Set invariant under the map $x mapsto -x$ that is a translation of symmetric sets
$begingroup$
Let $F subset mathbb{R}^n$ be an open bounded (non empty) convex set and assume that it is invariant under the map $x mapsto -x$ (which means that $F= { x : x in F } = { -x : x in F} = -F$).
Now assume that for any unit vector $nu in mathbb{R}^n$ there exists $d_{nu} in mathbb{R}$ such that $F= d_{nu} nu + E_{nu}$, where $E_{nu}$ is a set symmetric with respect to the hyperplane $nu^{perp}$. This fact means that for any direction $nu$ the set $F$ is the translation in the $nu$ direction of a set that is symmetric with respect to $nu^{perp}$.
I have to prove that $d_{nu}=0$ for each unit vector $nu in mathbb{R}^n$. That is, I have to prove that this set is invariant under reflection with respect to any hyperplane $nu^{perp}$.
MY IDEA: Fix an unit vector $nu in mathbb{R}^n$. Since $F=-F$ we have
$$ d_{nu} nu + E_{nu} = -( d_{nu} nu + E_{nu}) = - d_{nu} nu - E_{nu},$$
hence
$$E_{nu}= - E_{nu} - 2 d_{nu} nu .$$
This equality implies that, given $x in E_{nu}$ there exists $ y in E_{nu}$ such that $ x = -y -2 d_{nu} nu$, i.e. $y = -x - 2 d_{nu} nu.$ Since $ y in E_{nu}$ we have $ -x - 2 d_{nu} nu in E_{nu}$.
Thus for all $x in E_{nu}$ it holds $ -x - 2 d_{nu} nu in E_{nu}$. But now how can I infer that $d_{nu}=0$?
I think I should use the symmetry of $E_{nu}$ but I'm not sure and I don't know how to proceed!
EDIT: I have just noticed that since $F=-F$ and $F$ is convex then $0 in F$. Does this help? Moreover I've found this question: Set in $mathbb R^²$ with an axis of symmetry in every direction and I think that maybe we could extend the proof presented by zuggg in the case of a n-dimensional set (since we are assuming that $F$ is bounded).
Any advice would be really appreciated!
geometry euclidean-geometry symmetry invariance
asked Dec 4 '18 at 20:10
HermioneHermione
19619
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add a comment |
$begingroup$
Let $F subset mathbb{R}^n$ be an open bounded (non empty) convex set and assume that it is invariant under the map $x mapsto -x$ (which means that $F= { x : x in F } = { -x : x in F} = -F$).
Now assume that for any unit vector $nu in mathbb{R}^n$ there exists $d_{nu} in mathbb{R}$ such that $F= d_{nu} nu + E_{nu}$, where $E_{nu}$ is a set symmetric with respect to the hyperplane $nu^{perp}$. This fact means that for any direction $nu$ the set $F$ is the translation in the $nu$ direction of a set that is symmetric with respect to $nu^{perp}$.
I have to prove that $d_{nu}=0$ for each unit vector $nu in mathbb{R}^n$. That is, I have to prove that this set is invariant under reflection with respect to any hyperplane $nu^{perp}$.
MY IDEA: Fix an unit vector $nu in mathbb{R}^n$. Since $F=-F$ we have
$$ d_{nu} nu + E_{nu} = -( d_{nu} nu + E_{nu}) = - d_{nu} nu - E_{nu},$$
hence
$$E_{nu}= - E_{nu} - 2 d_{nu} nu .$$
This equality implies that, given $x in E_{nu}$ there exists $ y in E_{nu}$ such that $ x = -y -2 d_{nu} nu$, i.e. $y = -x - 2 d_{nu} nu.$ Since $ y in E_{nu}$ we have $ -x - 2 d_{nu} nu in E_{nu}$.
Thus for all $x in E_{nu}$ it holds $ -x - 2 d_{nu} nu in E_{nu}$. But now how can I infer that $d_{nu}=0$?
I think I should use the symmetry of $E_{nu}$ but I'm not sure and I don't know how to proceed!
EDIT: I have just noticed that since $F=-F$ and $F$ is convex then $0 in F$. Does this help? Moreover I've found this question: Set in $mathbb R^²$ with an axis of symmetry in every direction and I think that maybe we could extend the proof presented by zuggg in the case of a n-dimensional set (since we are assuming that $F$ is bounded).
Any advice would be really appreciated!
geometry euclidean-geometry symmetry invariance
asked Dec 4 '18 at 20:10
HermioneHermione
19619
$endgroup$
add a comment |
$begingroup$
Let $F subset mathbb{R}^n$ be an open bounded (non empty) convex set and assume that it is invariant under the map $x mapsto -x$ (which means that $F= { x : x in F } = { -x : x in F} = -F$).
