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Differentiable functions are dense in continous space?
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I am trying to solve this problem:
Given $X$ compact manifold, prove that any continuous function $f: X to S^p$ can be uniformly approximated by differentiable functions.
So, what I thought was this, it all comes down to prove for functions of the form $f: mathbbR^n to mathbbR$. Because, since $X$ is compact, you have a finite family $A=(U_1,phi_1),...,(U_m,phi_m)$ of local charts that cover $X$, lets say $X$ is an $n$-dimensional manifold, and since the $U_i$ are finite you can suppose the $tildeU_i=phi(U_i)$ disjoint.
Then the map $tildef: cup_1^mtildeU_i to S^p subset mathbbR^p+1$ given by $tildef(x)=f(phi^-1(x))$ is a continous mapping.
So, if I can uniformly approximate to $tildef$ by differentiable functions $g_i_i in mathbbN$, lets say close enough to $tildef$, such that they are nowhere zero and with a little subtlety you can assure that given $z=phi_i(x)$ and $y=phi_j(x)$ for $i,j in 1,...,m$, you have $g_l(z)=g_l(y) forall l in mathbbN$. (This is done by approximating locally in each $tildeU_i$)
Then taking $h_i=fracg_imidmid g_imidmid$, and $H_i=h_i circ phi_j(x)$ where $phi_j$ is an appropiate chart that depends on $x$, by the subtlety above this fucntions do not depend on the choice of chart, then these functions uniformly approximate to $f$.
To approximate $tildef$, all I really got to do is approximate its coordenate functions $tildef_i:cup_1^mtildeU_i to mathbbR $. Since the $tildeU_i$ are disjoint it actually comes to approximate $F_j=tildefrestriction_tildeU_j$, and just paste the approximations.
Keeping in mind that since $f(X)subset S^p$ is a bounded function I can work in the space of bounded functions: $mathbbR^nto mathbbR^p+1$.
All I need to do is prove the last part. Any ideas or errors you can find in this reasoning would be appreciated, thanks in advanced.
real-analysis differential-topology uniform-convergence
asked Jul 30 at 20:01
Bajo Fondo
376213
add a comment |Â
up vote
1
down vote
favorite
I am trying to solve this problem:
Given $X$ compact manifold, prove that any continuous function $f: X to S^p$ can be uniformly approximated by differentiable functions.
So, what I thought was this, it all comes down to prove for functions of the form $f: mathbbR^n to mathbbR$. Because, since $X$ is compact, you have a finite family $A=(U_1,phi_1),...,(U_m,phi_m)$ of local charts that cover $X$, lets say $X$ is an $n$-dimensional manifold, and since the $U_i$ are finite you can suppose the $tildeU_i=phi(U_i)$ disjoint.
Then the map $tildef: cup_1^mtildeU_i to S^p subset mathbbR^p+1$ given by $tildef(x)=f(phi^-1(x))$ is a continous mapping.
So, if I can uniformly approximate to $tildef$ by differentiable functions $g_i_i in mathbbN$, lets say close enough to $tildef$, such that they are nowhere zero and with a little subtlety you can assure that given $z=phi_i(x)$ and $y=phi_j(x)$ for $i,j in 1,...,m$, you have $g_l(z)=g_l(y) forall l in mathbbN$. (This is done by approximating locally in each $tildeU_i$)
Then taking $h_i=fracg_imidmid g_imidmid$, and $H_i=h_i circ phi_j(x)$ where $phi_j$ is an appropiate chart that depends on $x$, by the subtlety above this fucntions do not depend on the choice of chart, then these functions uniformly approximate to $f$.
To approximate $tildef$, all I really got to do is approximate its coordenate functions $tildef_i:cup_1^mtildeU_i to mathbbR $. Since the $tildeU_i$ are disjoint it actually comes to approximate $F_j=tildefrestriction_tildeU_j$, and just paste the approximations.
Keeping in mind that since $f(X)subset S^p$ is a bounded function I can work in the space of bounded functions: $mathbbR^nto mathbbR^p+1$.
All I need to do is prove the last part. Any ideas or errors you can find in this reasoning would be appreciated, thanks in advanced.
real-analysis differential-topology uniform-convergence
asked Jul 30 at 20:01
Bajo Fondo
376213
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to solve this problem:
Given $X$ compact manifold, prove that any continuous function $f: X to S^p$ can be uniformly approximated by differentiable functions.
