Approximation a finite set with an operator
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It is known that if $X$ is a compact manifold, then for every $epsilon>0$, there is $delta>0$ such that for every finite sequence ${(x_i, y_i)}_{i=0}^nsubset X$ with $x_ineq y_i$ and $d(x_i, y_i)<delta$, there is a homeomorphim $h:Xto X$ such that $d(h(x), x)<epsilon$ for all $xin X$ and $h(x_i)=y_i$ for all $0leq ileq n$.
Question. Above property is hold on Banach spaces?
this means that
if $X$ is a Banach space, Can I say that for every $epsilon>0$, there is $delta>0$ such that for every finite sequence ${(x_i, y_i)}_{i=0}^nsubset X$ with $x_ineq y_i$ and $||x_i-y_i||<delta$, there is a operator $h:Xto X$ such that $||h(x)-x||<epsilon$ for all $xin X$ and $h(x_i)=y_i$ for all $0leq ileq n$.
operator-theory dynamical-systems
asked Jan 17 at 10:53
user479859user479859
806
$endgroup$
add a comment |
$begingroup$
It is known that if $X$ is a compact manifold, then for every $epsilon>0$, there is $delta>0$ such that for every finite sequence ${(x_i, y_i)}_{i=0}^nsubset X$ with $x_ineq y_i$ and $d(x_i, y_i)<delta$, there is a homeomorphim $h:Xto X$ such that $d(h(x), x)<epsilon$ for all $xin X$ and $h(x_i)=y_i$ for all $0leq ileq n$.
Question. Above property is hold on Banach spaces?
this means that
if $X$ is a Banach space, Can I say that for every $epsilon>0$, there is $delta>0$ such that for every finite sequence ${(x_i, y_i)}_{i=0}^nsubset X$ with $x_ineq y_i$ and $||x_i-y_i||<delta$, there is a operator $h:Xto X$ such that $||h(x)-x||<epsilon$ for all $xin X$ and $h(x_i)=y_i$ for all $0leq ileq n$.
operator-theory dynamical-systems
asked Jan 17 at 10:53
user479859user479859
806
$endgroup$
$begingroup$
You mean $epsilon |x|$ when in your Banach?
$endgroup$
– Mindlack
Jan 17 at 10:59
add a comment |
$begingroup$
It is known that if $X$ is a compact manifold, then for every $epsilon>0$, there is $delta>0$ such that for every finite sequence ${(x_i, y_i)}_{i=0}^nsubset X$ with $x_ineq y_i$ and $d(x_i, y_i)<delta$, there is a homeomorphim $h:Xto X$ such that $d(h(x), x)<epsilon$ for all $xin X$ and $h(x_i)=y_i$ for all $0leq ileq n$.
Question. Above property is hold on Banach spaces?
this means that
if $X$ is a Banach space, Can I say that for every $epsilon>0$, there is $delta>0$ such that for every finite sequence ${(x_i, y_i)}_{i=0}^nsubset X$ with $x_ineq y_i$ and $||x_i-y_i||<delta$, there is a operator $h:Xto X$ such that $||h(x)-x||<epsilon$ for all $xin X$ and $h(x_i)=y_i$ for all $0leq ileq n$.
operator-theory dynamical-systems
asked Jan 17 at 10:53
user479859user479859
806
$endgroup$
It is known that if $X$ is a compact manifold, then for every $epsilon>0$, there is $delta>0$ such that for every finite sequence ${(x_i, y_i)}_{i=0}^nsubset X$ with $x_ineq y_i$ and $d(x_i, y_i)<delta$, there is a homeomorphim $h:Xto X$ such that $d(h(x), x)<epsilon$ for all $xin X$ and $h(x_i)=y_i$ for all $0leq ileq n$.
Question. Above property is hold on Banach spaces?
this means that
if $X$ is a Banach space, Can I say that for every $epsilon>0$, there is $delta>0$ such that for every finite sequence ${(x_i, y_i)}_{i=0}^nsubset X$ with $x_ineq y_i$ and $||x_i-y_i||<delta$, there is a operator $h:Xto X$ such that $||h(x)-x||<epsilon$ for all $xin X$ and $h(x_i)=y_i$ for all $0leq ileq n$.
operator-theory dynamical-systems
operator-theory dynamical-systems
asked Jan 17 at 10:53
user479859user479859
806
asked Jan 17 at 10:53
user479859user479859
806
asked Jan 17 at 10:53
user479859user479859
806
asked Jan 17 at 10:53
user479859user479859
806
asked Jan 17 at 10:53
user479859user479859
806
806
$begingroup$
You mean $epsilon |x|$ when in your Banach?