Now assume that for any unit vector $nu in mathbb{R}^n$ there exists $d_{nu} in mathbb{R}$ such that $F= d_{nu} nu + E_{nu}$, where $E_{nu}$ is a set symmetric with respect to the hyperplane $nu^{perp}$. This fact means that for any direction $nu$ the set $F$ is the translation in the $nu$ direction of a set that is symmetric with respect to $nu^{perp}$.
I have to prove that $d_{nu}=0$ for each unit vector $nu in mathbb{R}^n$. That is, I have to prove that this set is invariant under reflection with respect to any hyperplane $nu^{perp}$.
MY IDEA: Fix an unit vector $nu in mathbb{R}^n$. Since $F=-F$ we have
$$ d_{nu} nu + E_{nu} = -( d_{nu} nu + E_{nu}) = - d_{nu} nu - E_{nu},$$
hence
$$E_{nu}= - E_{nu} - 2 d_{nu} nu .$$
This equality implies that, given $x in E_{nu}$ there exists $ y in E_{nu}$ such that $ x = -y -2 d_{nu} nu$, i.e. $y = -x - 2 d_{nu} nu.$ Since $ y in E_{nu}$ we have $ -x - 2 d_{nu} nu in E_{nu}$.
Thus for all $x in E_{nu}$ it holds $ -x - 2 d_{nu} nu in E_{nu}$. But now how can I infer that $d_{nu}=0$?
I think I should use the symmetry of $E_{nu}$ but I'm not sure and I don't know how to proceed!
EDIT: I have just noticed that since $F=-F$ and $F$ is convex then $0 in F$. Does this help? Moreover I've found this question: Set in $mathbb R^²$ with an axis of symmetry in every direction and I think that maybe we could extend the proof presented by zuggg in the case of a n-dimensional set (since we are assuming that $F$ is bounded).
Any advice would be really appreciated!
geometry euclidean-geometry symmetry invariance
asked Dec 4 '18 at 20:10
HermioneHermione
19619
$endgroup$
Let $F subset mathbb{R}^n$ be an open bounded (non empty) convex set and assume that it is invariant under the map $x mapsto -x$ (which means that $F= { x : x in F } = { -x : x in F} = -F$).
Now assume that for any unit vector $nu in mathbb{R}^n$ there exists $d_{nu} in mathbb{R}$ such that $F= d_{nu} nu + E_{nu}$, where $E_{nu}$ is a set symmetric with respect to the hyperplane $nu^{perp}$. This fact means that for any direction $nu$ the set $F$ is the translation in the $nu$ direction of a set that is symmetric with respect to $nu^{perp}$.
I have to prove that $d_{nu}=0$ for each unit vector $nu in mathbb{R}^n$. That is, I have to prove that this set is invariant under reflection with respect to any hyperplane $nu^{perp}$.
MY IDEA: Fix an unit vector $nu in mathbb{R}^n$. Since $F=-F$ we have
$$ d_{nu} nu + E_{nu} = -( d_{nu} nu + E_{nu}) = - d_{nu} nu - E_{nu},$$
hence
$$E_{nu}= - E_{nu} - 2 d_{nu} nu .$$
This equality implies that, given $x in E_{nu}$ there exists $ y in E_{nu}$ such that $ x = -y -2 d_{nu} nu$, i.e. $y = -x - 2 d_{nu} nu.$ Since $ y in E_{nu}$ we have $ -x - 2 d_{nu} nu in E_{nu}$.
Thus for all $x in E_{nu}$ it holds $ -x - 2 d_{nu} nu in E_{nu}$. But now how can I infer that $d_{nu}=0$?
I think I should use the symmetry of $E_{nu}$ but I'm not sure and I don't know how to proceed!
EDIT: I have just noticed that since $F=-F$ and $F$ is convex then $0 in F$. Does this help? Moreover I've found this question: Set in $mathbb R^²$ with an axis of symmetry in every direction and I think that maybe we could extend the proof presented by zuggg in the case of a n-dimensional set (since we are assuming that $F$ is bounded).
Any advice would be really appreciated!
geometry euclidean-geometry symmetry invariance
geometry euclidean-geometry symmetry invariance
asked Dec 4 '18 at 20:10
HermioneHermione
19619
asked Dec 4 '18 at 20:10
HermioneHermione
19619
edited Dec 5 '18 at 13:28
Hermione
asked Dec 4 '18 at 20:10
HermioneHermione
19619
asked Dec 4 '18 at 20:10
HermioneHermione
19619
asked Dec 4 '18 at 20:10
HermioneHermione
19619
19619
add a comment |
add a comment |
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