If two spheres are isometric, does there exist a bijective isometry $T:Sto S$ with $|Tu-alpha Tv|_Y leq...
Let
$$(S,|cdot|) = {(x,y)in mathbb{R}^2: |(x,y)| =1},$$
that is, $S$ is the collection of all norm one vectors in $mathbb{R}^2$ with respect to the norm $|cdot|.$
Question: Let $|cdot|_X$ and $|cdot|_Y$ be two norms on $mathbb{R}^2$ be such that $(S,|cdot|_X)$ and $(S,|cdot|_Y)$ are isometric.
Does there exist a bijective isometry $T:(S,|cdot|_X)to (S,|cdot|_Y)$ such that
$$|Tu-alpha Tv|_Y leq |u-alpha v|_X$$
for all $u,vin (S,|cdot|_X)$ and all $alpha>0?$
Note that the norms $|cdot|_X$ and $|cdot|_Y$ may be distinct.
I tried $|(x,y)|_X = |x|+|y|,$ $|(x,y)|_Y = max{|x|,|y|}$ and
$$T(x,y) = begin{pmatrix}
1 & 1 \
-1 & 1
end{pmatrix}.$$
Note that $T$ is a rotation matrix.
Clearly $T$ is a bijective isometry and satisfies the inequality.
However, I do not know whether the same holds for general $|cdot|_X$ and $|cdot|_Y.$
real-analysis functional-analysis banach-spaces normed-spaces isometry
asked Jan 4 at 10:07
IdonknowIdonknow
2,350749113
|
show 3 more comments
Let
$$(S,|cdot|) = {(x,y)in mathbb{R}^2: |(x,y)| =1},$$
that is, $S$ is the collection of all norm one vectors in $mathbb{R}^2$ with respect to the norm $|cdot|.$
Question: Let $|cdot|_X$ and $|cdot|_Y$ be two norms on $mathbb{R}^2$ be such that $(S,|cdot|_X)$ and $(S,|cdot|_Y)$ are isometric.
Does there exist a bijective isometry $T:(S,|cdot|_X)to (S,|cdot|_Y)$ such that
$$|Tu-alpha Tv|_Y leq |u-alpha v|_X$$
for all $u,vin (S,|cdot|_X)$ and all $alpha>0?$
Note that the norms $|cdot|_X$ and $|cdot|_Y$ may be distinct.
I tried $|(x,y)|_X = |x|+|y|,$ $|(x,y)|_Y = max{|x|,|y|}$ and
$$T(x,y) = begin{pmatrix}
1 & 1 \
-1 & 1
end{pmatrix}.$$
Note that $T$ is a rotation matrix.
Clearly $T$ is a bijective isometry and satisfies the inequality.
However, I do not know whether the same holds for general $|cdot|_X$ and $|cdot|_Y.$
real-analysis functional-analysis banach-spaces normed-spaces isometry
asked Jan 4 at 10:07
IdonknowIdonknow
2,350749113
What do you mean by isometry? An isometry of $mathbb{R}^2$?
– 0x539
Jan 4 at 15:38
@0x539 Yes. Isometry of $mathbb{R}^2$.
– Idonknow
Jan 4 at 15:50
So if $|cdot|_2 = |cdot|_Y = frac12 |cdot|_X$ there could be no such isometry because the unit circle can't get mapped to itself?
– 0x539
Jan 4 at 15:52
Or are there supposed to be two unit circles $S_X$ and $S_Y$?
– 0x539
Jan 4 at 15:54
Yes. There are two unit circles
– Idonknow
Jan 4 at 16:13
|
show 3 more comments
Let
$$(S,|cdot|) = {(x,y)in mathbb{R}^2: |(x,y)| =1},$$
that is, $S$ is the collection of all norm one vectors in $mathbb{R}^2$ with respect to the norm $|cdot|.$
Question: Let $|cdot|_X$ and $|cdot|_Y$ be two norms on $mathbb{R}^2$ be such that $(S,|cdot|_X)$ and $(S,|cdot|_Y)$ are isometric.
Does there exist a bijective isometry $T:(S,|cdot|_X)to (S,|cdot|_Y)$ such that
$$|Tu-alpha Tv|_Y leq |u-alpha v|_X$$
for all $u,vin (S,|cdot|_X)$ and all $alpha>0?$
Note that the norms $|cdot|_X$ and $|cdot|_Y$ may be distinct.
I tried $|(x,y)|_X = |x|+|y|,$ $|(x,y)|_Y = max{|x|,|y|}$ and
$$T(x,y) = begin{pmatrix}
1 & 1 \
-1 & 1
end{pmatrix}.$$
Note that $T$ is a rotation matrix.
Clearly $T$ is a bijective isometry and satisfies the inequality.
