Stone–Weierstrass for maps $S^mto S^n$?
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In the middle of page 35 of Algebraic Topology by Tammo tom Dieck, the author remarks:
If $f:S^mto S^n$ is a continuous map, then there exists (by the theorem of
Stone–Weierstrass, say) a $C^infty$-map $g:S^mto S^n$ such that $|f(x)-g(x)|<2$, $forall, xin S^m$.
Here, of course, $S^n:={,xinmathbb{R}^{n+1}:|x|=1,}$.
The most general version of Stone–Weierstrass theorem that I know is about the density of a subalgebra of $C(X,mathbb{R})$ for $X$ a compact Hausdorff space. How does it apply here? Which version of the theorem is the author talking about?
analysis algebraic-topology
asked Dec 23 '18 at 5:09
ColescuColescu
3,26711137
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add a comment |
$begingroup$
In the middle of page 35 of Algebraic Topology by Tammo tom Dieck, the author remarks:
If $f:S^mto S^n$ is a continuous map, then there exists (by the theorem of
Stone–Weierstrass, say) a $C^infty$-map $g:S^mto S^n$ such that $|f(x)-g(x)|<2$, $forall, xin S^m$.
Here, of course, $S^n:={,xinmathbb{R}^{n+1}:|x|=1,}$.
The most general version of Stone–Weierstrass theorem that I know is about the density of a subalgebra of $C(X,mathbb{R})$ for $X$ a compact Hausdorff space. How does it apply here? Which version of the theorem is the author talking about?
analysis algebraic-topology
asked Dec 23 '18 at 5:09
ColescuColescu
3,26711137
$endgroup$
add a comment |
$begingroup$
In the middle of page 35 of Algebraic Topology by Tammo tom Dieck, the author remarks:
If $f:S^mto S^n$ is a continuous map, then there exists (by the theorem of
Stone–Weierstrass, say) a $C^infty$-map $g:S^mto S^n$ such that $|f(x)-g(x)|<2$, $forall, xin S^m$.
Here, of course, $S^n:={,xinmathbb{R}^{n+1}:|x|=1,}$.
The most general version of Stone–Weierstrass theorem that I know is about the density of a subalgebra of $C(X,mathbb{R})$ for $X$ a compact Hausdorff space. How does it apply here? Which version of the theorem is the author talking about?
analysis algebraic-topology
asked Dec 23 '18 at 5:09
ColescuColescu
3,26711137
$endgroup$
In the middle of page 35 of Algebraic Topology by Tammo tom Dieck, the author remarks:
If $f:S^mto S^n$ is a continuous map, then there exists (by the theorem of
Stone–Weierstrass, say) a $C^infty$-map $g:S^mto S^n$ such that $|f(x)-g(x)|<2$, $forall, xin S^m$.
Here, of course, $S^n:={,xinmathbb{R}^{n+1}:|x|=1,}$.
The most general version of Stone–Weierstrass theorem that I know is about the density of a subalgebra of $C(X,mathbb{R})$ for $X$ a compact Hausdorff space. How does it apply here? Which version of the theorem is the author talking about?
analysis algebraic-topology
analysis algebraic-topology
asked Dec 23 '18 at 5:09
ColescuColescu
3,26711137
asked Dec 23 '18 at 5:09
ColescuColescu
3,26711137
edited Dec 23 '18 at 5:16
Colescu
asked Dec 23 '18 at 5:09
ColescuColescu
3,26711137
asked Dec 23 '18 at 5:09
ColescuColescu
3,26711137
asked Dec 23 '18 at 5:09
ColescuColescu
3,26711137
3,26711137
add a comment |
add a comment |
1 Answer
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By the usual version of Stone-Weierstrass for maps $S^m to mathbb R$, you can approximate $f$ by a smooth $h: S^m to mathbb R^{n+1}$ (i.e. approximate each coordinate of $f$). But the map $p: mathbb R^{n+1}setminus{0} to S^n$ given by $p(x) = x/|x|$ is smooth, so take $g = p circ h$ if $|f - h| < 1$.
edited Dec 23 '18 at 5:21
Eric Wofsey
193k14221352
answered Dec 23 '18 at 5:18
Robert IsraelRobert Israel
332k23222481
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
By the usual version of Stone-Weierstrass for maps $S^m to mathbb R$, you can approximate $f$ by a smooth $h: S^m to mathbb R^{n+1}$ (i.e. approximate each coordinate of $f$). But the map $p: mathbb R^{n+1}setminus{0} to S^n$ given by $p(x) = x/|x|$ is smooth, so take $g = p circ h$ if $|f - h| < 1$.
edited Dec 23 '18 at 5:21
Eric Wofsey
193k14221352
answered Dec 23 '18 at 5:18
Robert IsraelRobert Israel
332k23222481
$endgroup$
add a comment |
$begingroup$
By the usual version of Stone-Weierstrass for maps $S^m to mathbb R$, you can approximate $f$ by a smooth $h: S^m to mathbb R^{n+1}$ (i.e. approximate each coordinate of $f$). But the map $p: mathbb R^{n+1}setminus{0} to S^n$ given by $p(x) = x/|x|$ is smooth, so take $g = p circ h$ if $|f - h| < 1$.
edited Dec 23 '18 at 5:21
Eric Wofsey
193k14221352
answered Dec 23 '18 at 5:18
Robert IsraelRobert Israel
332k23222481
$endgroup$
add a comment |
$begingroup$
By the usual version of Stone-Weierstrass for maps $S^m to mathbb R$, you can approximate $f$ by a smooth $h: S^m to mathbb R^{n+1}$ (i.e. approximate each coordinate of $f$). But the map $p: mathbb R^{n+1}setminus{0} to S^n$ given by $p(x) = x/|x|$ is smooth, so take $g = p circ h$ if $|f - h| < 1$.
edited Dec 23 '18 at 5:21
Eric Wofsey
193k14221352
answered Dec 23 '18 at 5:18
Robert IsraelRobert Israel
332k23222481
$endgroup$
By the usual version of Stone-Weierstrass for maps $S^m to mathbb R$, you can approximate $f$ by a smooth $h: S^m to mathbb R^{n+1}$ (i.e. approximate each coordinate of $f$). But the map $p: mathbb R^{n+1}setminus{0} to S^n$ given by $p(x) = x/|x|$ is smooth, so take $g = p circ h$ if $|f - h| < 1$.
edited Dec 23 '18 at 5:21
Eric Wofsey
193k14221352
answered Dec 23 '18 at 5:18
Robert IsraelRobert Israel
332k23222481
edited Dec 23 '18 at 5:21
Eric Wofsey
193k14221352
edited Dec 23 '18 at 5:21
Eric Wofsey
193k14221352
edited Dec 23 '18 at 5:21
Eric Wofsey
193k14221352
193k14221352
answered Dec 23 '18 at 5:18
Robert IsraelRobert Israel
332k23222481
answered Dec 23 '18 at 5:18
Robert IsraelRobert Israel
332k23222481
answered Dec 23 '18 at 5:18
Robert IsraelRobert Israel
332k23222481
332k23222481
add a comment |
add a comment |
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