Mixing
Proof that for every bounded open subset of a $n$-dimensional euclidean space, there exists a point in such...
$begingroup$
If $X=Bbb R^n$ is equipped with the usual topology $tau$ and the usual metric $d$, and we define the function $pi_U:Xto X$ for every bounded $Uintau $ as
$$pi_U(p)=int_U d(x,p)dx$$
I need to prove that for every bounded $Uintau$ there exists a $pin X$ such that for all $qin X$ it holds that $pi_U(p)<pi_U(q)$ i.e. that $pi_U$ has a unique point that attains its minimum value. I need to reach to a contradiction when defining two points with said property, but cannot find any, any help would be appreciated.
An example would be $pi_X$ where $X$ is the 2d disk or radius $r$(the set of points such that its norm is less than $r$) centered at $(0,0)$, then $(0,0)$ is a point with said property. Same with a rectangle of length $2r$ centered at the origin.
calculus metric-spaces
asked Jan 13 at 16:27
GarmekainGarmekain
1,386720
$endgroup$
|
show 10 more comments
$begingroup$
If $X=Bbb R^n$ is equipped with the usual topology $tau$ and the usual metric $d$, and we define the function $pi_U:Xto X$ for every bounded $Uintau $ as
$$pi_U(p)=int_U d(x,p)dx$$
I need to prove that for every bounded $Uintau$ there exists a $pin X$ such that for all $qin X$ it holds that $pi_U(p)<pi_U(q)$ i.e. that $pi_U$ has a unique point that attains its minimum value. I need to reach to a contradiction when defining two points with said property, but cannot find any, any help would be appreciated.
An example would be $pi_X$ where $X$ is the 2d disk or radius $r$(the set of points such that its norm is less than $r$) centered at $(0,0)$, then $(0,0)$ is a point with said property. Same with a rectangle of length $2r$ centered at the origin.
calculus metric-spaces
asked Jan 13 at 16:27
GarmekainGarmekain
1,386720
$endgroup$
$begingroup$
Are you integrating with respect to the Lebesgue measure? If that's the case then wouldn't we have $pi_K(p)=0$ for any $pinBbb R^n$ if $K$ is a finite set?
$endgroup$
– BigbearZzz
Jan 13 at 23:59
$begingroup$
@BigbearZzz But a finite set is not an open set in the usual topology.
$endgroup$
– Garmekain
Jan 14 at 0:05
$begingroup$
But didn't you specify that $U$ is compact? How can $U$ be both compact and open?
$endgroup$
– BigbearZzz
Jan 14 at 0:07
$begingroup$
@BigbearZzz $U$ should be closed, sorry.
$endgroup$
– Garmekain
Jan 14 at 0:08
1
$begingroup$
This is much easier if you integrate the square of the distance function.
$endgroup$
– Moishe Cohen
Jan 14 at 0:35
|
show 10 more comments
$begingroup$
If $X=Bbb R^n$ is equipped with the usual topology $tau$ and the usual metric $d$, and we define the function $pi_U:Xto X$ for every bounded $Uintau $ as
$$pi_U(p)=int_U d(x,p)dx$$
I need to prove that for every bounded $Uintau$ there exists a $pin X$ such that for all $qin X$ it holds that $pi_U(p)<pi_U(q)$ i.e. that $pi_U$ has a unique point that attains its minimum value. I need to reach to a contradiction when defining two points with said property, but cannot find any, any help would be appreciated.
An example would be $pi_X$ where $X$ is the 2d disk or radius $r$(the set of points such that its norm is less than $r$) centered at $(0,0)$, then $(0,0)$ is a point with said property. Same with a rectangle of length $2r$ centered at the origin.
calculus metric-spaces
asked Jan 13 at 16:27
GarmekainGarmekain
1,386720
$endgroup$
If $X=Bbb R^n$ is equipped with the usual topology $tau$ and the usual metric $d$, and we define the function $pi_U:Xto X$ for every bounded $Uintau $ as
$$pi_U(p)=int_U d(x,p)dx$$
I need to prove that for every bounded $Uintau$ there exists a $pin X$ such that for all $qin X$ it holds that $pi_U(p)<pi_U(q)$ i.e. that $pi_U$ has a unique point that attains its minimum value. I need to reach to a contradiction when defining two points with said property, but cannot find any, any help would be appreciated.
