Characterizing points by their distance to the unit ball
$begingroup$
Let $x,yinmathbb{R}^n$. Assume that, for all $z$ in the unit ball, $|x-z|=|y-z|=d_z$. From this we can deduce that $|x| = |y|$ since $0$ is in the unit ball. How can we show that $x=y$? I think it must be true but I cannot show it easily.
geometry functional-analysis analysis
asked Jan 15 at 17:30
Tony S.F.Tony S.F.
3,94621031
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add a comment |
$begingroup$
Let $x,yinmathbb{R}^n$. Assume that, for all $z$ in the unit ball, $|x-z|=|y-z|=d_z$. From this we can deduce that $|x| = |y|$ since $0$ is in the unit ball. How can we show that $x=y$? I think it must be true but I cannot show it easily.
geometry functional-analysis analysis
asked Jan 15 at 17:30
Tony S.F.Tony S.F.
3,94621031
$endgroup$
add a comment |
$begingroup$
Let $x,yinmathbb{R}^n$. Assume that, for all $z$ in the unit ball, $|x-z|=|y-z|=d_z$. From this we can deduce that $|x| = |y|$ since $0$ is in the unit ball. How can we show that $x=y$? I think it must be true but I cannot show it easily.
geometry functional-analysis analysis
asked Jan 15 at 17:30
Tony S.F.Tony S.F.
3,94621031
$endgroup$
Let $x,yinmathbb{R}^n$. Assume that, for all $z$ in the unit ball, $|x-z|=|y-z|=d_z$. From this we can deduce that $|x| = |y|$ since $0$ is in the unit ball. How can we show that $x=y$? I think it must be true but I cannot show it easily.
geometry functional-analysis analysis
geometry functional-analysis analysis
asked Jan 15 at 17:30
Tony S.F.Tony S.F.
3,94621031
asked Jan 15 at 17:30
Tony S.F.Tony S.F.
3,94621031
asked Jan 15 at 17:30
Tony S.F.Tony S.F.
3,94621031
asked Jan 15 at 17:30
Tony S.F.Tony S.F.
3,94621031
asked Jan 15 at 17:30
Tony S.F.Tony S.F.
3,94621031
3,94621031
add a comment |
add a comment |
1 Answer
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$begingroup$
You can use the fact that $bar x=x/|x|$ is the unique best approximation of $xin mathbb{R}^nsetminus B$ in $B$.
Since
$$
|y-bar x|=|x-bar x|<|x-z|=|y-z|
$$
for $zin Bsetminus{bar x}$ by assumption, $bar x$ is also a best approximation of $y$ in $B$. By uniqueness it follows that $bar x=bar y$. As you already know that $|x|=|y|$, you are are done.
The case $xin B$ is trivial since you only need to use $0=|x-x|=|y-x|$. Finally, if you work with the open instead of the closed unit ball, you can simply replace $B$ by a closed ball of smaller radius in the argument above.
answered Jan 15 at 17:54
MaoWaoMaoWao
3,963618
$endgroup$
$begingroup$
Thanks, this is what I was looking for. I had even started investigating $frac{x}{|x|}$ but didn't think about it being the projector onto the unit ball.
$endgroup$
– Tony S.F.
Jan 16 at 9:23
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use the fact that $bar x=x/|x|$ is the unique best approximation of $xin mathbb{R}^nsetminus B$ in $B$.
Since
$$
|y-bar x|=|x-bar x|<|x-z|=|y-z|
$$
for $zin Bsetminus{bar x}$ by assumption, $bar x$ is also a best approximation of $y$ in $B$. By uniqueness it follows that $bar x=bar y$. As you already know that $|x|=|y|$, you are are done.
The case $xin B$ is trivial since you only need to use $0=|x-x|=|y-x|$. Finally, if you work with the open instead of the closed unit ball, you can simply replace $B$ by a closed ball of smaller radius in the argument above.
answered Jan 15 at 17:54
MaoWaoMaoWao
3,963618
$endgroup$
$begingroup$
Thanks, this is what I was looking for. I had even started investigating $frac{x}{|x|}$ but didn't think about it being the projector onto the unit ball.
