Preserving Bijectivity under pointwise limit
$begingroup$
The following question I have tried to solve and have also tried to find reference for. A reference would be appreciated.
Suppose that $f_i, g_i :mathbb{R^n} to mathbb{R^n}$ are continuous functions such that $f_i circ g_i =g_i circ f_i = text{id}_mathbb{R^n}$ for each $i in mathbb{N}$. Let $Omega_{{f_i}}$ be the set of points for which ${f_i}_i$ converges pointwise to some function, say $f$. Similarly, let $Omega_{{g_i}}$ be the set of points for which ${g_i}_i$ converges pointwise to some function, say $g$. Then show that $$f circ g = text{id}_{Omega_{{g_i}}}$$ and $$g circ f = text{id}_{Omega_{{f_i}}}$$ specifically that such compositions make sense (i.e. that $Omega_{{f_i}} subseteq g(Omega_{{g_i}})$ and $Omega_{{g_i}} subseteq f(Omega_{{f_i}})$.
real-analysis convergence pointwise-convergence
asked Nov 8 '18 at 4:07
Better2BLuckyBetter2BLucky
586
$endgroup$
add a comment |
$begingroup$
The following question I have tried to solve and have also tried to find reference for. A reference would be appreciated.
Suppose that $f_i, g_i :mathbb{R^n} to mathbb{R^n}$ are continuous functions such that $f_i circ g_i =g_i circ f_i = text{id}_mathbb{R^n}$ for each $i in mathbb{N}$. Let $Omega_{{f_i}}$ be the set of points for which ${f_i}_i$ converges pointwise to some function, say $f$. Similarly, let $Omega_{{g_i}}$ be the set of points for which ${g_i}_i$ converges pointwise to some function, say $g$. Then show that $$f circ g = text{id}_{Omega_{{g_i}}}$$ and $$g circ f = text{id}_{Omega_{{f_i}}}$$ specifically that such compositions make sense (i.e. that $Omega_{{f_i}} subseteq g(Omega_{{g_i}})$ and $Omega_{{g_i}} subseteq f(Omega_{{f_i}})$.
real-analysis convergence pointwise-convergence
asked Nov 8 '18 at 4:07
Better2BLuckyBetter2BLucky
586
$endgroup$
add a comment |
$begingroup$
The following question I have tried to solve and have also tried to find reference for. A reference would be appreciated.
Suppose that $f_i, g_i :mathbb{R^n} to mathbb{R^n}$ are continuous functions such that $f_i circ g_i =g_i circ f_i = text{id}_mathbb{R^n}$ for each $i in mathbb{N}$. Let $Omega_{{f_i}}$ be the set of points for which ${f_i}_i$ converges pointwise to some function, say $f$. Similarly, let $Omega_{{g_i}}$ be the set of points for which ${g_i}_i$ converges pointwise to some function, say $g$. Then show that $$f circ g = text{id}_{Omega_{{g_i}}}$$ and $$g circ f = text{id}_{Omega_{{f_i}}}$$ specifically that such compositions make sense (i.e. that $Omega_{{f_i}} subseteq g(Omega_{{g_i}})$ and $Omega_{{g_i}} subseteq f(Omega_{{f_i}})$.
real-analysis convergence pointwise-convergence
asked Nov 8 '18 at 4:07
Better2BLuckyBetter2BLucky
586
$endgroup$
The following question I have tried to solve and have also tried to find reference for. A reference would be appreciated.
Suppose that $f_i, g_i :mathbb{R^n} to mathbb{R^n}$ are continuous functions such that $f_i circ g_i =g_i circ f_i = text{id}_mathbb{R^n}$ for each $i in mathbb{N}$. Let $Omega_{{f_i}}$ be the set of points for which ${f_i}_i$ converges pointwise to some function, say $f$. Similarly, let $Omega_{{g_i}}$ be the set of points for which ${g_i}_i$ converges pointwise to some function, say $g$. Then show that $$f circ g = text{id}_{Omega_{{g_i}}}$$ and $$g circ f = text{id}_{Omega_{{f_i}}}$$ specifically that such compositions make sense (i.e. that $Omega_{{f_i}} subseteq g(Omega_{{g_i}})$ and $Omega_{{g_i}} subseteq f(Omega_{{f_i}})$.