$endgroup$
– Mindlack
Jan 17 at 10:59
add a comment |
$begingroup$
You mean $epsilon |x|$ when in your Banach?
$endgroup$
– Mindlack
Jan 17 at 10:59
$begingroup$
You mean $epsilon |x|$ when in your Banach?
$endgroup$
– Mindlack
Jan 17 at 10:59
$begingroup$
You mean $epsilon |x|$ when in your Banach?
$endgroup$
– Mindlack
Jan 17 at 10:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Surely not, if you mean bounded linear map when you say operator. First note that (as commented by Mindlack) $|h(x)-x|<epsilon$ for all $x$ means $h$ is the identity so you should replace it with $|h(x)-x|leq epsilon |x|$ for all $x$. Even then the result is false. Consider $mathbb R$ as a one-dimensional Banach space. In this case $h(x)=cx$ for some real number $c$. If $|x_i-y_i|<delta$ you want $cx_i=y_i$ for all $i$ which can hold only when $frac {y_i} {x_i}$ is same for all $i$.
answered Jan 17 at 12:03
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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votes
$begingroup$
Surely not, if you mean bounded linear map when you say operator. First note that (as commented by Mindlack) $|h(x)-x|<epsilon$ for all $x$ means $h$ is the identity so you should replace it with $|h(x)-x|leq epsilon |x|$ for all $x$. Even then the result is false. Consider $mathbb R$ as a one-dimensional Banach space. In this case $h(x)=cx$ for some real number $c$. If $|x_i-y_i|<delta$ you want $cx_i=y_i$ for all $i$ which can hold only when $frac {y_i} {x_i}$ is same for all $i$.
answered Jan 17 at 12:03
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
add a comment |
$begingroup$
Surely not, if you mean bounded linear map when you say operator. First note that (as commented by Mindlack) $|h(x)-x|<epsilon$ for all $x$ means $h$ is the identity so you should replace it with $|h(x)-x|leq epsilon |x|$ for all $x$. Even then the result is false. Consider $mathbb R$ as a one-dimensional Banach space. In this case $h(x)=cx$ for some real number $c$. If $|x_i-y_i|<delta$ you want $cx_i=y_i$ for all $i$ which can hold only when $frac {y_i} {x_i}$ is same for all $i$.
answered Jan 17 at 12:03
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
add a comment |
$begingroup$
Surely not, if you mean bounded linear map when you say operator. First note that (as commented by Mindlack) $|h(x)-x|<epsilon$ for all $x$ means $h$ is the identity so you should replace it with $|h(x)-x|leq epsilon |x|$ for all $x$. Even then the result is false. Consider $mathbb R$ as a one-dimensional Banach space. In this case $h(x)=cx$ for some real number $c$. If $|x_i-y_i|<delta$ you want $cx_i=y_i$ for all $i$ which can hold only when $frac {y_i} {x_i}$ is same for all $i$.
answered Jan 17 at 12:03
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
Surely not, if you mean bounded linear map when you say operator. First note that (as commented by Mindlack) $|h(x)-x|<epsilon$ for all $x$ means $h$ is the identity so you should replace it with $|h(x)-x|leq epsilon |x|$ for all $x$. Even then the result is false. Consider $mathbb R$ as a one-dimensional Banach space. In this case $h(x)=cx$ for some real number $c$. If $|x_i-y_i|<delta$ you want $cx_i=y_i$ for all $i$ which can hold only when $frac {y_i} {x_i}$ is same for all $i$.
answered Jan 17 at 12:03
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
edited Jan 17 at 12:30
answered Jan 17 at 12:03
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
answered Jan 17 at 12:03
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
answered Jan 17 at 12:03
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
59.2k42161
add a comment |
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$begingroup$
You mean $epsilon |x|$ when in your Banach?
$endgroup$
– Mindlack
Jan 17 at 10:59