An example would be $pi_X$ where $X$ is the 2d disk or radius $r$(the set of points such that its norm is less than $r$) centered at $(0,0)$, then $(0,0)$ is a point with said property. Same with a rectangle of length $2r$ centered at the origin.
calculus metric-spaces
calculus metric-spaces
asked Jan 13 at 16:27
GarmekainGarmekain
1,386720
asked Jan 13 at 16:27
GarmekainGarmekain
1,386720
edited Jan 15 at 0:58
Garmekain
asked Jan 13 at 16:27
GarmekainGarmekain
1,386720
asked Jan 13 at 16:27
GarmekainGarmekain
1,386720
asked Jan 13 at 16:27
GarmekainGarmekain
1,386720
1,386720
$begingroup$
Are you integrating with respect to the Lebesgue measure? If that's the case then wouldn't we have $pi_K(p)=0$ for any $pinBbb R^n$ if $K$ is a finite set?
$endgroup$
– BigbearZzz
Jan 13 at 23:59
$begingroup$
@BigbearZzz But a finite set is not an open set in the usual topology.
$endgroup$
– Garmekain
Jan 14 at 0:05
$begingroup$
But didn't you specify that $U$ is compact? How can $U$ be both compact and open?
$endgroup$
– BigbearZzz
Jan 14 at 0:07
$begingroup$
@BigbearZzz $U$ should be closed, sorry.
$endgroup$
– Garmekain
Jan 14 at 0:08
1
$begingroup$
This is much easier if you integrate the square of the distance function.
$endgroup$
– Moishe Cohen
Jan 14 at 0:35
|
show 10 more comments
$begingroup$
Are you integrating with respect to the Lebesgue measure? If that's the case then wouldn't we have $pi_K(p)=0$ for any $pinBbb R^n$ if $K$ is a finite set?
$endgroup$
– BigbearZzz
Jan 13 at 23:59
$begingroup$
@BigbearZzz But a finite set is not an open set in the usual topology.
$endgroup$
– Garmekain
Jan 14 at 0:05
$begingroup$
But didn't you specify that $U$ is compact? How can $U$ be both compact and open?
$endgroup$
– BigbearZzz
Jan 14 at 0:07
$begingroup$
@BigbearZzz $U$ should be closed, sorry.
$endgroup$
– Garmekain
Jan 14 at 0:08
1
$begingroup$
This is much easier if you integrate the square of the distance function.
$endgroup$
– Moishe Cohen
Jan 14 at 0:35
$begingroup$
Are you integrating with respect to the Lebesgue measure? If that's the case then wouldn't we have $pi_K(p)=0$ for any $pinBbb R^n$ if $K$ is a finite set?
$endgroup$
– BigbearZzz
Jan 13 at 23:59
$begingroup$
Are you integrating with respect to the Lebesgue measure? If that's the case then wouldn't we have $pi_K(p)=0$ for any $pinBbb R^n$ if $K$ is a finite set?
$endgroup$
– BigbearZzz
Jan 13 at 23:59
$begingroup$
@BigbearZzz But a finite set is not an open set in the usual topology.
$endgroup$
– Garmekain
Jan 14 at 0:05
$begingroup$
@BigbearZzz But a finite set is not an open set in the usual topology.
$endgroup$
– Garmekain
Jan 14 at 0:05
$begingroup$
But didn't you specify that $U$ is compact? How can $U$ be both compact and open?
$endgroup$
– BigbearZzz
Jan 14 at 0:07
$begingroup$
But didn't you specify that $U$ is compact? How can $U$ be both compact and open?
$endgroup$
– BigbearZzz
Jan 14 at 0:07
$begingroup$
@BigbearZzz $U$ should be closed, sorry.
$endgroup$
– Garmekain
Jan 14 at 0:08
$begingroup$
@BigbearZzz $U$ should be closed, sorry.