real-analysis convergence pointwise-convergence
real-analysis convergence pointwise-convergence
asked Nov 8 '18 at 4:07
Better2BLuckyBetter2BLucky
586
asked Nov 8 '18 at 4:07
Better2BLuckyBetter2BLucky
586
asked Nov 8 '18 at 4:07
Better2BLuckyBetter2BLucky
586
asked Nov 8 '18 at 4:07
Better2BLuckyBetter2BLucky
586
asked Nov 8 '18 at 4:07
Better2BLuckyBetter2BLucky
586
586
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Counterexample on $mathbb R^1:$ For $m=1,2,dots,$ define
$$f_m(x)= begin{cases}x,& xle 0\x^m,& 0<x<1\ x,& xge 1\end{cases}$$
Then each $f_m$ is a homeomorphism of $mathbb R$ onto $mathbb R.$ Set $g_m= (f_m)^{-1},$ i.e.,
$$g_m(x)= begin{cases}x,& xle 0\x^{1/m},& 0<x<1\ x,& xge 1\end{cases}$$
Check that $f_m$ converges pointwise everywhere to the function $f$ that is $0$ on $(0,1)$ and is the identity elsewhere. Similarly, $g_m$ converges pointwise everywhere to the function $g$ that is $1$ on $(0,1)$ and is the identity elsewhere. We have $fcirc g=1$ on $(0,1),$ and $gcirc f=0$ on $(0,1),$ showing the desired conclusion does not hold.
answered Dec 1 '18 at 18:57
zhw.zhw.
71.8k43075
$endgroup$
$begingroup$
Typo: $g circ f = 0$ on $(0,1)$.
$endgroup$
– Paul Frost
Dec 2 '18 at 9:31
$begingroup$
@PaulFrost Indeed. Thanks. Corrected.
$endgroup$
– zhw.
Dec 2 '18 at 16:57
add a comment |
$begingroup$
It is not true.
Let us first consider the following sequence of piecewise linear homeomorphisms on the unit interval $I = [0,1]$:
$$h_i(t) =
begin{cases}
frac{t}{i} & t le frac{1}{2} \
frac{1}{2i}+2(1-frac{1}{2i})(t-frac{1}{2}) & t ge frac{1}{2}
end{cases}
$$
The inverse homeomorphisms are given by
$$h_i^{-1} =
begin{cases}
ti & t le frac{1}{2i} \
frac{1}{2}+frac{1}{2(1-frac{1}{2i})}(t-frac{1}{2i}) & t ge frac{1}{2i}
end{cases}
$$
Obviosuly $(h_i)$ converges (even uniformly) to the continuous surjection
$$h : I to I, h(t) =
begin{cases}
0 & t le frac{1}{2} \
2(t-frac{1}{2}) & t ge frac{1}{2}
end{cases}
$$
The sequence $(h_i^{-1})$ also converges pointwise on $I$. Its limit is
$$h^- : I to I, h^-(t) =
begin{cases}
0 & t = 0 \
frac{1}{2}+frac{1}{2}t & t > 0
end{cases}
$$
This is a non-continuous injection. We have $h circ h^- = id$, but $h^- circ h ne id$.
To get examples on $mathbb{R}^n$, define
$$f_i(x) =
begin{cases}
x & lVert x rVert ge 1 \
h_i(lVert x rVert)x & lVert x rVert le 1
end{cases}
$$
answered Dec 1 '18 at 16:29
Paul FrostPaul Frost
9,8453732
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Counterexample on $mathbb R^1:$ For $m=1,2,dots,$ define
$$f_m(x)= begin{cases}x,& xle 0\x^m,& 0<x<1\ x,& xge 1\end{cases}$$
Then each $f_m$ is a homeomorphism of $mathbb R$ onto $mathbb R.$ Set $g_m= (f_m)^{-1},$ i.e.,
$$g_m(x)= begin{cases}x,& xle 0\x^{1/m},& 0<x<1\ x,& xge 1\end{cases}$$
Check that $f_m$ converges pointwise everywhere to the function $f$ that is $0$ on $(0,1)$ and is the identity elsewhere. Similarly, $g_m$ converges pointwise everywhere to the function $g$ that is $1$ on $(0,1)$ and is the identity elsewhere. We have $fcirc g=1$ on $(0,1),$ and $gcirc f=0$ on $(0,1),$ showing the desired conclusion does not hold.
answered Dec 1 '18 at 18:57
zhw.zhw.
