A$. The claim then follows from the fact that contractions do not increase length (or area, etc).
The proofs of 2 and 3 are easier to carry out yourself than to watch/read someone else do it.
$endgroup$
Given any convex closed subset $A$ of a Euclidean space $mathbb R^n$, we can define the nearest-point projection $p_Acolonmathbb R^nto A$ by sending each point $xinmathbb R^n$ into $yin A$ that attains the minimum $min_{yin A}|x-y|$. Due to the convexity of $A$, $p_A$ is well-defined. This concept generalizes the orthogonal projection onto a linear subspace.
A notable property of orthogonal projections is shared by $p_A$: it is a contraction in the sense that $|p_A(x)-p_A(y)|le |x-y|$ for all $x,yin mathbb R^n$.
It remains to show that $p_A$ maps $partial B$ onto $partial A$. The claim then follows from the fact that contractions do not increase length (or area, etc).
The proofs of 2 and 3 are easier to carry out yourself than to watch/read someone else do it.
answered Dec 20 '12 at 17:44
user53153
add a comment |
add a comment |
$begingroup$
First, show that for a general half-space $H$, $Bcap H$ has smaller boundary than $B$.
If $A$ is sufficiently smooth, you can write $A$ as the intersection of $B$ with a countable family of half-spaces, say
$$A=Bcap(bigcap_{k=1}^infty H_k).$$
From this (and lower semicontinuity of the perimeter functional, see any text of sets of finite perimeter), it follows that $A$ has smaller boundary than $B$.
edited Feb 2 at 23:16
user26857
39.5k124284
answered Dec 20 '12 at 10:30
MartijnMartijn
301129
$endgroup$
$begingroup$
@Martjin. I know an elementary argument to show that the perimeter of a convex poligon 'inside' a simple closed curve is smaller than the length of the curve. I think this argument could be passed to the limit to show it for an arbitrary convex curve.
$endgroup$
– Sgernesto
Dec 20 '12 at 12:43
$begingroup$
In any case I think that an analogous statement is true in any dimension.
$endgroup$
– Sgernesto
Dec 20 '12 at 12:44
$begingroup$
@Sgernesto: probably, this is the same argument
$endgroup$
– Martijn
Dec 20 '12 at 15:32
add a comment |
$begingroup$
First, show that for a general half-space $H$, $Bcap H$ has smaller boundary than $B$.
If $A$ is sufficiently smooth, you can write $A$ as the intersection of $B$ with a countable family of half-spaces, say
$$A=Bcap(bigcap_{k=1}^infty H_k).$$
From this (and lower semicontinuity of the perimeter functional, see any text of sets of finite perimeter), it follows that $A$ has smaller boundary than $B$.
edited Feb 2 at 23:16
user26857
39.5k124284
answered Dec 20 '12 at 10:30
MartijnMartijn
301129
$endgroup$
$begingroup$
@Martjin. I know an elementary argument to show that the perimeter of a convex poligon 'inside' a simple closed curve is smaller than the length of the curve. I think this argument could be passed to the limit to show it for an arbitrary convex curve.
$endgroup$
– Sgernesto
Dec 20 '12 at 12:43
$begingroup$
In any case I think that an analogous statement is true in any dimension.
$endgroup$
– Sgernesto
Dec 20 '12 at 12:44
$begingroup$
@Sgernesto: probably, this is the same argument
$endgroup$
– Martijn
Dec 20 '12 at 15:32
add a comment |
$begingroup$
First, show that for a general half-space $H$, $Bcap H$ has smaller boundary than $B$.
If $A$ is sufficiently smooth, you can write $A$ as the intersection of $B$ with a countable family of half-spaces, say
$$A=Bcap(bigcap_{k=1}^infty H_k).$$
From this (and lower semicontinuity of the perimeter functional, see any text of sets of finite perimeter), it follows that $A$ has smaller boundary than $B$.
edited Feb 2 at 23:16
user26857
39.5k124284
answered Dec 20 '12 at 10:30
MartijnMartijn
301129
$endgroup$
First, show that for a general half-space $H$, $Bcap H$ has smaller boundary than $B$.
If $A$ is sufficiently smooth, you can write $A$ as the intersection of $B$ with a countable family of half-spaces, say
$$A=Bcap(bigcap_{k=1}^infty H_k).$$
From this (and lower semicontinuity of the perimeter functional, see any text of sets of finite perimeter), it follows that $A$ has smaller boundary than $B$.
edited Feb 2 at 23:16
user26857
39.5k124284
answered Dec 20 '12 at 10:30
MartijnMartijn
301129
edited Feb 2 at 23:16
user26857
39.5k124284
edited Feb 2 at 23:16
user26857
39.5k124284
edited Feb 2 at 23:16
user26857
39.5k124284
39.5k124284
answered Dec 20 '12 at 10:30
MartijnMartijn
301129
answered Dec 20 '12 at 10:30
MartijnMartijn
301129
answered Dec 20 '12 at 10:30
MartijnMartijn
301129
301129
$begingroup$
@Martjin. I know an elementary argument to show that the perimeter of a convex poligon 'inside' a simple closed curve is smaller than the length of the curve. I think this argument could be passed to the limit to show it for an arbitrary convex curve.
$endgroup$
– Sgernesto
Dec 20 '12 at 12:43
$begingroup$
In any case I think that an analogous statement is true in any dimension.
$endgroup$
– Sgernesto
Dec 20 '12 at 12:44
$begingroup$
@Sgernesto: probably, this is the same argument
$endgroup$
– Martijn
Dec 20 '12 at 15:32
add a comment |
$begingroup$
@Martjin. I know an elementary argument to show that the perimeter of a convex poligon 'inside' a simple closed curve is smaller than the length of the curve. I think this argument could be passed to the limit to show it for an arbitrary convex curve.
$endgroup$
– Sgernesto
Dec 20 '12 at 12:43
$begingroup$
In any case I think that an analogous statement is true in any dimension.
$endgroup$
– Sgernesto
Dec 20 '12 at 12:44
$begingroup$
@Sgernesto: probably, this is the same argument
$endgroup$
– Martijn
Dec 20 '12 at 15:32
$begingroup$
@Martjin. I know an elementary argument to show that the perimeter of a convex poligon 'inside' a simple closed curve is smaller than the length of the curve. I think this argument could be passed to the limit to show it for an arbitrary convex curve.
$endgroup$
– Sgernesto
Dec 20 '12 at 12:43
$begingroup$
@Martjin. I know an elementary argument to show that the perimeter of a convex poligon 'inside' a simple closed curve is smaller than the length of the curve. I think this argument could be passed to the limit to show it for an arbitrary convex curve.
$endgroup$
– Sgernesto
Dec 20 '12 at 12:43
$begingroup$
In any case I think that an analogous statement is true in any dimension.
$endgroup$
– Sgernesto
Dec 20 '12 at 12:44
$begingroup$
In any case I think that an analogous statement is true in any dimension.
$endgroup$
– Sgernesto
Dec 20 '12 at 12:44
$begingroup$
@Sgernesto: probably, this is the same argument
$endgroup$
– Martijn
Dec 20 '12 at 15:32
$begingroup$
@Sgernesto: probably, this is the same argument
$endgroup$
– Martijn
Dec 20 '12 at 15:32
add a comment |
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