Mixing
If $f$ is $C^1$ s. t. $Vert f(x) - f(y) Vert geq k Vert x - y Vert$, then $f$ is a diffeomorphism of...
$begingroup$
Let $f:mathbb{R}^{n} to mathbb{R}^{n}$ be a function of class $C^1$ and suppose that there is $k>0$ such that
$$Vert f(x) - f(y) Vert geq k Vert x - y Vert$$
for any $x,y in mathbb{R}^{n}$.
(a) Prove that $f$ is injective and $f(mathbb{R}^{n})$ is closed.
(b) Prove that $f'$ is invertible.
(c) Prove that $f(mathbb{R}^{n})$ is open. Conclude that $f$ is a $C^1$ diffeomorphism of $mathbb{R}^{n}$.
My attempt.
(a) If $f(x) = f(y)$, then $Vert x - y Vert = 0$, that is, $x=y$. Take $(f(x_{n}))$ such that $f(x_{n}) to p$. By hypothesis, $(x_{n})$ is Cauchy and by continuity, $f(x_{n}) to f(q) = p$ where $x_{n} to q$. Thus, $f(mathbb{R}^{n})$ is closed.
(b) I dont know how to prove.
(c) I proved in a previous question that
if $f: U to mathbb{R}^{n}$ with $U subset mathbb{R}^{n}$ is $C^1$ with $det Df neq 0$, then $f$ is an open map.
Suppose (b), $f$ is an open map and, therefore, $f(mathbb{R}^{n})$ is open. So, $f(mathbb{R}^{n})$ is non-empty clopen set, that is, $f(mathbb{R}^{n}) = mathbb{R}^{n}$. By Inverse Function Theorem, $f$ is a local diffeomorphism. Note that $f: mathbb{R}^{n} to mathbb{R}^{n}$ is bijective and, since $f$ is a local diffeomorphism, $f^{-1}$ is differentiable for each $x$. Then $f$ is a $C^1$ diffeormorphism of $mathbb{R}^{n}$.
Is (a) and (c) corrects? I need help with item (b).
real-analysis derivatives inverse-function-theorem
asked Jan 23 at 20:56
Lucas CorrêaLucas Corrêa
1,6231321
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}^{n} to mathbb{R}^{n}$ be a function of class $C^1$ and suppose that there is $k>0$ such that
$$Vert f(x) - f(y) Vert geq k Vert x - y Vert$$
for any $x,y in mathbb{R}^{n}$.
(a) Prove that $f$ is injective and $f(mathbb{R}^{n})$ is closed.
(b) Prove that $f'$ is invertible.
(c) Prove that $f(mathbb{R}^{n})$ is open. Conclude that $f$ is a $C^1$ diffeomorphism of $mathbb{R}^{n}$.
My attempt.
(a) If $f(x) = f(y)$, then $Vert x - y Vert = 0$, that is, $x=y$. Take $(f(x_{n}))$ such that $f(x_{n}) to p$. By hypothesis, $(x_{n})$ is Cauchy and by continuity, $f(x_{n}) to f(q) = p$ where $x_{n} to q$. Thus, $f(mathbb{R}^{n})$ is closed.
(b) I dont know how to prove.
(c) I proved in a previous question that
if $f: U to mathbb{R}^{n}$ with $U subset mathbb{R}^{n}$ is $C^1$ with $det Df neq 0$, then $f$ is an open map.
Suppose (b), $f$ is an open map and, therefore, $f(mathbb{R}^{n})$ is open. So, $f(mathbb{R}^{n})$ is non-empty clopen set, that is, $f(mathbb{R}^{n}) = mathbb{R}^{n}$. By Inverse Function Theorem, $f$ is a local diffeomorphism. Note that $f: mathbb{R}^{n} to mathbb{R}^{n}$ is bijective and, since $f$ is a local diffeomorphism, $f^{-1}$ is differentiable for each $x$. Then $f$ is a $C^1$ diffeormorphism of $mathbb{R}^{n}$.
Is (a) and (c) corrects? I need help with item (b).
real-analysis derivatives inverse-function-theorem
asked Jan 23 at 20:56
Lucas CorrêaLucas Corrêa
1,6231321
$endgroup$
2
$begingroup$
Parts (a) and (c) look good. Re part (b), the question should probably say to prove that $Df$ is invertible.
$endgroup$
– Jordan Green
Jan 23 at 21:33
1
$begingroup$
For part (b), what if there is a point $x$ so that $Df(x)$ is not invertible?
$endgroup$
– Jordan Green
Jan 23 at 21:34
$begingroup$
If $Df$ is not invertible for some $x$, there is a unity vector $v$ such that $D_{x}f(v) = 0$. But, $D_{x}f(v) = lim_{t to 0}frac{Vert f(x+tv) - f(x)Vert}{|t|} = 0$, that is, for $epsilon >0$ there is $delta > 0$ such that $|t|<delta$ implies $Vert f(x+tv) - f(x)Vert < k|t| = kVert tv Vert = kVert x+tv - x$, a contradiction.
$endgroup$
– Lucas Corrêa
Jan 23 at 22:03
add a comment |
$begingroup$
Let $f:mathbb{R}^{n} to mathbb{R}^{n}$ be a function of class $C^1$ and suppose that there is $k>0$ such that
$$Vert f(x) - f(y) Vert geq k Vert x - y Vert$$
for any $x,y in mathbb{R}^{n}$.