71.8k43075
$endgroup$
$begingroup$
Typo: $g circ f = 0$ on $(0,1)$.
$endgroup$
– Paul Frost
Dec 2 '18 at 9:31
$begingroup$
@PaulFrost Indeed. Thanks. Corrected.
$endgroup$
– zhw.
Dec 2 '18 at 16:57
add a comment |
$begingroup$
Counterexample on $mathbb R^1:$ For $m=1,2,dots,$ define
$$f_m(x)= begin{cases}x,& xle 0\x^m,& 0<x<1\ x,& xge 1\end{cases}$$
Then each $f_m$ is a homeomorphism of $mathbb R$ onto $mathbb R.$ Set $g_m= (f_m)^{-1},$ i.e.,
$$g_m(x)= begin{cases}x,& xle 0\x^{1/m},& 0<x<1\ x,& xge 1\end{cases}$$
Check that $f_m$ converges pointwise everywhere to the function $f$ that is $0$ on $(0,1)$ and is the identity elsewhere. Similarly, $g_m$ converges pointwise everywhere to the function $g$ that is $1$ on $(0,1)$ and is the identity elsewhere. We have $fcirc g=1$ on $(0,1),$ and $gcirc f=0$ on $(0,1),$ showing the desired conclusion does not hold.
answered Dec 1 '18 at 18:57
zhw.zhw.
71.8k43075
$endgroup$
$begingroup$
Typo: $g circ f = 0$ on $(0,1)$.
$endgroup$
– Paul Frost
Dec 2 '18 at 9:31
$begingroup$
@PaulFrost Indeed. Thanks. Corrected.
$endgroup$
– zhw.
Dec 2 '18 at 16:57
add a comment |
$begingroup$
Counterexample on $mathbb R^1:$ For $m=1,2,dots,$ define
$$f_m(x)= begin{cases}x,& xle 0\x^m,& 0<x<1\ x,& xge 1\end{cases}$$
Then each $f_m$ is a homeomorphism of $mathbb R$ onto $mathbb R.$ Set $g_m= (f_m)^{-1},$ i.e.,
$$g_m(x)= begin{cases}x,& xle 0\x^{1/m},& 0<x<1\ x,& xge 1\end{cases}$$
Check that $f_m$ converges pointwise everywhere to the function $f$ that is $0$ on $(0,1)$ and is the identity elsewhere. Similarly, $g_m$ converges pointwise everywhere to the function $g$ that is $1$ on $(0,1)$ and is the identity elsewhere. We have $fcirc g=1$ on $(0,1),$ and $gcirc f=0$ on $(0,1),$ showing the desired conclusion does not hold.
answered Dec 1 '18 at 18:57
zhw.zhw.
71.8k43075
$endgroup$
Counterexample on $mathbb R^1:$ For $m=1,2,dots,$ define
$$f_m(x)= begin{cases}x,& xle 0\x^m,& 0<x<1\ x,& xge 1\end{cases}$$
Then each $f_m$ is a homeomorphism of $mathbb R$ onto $mathbb R.$ Set $g_m= (f_m)^{-1},$ i.e.,
$$g_m(x)= begin{cases}x,& xle 0\x^{1/m},& 0<x<1\ x,& xge 1\end{cases}$$
Check that $f_m$ converges pointwise everywhere to the function $f$ that is $0$ on $(0,1)$ and is the identity elsewhere. Similarly, $g_m$ converges pointwise everywhere to the function $g$ that is $1$ on $(0,1)$ and is the identity elsewhere. We have $fcirc g=1$ on $(0,1),$ and $gcirc f=0$ on $(0,1),$ showing the desired conclusion does not hold.
answered Dec 1 '18 at 18:57
zhw.zhw.