$endgroup$
– Garmekain
Jan 14 at 0:08
1
1
$begingroup$
This is much easier if you integrate the square of the distance function.
$endgroup$
– Moishe Cohen
Jan 14 at 0:35
$begingroup$
This is much easier if you integrate the square of the distance function.
$endgroup$
– Moishe Cohen
Jan 14 at 0:35
|
show 10 more comments
1 Answer
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$begingroup$
This is not true in the general case. Suppose $n=1$ and $U=(-3,-1)cup(1,3)$, clearly $U$ is an open bounded subset of $Bbb R$. Then every value $-1<y<1$ of $int_Ud(x,y)dx$ will yield the minimum value of the function, $8$, so $pi_{(-3,-1)cup(1,3)}$ has not a unique value that attains its minimum value, there are actually uncountable many of them.
answered Jan 17 at 13:35
GarmekainGarmekain
1,386720
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is not true in the general case. Suppose $n=1$ and $U=(-3,-1)cup(1,3)$, clearly $U$ is an open bounded subset of $Bbb R$. Then every value $-1<y<1$ of $int_Ud(x,y)dx$ will yield the minimum value of the function, $8$, so $pi_{(-3,-1)cup(1,3)}$ has not a unique value that attains its minimum value, there are actually uncountable many of them.
answered Jan 17 at 13:35
GarmekainGarmekain
1,386720
$endgroup$
add a comment |
$begingroup$
This is not true in the general case. Suppose $n=1$ and $U=(-3,-1)cup(1,3)$, clearly $U$ is an open bounded subset of $Bbb R$. Then every value $-1<y<1$ of $int_Ud(x,y)dx$ will yield the minimum value of the function, $8$, so $pi_{(-3,-1)cup(1,3)}$ has not a unique value that attains its minimum value, there are actually uncountable many of them.
answered Jan 17 at 13:35
GarmekainGarmekain
1,386720
$endgroup$
add a comment |
$begingroup$
This is not true in the general case. Suppose $n=1$ and $U=(-3,-1)cup(1,3)$, clearly $U$ is an open bounded subset of $Bbb R$. Then every value $-1<y<1$ of $int_Ud(x,y)dx$ will yield the minimum value of the function, $8$, so $pi_{(-3,-1)cup(1,3)}$ has not a unique value that attains its minimum value, there are actually uncountable many of them.
answered Jan 17 at 13:35
GarmekainGarmekain
1,386720
$endgroup$
This is not true in the general case. Suppose $n=1$ and $U=(-3,-1)cup(1,3)$, clearly $U$ is an open bounded subset of $Bbb R$. Then every value $-1<y<1$ of $int_Ud(x,y)dx$ will yield the minimum value of the function, $8$, so $pi_{(-3,-1)cup(1,3)}$ has not a unique value that attains its minimum value, there are actually uncountable many of them.
answered Jan 17 at 13:35
GarmekainGarmekain
1,386720
answered Jan 17 at 13:35
GarmekainGarmekain
1,386720
answered Jan 17 at 13:35
GarmekainGarmekain
1,386720
answered Jan 17 at 13:35
GarmekainGarmekain
1,386720
1,386720
add a comment |
add a comment |
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$begingroup$
Are you integrating with respect to the Lebesgue measure? If that's the case then wouldn't we have $pi_K(p)=0$ for any $pinBbb R^n$ if $K$ is a finite set?
$endgroup$
– BigbearZzz
Jan 13 at 23:59
$begingroup$
@BigbearZzz But a finite set is not an open set in the usual topology.
$endgroup$
– Garmekain
Jan 14 at 0:05
$begingroup$
But didn't you specify that $U$ is compact? How can $U$ be both compact and open?
$endgroup$
– BigbearZzz
Jan 14 at 0:07
$begingroup$
@BigbearZzz $U$ should be closed, sorry.
$endgroup$
– Garmekain
Jan 14 at 0:08
1
$begingroup$
This is much easier if you integrate the square of the distance function.
$endgroup$
– Moishe Cohen
Jan 14 at 0:35