$endgroup$
– Tony S.F.
Jan 16 at 9:23
add a comment |
$begingroup$
You can use the fact that $bar x=x/|x|$ is the unique best approximation of $xin mathbb{R}^nsetminus B$ in $B$.
Since
$$
|y-bar x|=|x-bar x|<|x-z|=|y-z|
$$
for $zin Bsetminus{bar x}$ by assumption, $bar x$ is also a best approximation of $y$ in $B$. By uniqueness it follows that $bar x=bar y$. As you already know that $|x|=|y|$, you are are done.
The case $xin B$ is trivial since you only need to use $0=|x-x|=|y-x|$. Finally, if you work with the open instead of the closed unit ball, you can simply replace $B$ by a closed ball of smaller radius in the argument above.
answered Jan 15 at 17:54
MaoWaoMaoWao
3,963618
$endgroup$
$begingroup$
Thanks, this is what I was looking for. I had even started investigating $frac{x}{|x|}$ but didn't think about it being the projector onto the unit ball.
$endgroup$
– Tony S.F.
Jan 16 at 9:23
add a comment |
$begingroup$
You can use the fact that $bar x=x/|x|$ is the unique best approximation of $xin mathbb{R}^nsetminus B$ in $B$.
Since
$$
|y-bar x|=|x-bar x|<|x-z|=|y-z|
$$
for $zin Bsetminus{bar x}$ by assumption, $bar x$ is also a best approximation of $y$ in $B$. By uniqueness it follows that $bar x=bar y$. As you already know that $|x|=|y|$, you are are done.
The case $xin B$ is trivial since you only need to use $0=|x-x|=|y-x|$. Finally, if you work with the open instead of the closed unit ball, you can simply replace $B$ by a closed ball of smaller radius in the argument above.
answered Jan 15 at 17:54
MaoWaoMaoWao
3,963618
$endgroup$
You can use the fact that $bar x=x/|x|$ is the unique best approximation of $xin mathbb{R}^nsetminus B$ in $B$.
Since
$$
|y-bar x|=|x-bar x|<|x-z|=|y-z|
$$
for $zin Bsetminus{bar x}$ by assumption, $bar x$ is also a best approximation of $y$ in $B$. By uniqueness it follows that $bar x=bar y$. As you already know that $|x|=|y|$, you are are done.
The case $xin B$ is trivial since you only need to use $0=|x-x|=|y-x|$. Finally, if you work with the open instead of the closed unit ball, you can simply replace $B$ by a closed ball of smaller radius in the argument above.
answered Jan 15 at 17:54
MaoWaoMaoWao
3,963618
answered Jan 15 at 17:54
MaoWaoMaoWao
3,963618
answered Jan 15 at 17:54
MaoWaoMaoWao
3,963618
answered Jan 15 at 17:54
MaoWaoMaoWao
3,963618
3,963618
$begingroup$
Thanks, this is what I was looking for. I had even started investigating $frac{x}{|x|}$ but didn't think about it being the projector onto the unit ball.
$endgroup$
– Tony S.F.
Jan 16 at 9:23
add a comment |
$begingroup$
Thanks, this is what I was looking for. I had even started investigating $frac{x}{|x|}$ but didn't think about it being the projector onto the unit ball.
$endgroup$
– Tony S.F.
Jan 16 at 9:23
$begingroup$
Thanks, this is what I was looking for. I had even started investigating $frac{x}{|x|}$ but didn't think about it being the projector onto the unit ball.
$endgroup$
– Tony S.F.
Jan 16 at 9:23
$begingroup$
Thanks, this is what I was looking for. I had even started investigating $frac{x}{|x|}$ but didn't think about it being the projector onto the unit ball.
$endgroup$
– Tony S.F.
Jan 16 at 9:23
add a comment |
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