71.8k43075
edited Dec 2 '18 at 19:21
answered Dec 1 '18 at 18:57
zhw.zhw.
71.8k43075
answered Dec 1 '18 at 18:57
zhw.zhw.
71.8k43075
answered Dec 1 '18 at 18:57
zhw.zhw.
71.8k43075
71.8k43075
$begingroup$
Typo: $g circ f = 0$ on $(0,1)$.
$endgroup$
– Paul Frost
Dec 2 '18 at 9:31
$begingroup$
@PaulFrost Indeed. Thanks. Corrected.
$endgroup$
– zhw.
Dec 2 '18 at 16:57
add a comment |
$begingroup$
Typo: $g circ f = 0$ on $(0,1)$.
$endgroup$
– Paul Frost
Dec 2 '18 at 9:31
$begingroup$
@PaulFrost Indeed. Thanks. Corrected.
$endgroup$
– zhw.
Dec 2 '18 at 16:57
$begingroup$
Typo: $g circ f = 0$ on $(0,1)$.
$endgroup$
– Paul Frost
Dec 2 '18 at 9:31
$begingroup$
Typo: $g circ f = 0$ on $(0,1)$.
$endgroup$
– Paul Frost
Dec 2 '18 at 9:31
$begingroup$
@PaulFrost Indeed. Thanks. Corrected.
$endgroup$
– zhw.
Dec 2 '18 at 16:57
$begingroup$
@PaulFrost Indeed. Thanks. Corrected.
$endgroup$
– zhw.
Dec 2 '18 at 16:57
add a comment |
$begingroup$
It is not true.
Let us first consider the following sequence of piecewise linear homeomorphisms on the unit interval $I = [0,1]$:
$$h_i(t) =
begin{cases}
frac{t}{i} & t le frac{1}{2} \
frac{1}{2i}+2(1-frac{1}{2i})(t-frac{1}{2}) & t ge frac{1}{2}
end{cases}
$$
The inverse homeomorphisms are given by
$$h_i^{-1} =
begin{cases}
ti & t le frac{1}{2i} \
frac{1}{2}+frac{1}{2(1-frac{1}{2i})}(t-frac{1}{2i}) & t ge frac{1}{2i}
end{cases}
$$
Obviosuly $(h_i)$ converges (even uniformly) to the continuous surjection
$$h : I to I, h(t) =
begin{cases}
0 & t le frac{1}{2} \
2(t-frac{1}{2}) & t ge frac{1}{2}
end{cases}
$$
The sequence $(h_i^{-1})$ also converges pointwise on $I$. Its limit is
$$h^- : I to I, h^-(t) =
begin{cases}
0 & t = 0 \
frac{1}{2}+frac{1}{2}t & t > 0
end{cases}
$$
This is a non-continuous injection. We have $h circ h^- = id$, but $h^- circ h ne id$.
To get examples on $mathbb{R}^n$, define
$$f_i(x) =
begin{cases}
x & lVert x rVert ge 1 \
h_i(lVert x rVert)x & lVert x rVert le 1
end{cases}
$$
answered Dec 1 '18 at 16:29
Paul FrostPaul Frost
9,8453732
$endgroup$
add a comment |
$begingroup$
It is not true.
Let us first consider the following sequence of piecewise linear homeomorphisms on the unit interval $I = [0,1]$:
$$h_i(t) =
begin{cases}
frac{t}{i} & t le frac{1}{2} \
frac{1}{2i}+2(1-frac{1}{2i})(t-frac{1}{2}) & t ge frac{1}{2}
end{cases}
$$
The inverse homeomorphisms are given by
$$h_i^{-1} =
begin{cases}
ti & t le frac{1}{2i} \
frac{1}{2}+frac{1}{2(1-frac{1}{2i})}(t-frac{1}{2i}) & t ge frac{1}{2i}
end{cases}
$$
Obviosuly $(h_i)$ converges (even uniformly) to the continuous surjection
$$h : I to I, h(t) =
begin{cases}
0 & t le frac{1}{2} \
2(t-frac{1}{2}) & t ge frac{1}{2}
end{cases}
$$
The sequence $(h_i^{-1})$ also converges pointwise on $I$. Its limit is
$$h^- : I to I, h^-(t) =
begin{cases}
0 & t = 0 \
frac{1}{2}+frac{1}{2}t & t > 0
end{cases}
$$
This is a non-continuous injection. We have $h circ h^- = id$, but $h^- circ h ne id$.