So, what I thought was this, it all comes down to prove for functions of the form $f: mathbbR^n to mathbbR$. Because, since $X$ is compact, you have a finite family $A=(U_1,phi_1),...,(U_m,phi_m)$ of local charts that cover $X$, lets say $X$ is an $n$-dimensional manifold, and since the $U_i$ are finite you can suppose the $tildeU_i=phi(U_i)$ disjoint.
Then the map $tildef: cup_1^mtildeU_i to S^p subset mathbbR^p+1$ given by $tildef(x)=f(phi^-1(x))$ is a continous mapping.
So, if I can uniformly approximate to $tildef$ by differentiable functions $g_i_i in mathbbN$, lets say close enough to $tildef$, such that they are nowhere zero and with a little subtlety you can assure that given $z=phi_i(x)$ and $y=phi_j(x)$ for $i,j in 1,...,m$, you have $g_l(z)=g_l(y) forall l in mathbbN$. (This is done by approximating locally in each $tildeU_i$)
Then taking $h_i=fracg_imidmid g_imidmid$, and $H_i=h_i circ phi_j(x)$ where $phi_j$ is an appropiate chart that depends on $x$, by the subtlety above this fucntions do not depend on the choice of chart, then these functions uniformly approximate to $f$.
To approximate $tildef$, all I really got to do is approximate its coordenate functions $tildef_i:cup_1^mtildeU_i to mathbbR $. Since the $tildeU_i$ are disjoint it actually comes to approximate $F_j=tildefrestriction_tildeU_j$, and just paste the approximations.
Keeping in mind that since $f(X)subset S^p$ is a bounded function I can work in the space of bounded functions: $mathbbR^nto mathbbR^p+1$.
All I need to do is prove the last part. Any ideas or errors you can find in this reasoning would be appreciated, thanks in advanced.
real-analysis differential-topology uniform-convergence
asked Jul 30 at 20:01
Bajo Fondo
376213
I am trying to solve this problem:
Given $X$ compact manifold, prove that any continuous function $f: X to S^p$ can be uniformly approximated by differentiable functions.
So, what I thought was this, it all comes down to prove for functions of the form $f: mathbbR^n to mathbbR$. Because, since $X$ is compact, you have a finite family $A=(U_1,phi_1),...,(U_m,phi_m)$ of local charts that cover $X$, lets say $X$ is an $n$-dimensional manifold, and since the $U_i$ are finite you can suppose the $tildeU_i=phi(U_i)$ disjoint.
Then the map $tildef: cup_1^mtildeU_i to S^p subset mathbbR^p+1$ given by $tildef(x)=f(phi^-1(x))$ is a continous mapping.
So, if I can uniformly approximate to $tildef$ by differentiable functions $g_i_i in mathbbN$, lets say close enough to $tildef$, such that they are nowhere zero and with a little subtlety you can assure that given $z=phi_i(x)$ and $y=phi_j(x)$ for $i,j in 1,...,m$, you have $g_l(z)=g_l(y) forall l in mathbbN$. (This is done by approximating locally in each $tildeU_i$)
Then taking $h_i=fracg_imidmid g_imidmid$, and $H_i=h_i circ phi_j(x)$ where $phi_j$ is an appropiate chart that depends on $x$, by the subtlety above this fucntions do not depend on the choice of chart, then these functions uniformly approximate to $f$.
To approximate $tildef$, all I really got to do is approximate its coordenate functions $tildef_i:cup_1^mtildeU_i to mathbbR $. Since the $tildeU_i$ are disjoint it actually comes to approximate $F_j=tildefrestriction_tildeU_j$, and just paste the approximations.
Keeping in mind that since $f(X)subset S^p$ is a bounded function I can work in the space of bounded functions: $mathbbR^nto mathbbR^p+1$.
