Mixing
If $psi(x)=int_U d(x,y)dy$, proof that there is a unique $xin Bbb R$ such that $psi(x)$ is minimal?
$begingroup$
Let $(Bbb R,d)$ be a metric space and let $psi:Bbb Rto[0,infty)$ be defined as $psi(x)=int_Ud(x,y)dy$ given a subset $Usubseteq Bbb R$ that is a bounded member of the usual topology on $Bbb R$.
I'm trying to prove that there is a unique $xinBbb R$ such that $psi(x)$ is minimum, i.e. for all other $yinBbb R$, we have that $psi(x)<psi(y)$, and not just $psi(x)leqpsi(y)$.
If $psi$ were injective then we are done, but it's not always the case. I need to check injectiveness just for the argument that just gives the minimal, but I don't know to do that. Any help on that or any other methods would be appreciated. Thanks.
functions metric-spaces
asked Jan 2 at 13:12
GarmekainGarmekain
1,304720
$endgroup$
add a comment |
$begingroup$
Let $(Bbb R,d)$ be a metric space and let $psi:Bbb Rto[0,infty)$ be defined as $psi(x)=int_Ud(x,y)dy$ given a subset $Usubseteq Bbb R$ that is a bounded member of the usual topology on $Bbb R$.
I'm trying to prove that there is a unique $xinBbb R$ such that $psi(x)$ is minimum, i.e. for all other $yinBbb R$, we have that $psi(x)<psi(y)$, and not just $psi(x)leqpsi(y)$.
If $psi$ were injective then we are done, but it's not always the case. I need to check injectiveness just for the argument that just gives the minimal, but I don't know to do that. Any help on that or any other methods would be appreciated. Thanks.
functions metric-spaces
asked Jan 2 at 13:12
GarmekainGarmekain
1,304720
$endgroup$
$begingroup$
What is the connection between $d$ and the standard metric on ${mathbb R}$?
$endgroup$
– Christian Blatter
Jan 2 at 13:21
$begingroup$
You're trying to prove a false statement. A counterexample is $d$, the discrete metric ($d(x,y) = 1$ for $x not = y$), and $U = {0,1}$.
$endgroup$
– mathworker21
Jan 2 at 13:25
add a comment |
$begingroup$
Let $(Bbb R,d)$ be a metric space and let $psi:Bbb Rto[0,infty)$ be defined as $psi(x)=int_Ud(x,y)dy$ given a subset $Usubseteq Bbb R$ that is a bounded member of the usual topology on $Bbb R$.
I'm trying to prove that there is a unique $xinBbb R$ such that $psi(x)$ is minimum, i.e. for all other $yinBbb R$, we have that $psi(x)<psi(y)$, and not just $psi(x)leqpsi(y)$.
If $psi$ were injective then we are done, but it's not always the case. I need to check injectiveness just for the argument that just gives the minimal, but I don't know to do that. Any help on that or any other methods would be appreciated. Thanks.
functions metric-spaces
asked Jan 2 at 13:12
GarmekainGarmekain
1,304720
$endgroup$
Let $(Bbb R,d)$ be a metric space and let $psi:Bbb Rto[0,infty)$ be defined as $psi(x)=int_Ud(x,y)dy$ given a subset $Usubseteq Bbb R$ that is a bounded member of the usual topology on $Bbb R$.
I'm trying to prove that there is a unique $xinBbb R$ such that $psi(x)$ is minimum, i.e. for all other $yinBbb R$, we have that $psi(x)<psi(y)$, and not just $psi(x)leqpsi(y)$.
If $psi$ were injective then we are done, but it's not always the case. I need to check injectiveness just for the argument that just gives the minimal, but I don't know to do that. Any help on that or any other methods would be appreciated. Thanks.
functions metric-spaces
functions metric-spaces
asked Jan 2 at 13:12
GarmekainGarmekain
1,304720
asked Jan 2 at 13:12
GarmekainGarmekain
1,304720
asked Jan 2 at 13:12
GarmekainGarmekain
1,304720
asked Jan 2 at 13:12
GarmekainGarmekain
1,304720
asked Jan 2 at 13:12
GarmekainGarmekain
1,304720
1,304720
$begingroup$
What is the connection between $d$ and the standard metric on ${mathbb R}$?
$endgroup$
– Christian Blatter
Jan 2 at 13:21
$begingroup$
You're trying to prove a false statement. A counterexample is $d$, the discrete metric ($d(x,y) = 1$ for $x not = y$), and $U = {0,1}$.
$endgroup$
– mathworker21
Jan 2 at 13:25
add a comment |
$begingroup$
What is the connection between $d$ and the standard metric on ${mathbb R}$?
$endgroup$
– Christian Blatter
Jan 2 at 13:21
$begingroup$
You're trying to prove a false statement. A counterexample is $d$, the discrete metric ($d(x,y) = 1$ for $x not = y$), and $U = {0,1}$.
$endgroup$
– mathworker21
Jan 2 at 13:25
$begingroup$
What is the connection between $d$ and the standard metric on ${mathbb R}$?
$endgroup$
– Christian Blatter
Jan 2 at 13:21
$begingroup$
What is the connection between $d$ and the standard metric on ${mathbb R}$?
$endgroup$
– Christian Blatter
Jan 2 at 13:21
$begingroup$
You're trying to prove a false statement. A counterexample is $d$, the discrete metric ($d(x,y) = 1$ for $x not = y$), and $U = {0,1}$.
$endgroup$
– mathworker21
Jan 2 at 13:25
$begingroup$
You're trying to prove a false statement. A counterexample is $d$, the discrete metric ($d(x,y) = 1$ for $x not = y$), and $U = {0,1}$.
$endgroup$
– mathworker21
Jan 2 at 13:25
add a comment |
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$begingroup$
What is the connection between $d$ and the standard metric on ${mathbb R}$?
$endgroup$
– Christian Blatter
Jan 2 at 13:21
$begingroup$
You're trying to prove a false statement. A counterexample is $d$, the discrete metric ($d(x,y) = 1$ for $x not = y$), and $U = {0,1}$.
$endgroup$
– mathworker21
Jan 2 at 13:25