To get examples on $mathbb{R}^n$, define
$$f_i(x) =
begin{cases}
x & lVert x rVert ge 1 \
h_i(lVert x rVert)x & lVert x rVert le 1
end{cases}
$$
answered Dec 1 '18 at 16:29
Paul FrostPaul Frost
9,8453732
$endgroup$
add a comment |
$begingroup$
It is not true.
Let us first consider the following sequence of piecewise linear homeomorphisms on the unit interval $I = [0,1]$:
$$h_i(t) =
begin{cases}
frac{t}{i} & t le frac{1}{2} \
frac{1}{2i}+2(1-frac{1}{2i})(t-frac{1}{2}) & t ge frac{1}{2}
end{cases}
$$
The inverse homeomorphisms are given by
$$h_i^{-1} =
begin{cases}
ti & t le frac{1}{2i} \
frac{1}{2}+frac{1}{2(1-frac{1}{2i})}(t-frac{1}{2i}) & t ge frac{1}{2i}
end{cases}
$$
Obviosuly $(h_i)$ converges (even uniformly) to the continuous surjection
$$h : I to I, h(t) =
begin{cases}
0 & t le frac{1}{2} \
2(t-frac{1}{2}) & t ge frac{1}{2}
end{cases}
$$
The sequence $(h_i^{-1})$ also converges pointwise on $I$. Its limit is
$$h^- : I to I, h^-(t) =
begin{cases}
0 & t = 0 \
frac{1}{2}+frac{1}{2}t & t > 0
end{cases}
$$
This is a non-continuous injection. We have $h circ h^- = id$, but $h^- circ h ne id$.
To get examples on $mathbb{R}^n$, define
$$f_i(x) =
begin{cases}
x & lVert x rVert ge 1 \
h_i(lVert x rVert)x & lVert x rVert le 1
end{cases}
$$
answered Dec 1 '18 at 16:29
Paul FrostPaul Frost
9,8453732
$endgroup$
It is not true.
Let us first consider the following sequence of piecewise linear homeomorphisms on the unit interval $I = [0,1]$:
$$h_i(t) =
begin{cases}
frac{t}{i} & t le frac{1}{2} \
frac{1}{2i}+2(1-frac{1}{2i})(t-frac{1}{2}) & t ge frac{1}{2}
end{cases}
$$
The inverse homeomorphisms are given by
$$h_i^{-1} =
begin{cases}
ti & t le frac{1}{2i} \
frac{1}{2}+frac{1}{2(1-frac{1}{2i})}(t-frac{1}{2i}) & t ge frac{1}{2i}
end{cases}
$$
Obviosuly $(h_i)$ converges (even uniformly) to the continuous surjection
$$h : I to I, h(t) =
begin{cases}
0 & t le frac{1}{2} \
2(t-frac{1}{2}) & t ge frac{1}{2}
end{cases}
$$
The sequence $(h_i^{-1})$ also converges pointwise on $I$. Its limit is
$$h^- : I to I, h^-(t) =
begin{cases}
0 & t = 0 \
frac{1}{2}+frac{1}{2}t & t > 0
end{cases}
$$
This is a non-continuous injection. We have $h circ h^- = id$, but $h^- circ h ne id$.
To get examples on $mathbb{R}^n$, define
$$f_i(x) =
begin{cases}
x & lVert x rVert ge 1 \
h_i(lVert x rVert)x & lVert x rVert le 1
end{cases}
$$
answered Dec 1 '18 at 16:29
Paul FrostPaul Frost
9,8453732
answered Dec 1 '18 at 16:29
Paul FrostPaul Frost
9,8453732
answered Dec 1 '18 at 16:29
Paul FrostPaul Frost
9,8453732
answered Dec 1 '18 at 16:29
Paul FrostPaul Frost
9,8453732
9,8453732
add a comment |
add a comment |
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