All I need to do is prove the last part. Any ideas or errors you can find in this reasoning would be appreciated, thanks in advanced.
real-analysis differential-topology uniform-convergence
asked Jul 30 at 20:01
Bajo Fondo
376213
asked Jul 30 at 20:01
Bajo Fondo
376213
asked Jul 30 at 20:01
Bajo Fondo
376213
asked Jul 30 at 20:01
Bajo Fondo
376213
376213
add a comment |Â
add a comment |Â
1 Answer
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oldest
votes
up vote
1
down vote
accepted
I think that using the Weierstrass Approximation Theorem should give the desired result: https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem
However, I think you should proceed a little differently in the beginning. Namely, cover $X$ by compact sets $C_1, dots , C_k$ such that each $C_i$ is homeomorphic to a closed rectangle in $R^n$ and moreover each $C_i$ is contained entirely within some chart. Then apply the Weierstrass approximation theorem. (I think it is necessary to cover by compact sets because the Weierstrass approximation theorem requires the interval to be closed.) Since there are finitely many components (manifolds are finite-dimensional) and finitely many $C_i,$ you will get a uniform approximation (since you can pick the maximum of some finitely many big N's.
answered Jul 30 at 20:23
Mishel Skenderi
1398
Ok.. will do. Thanks.
– Bajo Fondo
Jul 30 at 20:27
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I think that using the Weierstrass Approximation Theorem should give the desired result: https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem
However, I think you should proceed a little differently in the beginning. Namely, cover $X$ by compact sets $C_1, dots , C_k$ such that each $C_i$ is homeomorphic to a closed rectangle in $R^n$ and moreover each $C_i$ is contained entirely within some chart. Then apply the Weierstrass approximation theorem. (I think it is necessary to cover by compact sets because the Weierstrass approximation theorem requires the interval to be closed.) Since there are finitely many components (manifolds are finite-dimensional) and finitely many $C_i,$ you will get a uniform approximation (since you can pick the maximum of some finitely many big N's.
answered Jul 30 at 20:23
Mishel Skenderi
1398
Ok.. will do. Thanks.
– Bajo Fondo
Jul 30 at 20:27
add a comment |Â
up vote
1
down vote
accepted
I think that using the Weierstrass Approximation Theorem should give the desired result: https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem
However, I think you should proceed a little differently in the beginning. Namely, cover $X$ by compact sets $C_1, dots , C_k$ such that each $C_i$ is homeomorphic to a closed rectangle in $R^n$ and moreover each $C_i$ is contained entirely within some chart. Then apply the Weierstrass approximation theorem. (I think it is necessary to cover by compact sets because the Weierstrass approximation theorem requires the interval to be closed.) Since there are finitely many components (manifolds are finite-dimensional) and finitely many $C_i,$ you will get a uniform approximation (since you can pick the maximum of some finitely many big N's.
answered Jul 30 at 20:23
Mishel Skenderi
1398
Ok.. will do. Thanks.
– Bajo Fondo
Jul 30 at 20:27
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I think that using the Weierstrass Approximation Theorem should give the desired result: https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem
However, I think you should proceed a little differently in the beginning. Namely, cover $X$ by compact sets $C_1, dots , C_k$ such that each $C_i$ is homeomorphic to a closed rectangle in $R^n$ and moreover each $C_i$ is contained entirely within some chart. Then apply the Weierstrass approximation theorem. (I think it is necessary to cover by compact sets because the Weierstrass approximation theorem requires the interval to be closed.) Since there are finitely many components (manifolds are finite-dimensional) and finitely many $C_i,$ you will get a uniform approximation (since you can pick the maximum of some finitely many big N's.
answered Jul 30 at 20:23
Mishel Skenderi
1398
I think that using the Weierstrass Approximation Theorem should give the desired result: https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem
However, I think you should proceed a little differently in the beginning. Namely, cover $X$ by compact sets $C_1, dots , C_k$ such that each $C_i$ is homeomorphic to a closed rectangle in $R^n$ and moreover each $C_i$ is contained entirely within some chart. Then apply the Weierstrass approximation theorem. (I think it is necessary to cover by compact sets because the Weierstrass approximation theorem requires the interval to be closed.) Since there are finitely many components (manifolds are finite-dimensional) and finitely many $C_i,$ you will get a uniform approximation (since you can pick the maximum of some finitely many big N's.
answered Jul 30 at 20:23
Mishel Skenderi
1398
answered Jul 30 at 20:23
Mishel Skenderi
1398
answered Jul 30 at 20:23
Mishel Skenderi
1398
answered Jul 30 at 20:23
Mishel Skenderi
1398
1398
Ok.. will do. Thanks.
– Bajo Fondo
Jul 30 at 20:27
add a comment |Â
Ok.. will do. Thanks.
– Bajo Fondo
Jul 30 at 20:27
Ok.. will do. Thanks.
– Bajo Fondo
Jul 30 at 20:27
Ok.. will do. Thanks.
– Bajo Fondo
Jul 30 at 20:27
add a comment |Â
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