However, I do not know whether the same holds for general $|cdot|_X$ and $|cdot|_Y.$
real-analysis functional-analysis banach-spaces normed-spaces isometry
asked Jan 4 at 10:07
IdonknowIdonknow
2,350749113
Let
$$(S,|cdot|) = {(x,y)in mathbb{R}^2: |(x,y)| =1},$$
that is, $S$ is the collection of all norm one vectors in $mathbb{R}^2$ with respect to the norm $|cdot|.$
Question: Let $|cdot|_X$ and $|cdot|_Y$ be two norms on $mathbb{R}^2$ be such that $(S,|cdot|_X)$ and $(S,|cdot|_Y)$ are isometric.
Does there exist a bijective isometry $T:(S,|cdot|_X)to (S,|cdot|_Y)$ such that
$$|Tu-alpha Tv|_Y leq |u-alpha v|_X$$
for all $u,vin (S,|cdot|_X)$ and all $alpha>0?$
Note that the norms $|cdot|_X$ and $|cdot|_Y$ may be distinct.
I tried $|(x,y)|_X = |x|+|y|,$ $|(x,y)|_Y = max{|x|,|y|}$ and
$$T(x,y) = begin{pmatrix}
1 & 1 \
-1 & 1
end{pmatrix}.$$
Note that $T$ is a rotation matrix.
Clearly $T$ is a bijective isometry and satisfies the inequality.
However, I do not know whether the same holds for general $|cdot|_X$ and $|cdot|_Y.$
real-analysis functional-analysis banach-spaces normed-spaces isometry
real-analysis functional-analysis banach-spaces normed-spaces isometry
asked Jan 4 at 10:07
IdonknowIdonknow
2,350749113
asked Jan 4 at 10:07
IdonknowIdonknow
2,350749113
edited 2 days ago
Idonknow
asked Jan 4 at 10:07
IdonknowIdonknow
2,350749113
asked Jan 4 at 10:07
IdonknowIdonknow
2,350749113
asked Jan 4 at 10:07
IdonknowIdonknow
2,350749113
2,350749113
What do you mean by isometry? An isometry of $mathbb{R}^2$?
– 0x539
Jan 4 at 15:38
@0x539 Yes. Isometry of $mathbb{R}^2$.
– Idonknow
Jan 4 at 15:50
So if $|cdot|_2 = |cdot|_Y = frac12 |cdot|_X$ there could be no such isometry because the unit circle can't get mapped to itself?
– 0x539
Jan 4 at 15:52
Or are there supposed to be two unit circles $S_X$ and $S_Y$?
– 0x539
Jan 4 at 15:54
Yes. There are two unit circles
– Idonknow
Jan 4 at 16:13
|
show 3 more comments
What do you mean by isometry? An isometry of $mathbb{R}^2$?
– 0x539
Jan 4 at 15:38
@0x539 Yes. Isometry of $mathbb{R}^2$.
– Idonknow
Jan 4 at 15:50
So if $|cdot|_2 = |cdot|_Y = frac12 |cdot|_X$ there could be no such isometry because the unit circle can't get mapped to itself?
– 0x539
Jan 4 at 15:52
Or are there supposed to be two unit circles $S_X$ and $S_Y$?
– 0x539
Jan 4 at 15:54
Yes. There are two unit circles
– Idonknow
Jan 4 at 16:13
What do you mean by isometry? An isometry of $mathbb{R}^2$?
– 0x539
Jan 4 at 15:38
What do you mean by isometry? An isometry of $mathbb{R}^2$?
– 0x539
Jan 4 at 15:38
@0x539 Yes. Isometry of $mathbb{R}^2$.
– Idonknow
Jan 4 at 15:50
@0x539 Yes. Isometry of $mathbb{R}^2$.
– Idonknow
Jan 4 at 15:50
So if $|cdot|_2 = |cdot|_Y = frac12 |cdot|_X$ there could be no such isometry because the unit circle can't get mapped to itself?
– 0x539
Jan 4 at 15:52
So if $|cdot|_2 = |cdot|_Y = frac12 |cdot|_X$ there could be no such isometry because the unit circle can't get mapped to itself?
– 0x539
Jan 4 at 15:52
Or are there supposed to be two unit circles $S_X$ and $S_Y$?
– 0x539
Jan 4 at 15:54
Or are there supposed to be two unit circles $S_X$ and $S_Y$?
– 0x539
Jan 4 at 15:54
Yes. There are two unit circles
– Idonknow
Jan 4 at 16:13
Yes. There are two unit circles
– Idonknow
Jan 4 at 16:13
|
show 3 more comments
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What do you mean by isometry? An isometry of $mathbb{R}^2$?
– 0x539
Jan 4 at 15:38
@0x539 Yes. Isometry of $mathbb{R}^2$.
– Idonknow
Jan 4 at 15:50
So if $|cdot|_2 = |cdot|_Y = frac12 |cdot|_X$ there could be no such isometry because the unit circle can't get mapped to itself?
– 0x539
Jan 4 at 15:52
Or are there supposed to be two unit circles $S_X$ and $S_Y$?
– 0x539
Jan 4 at 15:54
Yes. There are two unit circles
– Idonknow
Jan 4 at 16:13