(a) Prove that $f$ is injective and $f(mathbb{R}^{n})$ is closed.
(b) Prove that $f'$ is invertible.
(c) Prove that $f(mathbb{R}^{n})$ is open. Conclude that $f$ is a $C^1$ diffeomorphism of $mathbb{R}^{n}$.
My attempt.
(a) If $f(x) = f(y)$, then $Vert x - y Vert = 0$, that is, $x=y$. Take $(f(x_{n}))$ such that $f(x_{n}) to p$. By hypothesis, $(x_{n})$ is Cauchy and by continuity, $f(x_{n}) to f(q) = p$ where $x_{n} to q$. Thus, $f(mathbb{R}^{n})$ is closed.
(b) I dont know how to prove.
(c) I proved in a previous question that
if $f: U to mathbb{R}^{n}$ with $U subset mathbb{R}^{n}$ is $C^1$ with $det Df neq 0$, then $f$ is an open map.
Suppose (b), $f$ is an open map and, therefore, $f(mathbb{R}^{n})$ is open. So, $f(mathbb{R}^{n})$ is non-empty clopen set, that is, $f(mathbb{R}^{n}) = mathbb{R}^{n}$. By Inverse Function Theorem, $f$ is a local diffeomorphism. Note that $f: mathbb{R}^{n} to mathbb{R}^{n}$ is bijective and, since $f$ is a local diffeomorphism, $f^{-1}$ is differentiable for each $x$. Then $f$ is a $C^1$ diffeormorphism of $mathbb{R}^{n}$.
Is (a) and (c) corrects? I need help with item (b).
real-analysis derivatives inverse-function-theorem
asked Jan 23 at 20:56
Lucas CorrêaLucas Corrêa
1,6231321
$endgroup$
Let $f:mathbb{R}^{n} to mathbb{R}^{n}$ be a function of class $C^1$ and suppose that there is $k>0$ such that
$$Vert f(x) - f(y) Vert geq k Vert x - y Vert$$
for any $x,y in mathbb{R}^{n}$.
(a) Prove that $f$ is injective and $f(mathbb{R}^{n})$ is closed.
(b) Prove that $f'$ is invertible.
(c) Prove that $f(mathbb{R}^{n})$ is open. Conclude that $f$ is a $C^1$ diffeomorphism of $mathbb{R}^{n}$.
My attempt.
(a) If $f(x) = f(y)$, then $Vert x - y Vert = 0$, that is, $x=y$. Take $(f(x_{n}))$ such that $f(x_{n}) to p$. By hypothesis, $(x_{n})$ is Cauchy and by continuity, $f(x_{n}) to f(q) = p$ where $x_{n} to q$. Thus, $f(mathbb{R}^{n})$ is closed.
(b) I dont know how to prove.
(c) I proved in a previous question that
if $f: U to mathbb{R}^{n}$ with $U subset mathbb{R}^{n}$ is $C^1$ with $det Df neq 0$, then $f$ is an open map.
Suppose (b), $f$ is an open map and, therefore, $f(mathbb{R}^{n})$ is open. So, $f(mathbb{R}^{n})$ is non-empty clopen set, that is, $f(mathbb{R}^{n}) = mathbb{R}^{n}$. By Inverse Function Theorem, $f$ is a local diffeomorphism. Note that $f: mathbb{R}^{n} to mathbb{R}^{n}$ is bijective and, since $f$ is a local diffeomorphism, $f^{-1}$ is differentiable for each $x$. Then $f$ is a $C^1$ diffeormorphism of $mathbb{R}^{n}$.
Is (a) and (c) corrects? I need help with item (b).
real-analysis derivatives inverse-function-theorem
real-analysis derivatives inverse-function-theorem
asked Jan 23 at 20:56
Lucas CorrêaLucas Corrêa
1,6231321
asked Jan 23 at 20:56
Lucas CorrêaLucas Corrêa
1,6231321
asked Jan 23 at 20:56
Lucas CorrêaLucas Corrêa
1,6231321
asked Jan 23 at 20:56
Lucas CorrêaLucas Corrêa
1,6231321
asked Jan 23 at 20:56
Lucas CorrêaLucas Corrêa
1,6231321
1,6231321
2
$begingroup$
Parts (a) and (c) look good. Re part (b), the question should probably say to prove that $Df$ is invertible.
$endgroup$
– Jordan Green
Jan 23 at 21:33
1
$begingroup$
For part (b), what if there is a point $x$ so that $Df(x)$ is not invertible?
$endgroup$
– Jordan Green
Jan 23 at 21:34
$begingroup$
If $Df$ is not invertible for some $x$, there is a unity vector $v$ such that $D_{x}f(v) = 0$. But, $D_{x}f(v) = lim_{t to 0}frac{Vert f(x+tv) - f(x)Vert}{|t|} = 0$, that is, for $epsilon >0$ there is $delta > 0$ such that $|t|<delta$ implies $Vert f(x+tv) - f(x)Vert < k|t| = kVert tv Vert = kVert x+tv - x$, a contradiction.
$endgroup$
– Lucas Corrêa
Jan 23 at 22:03
add a comment |
2
$begingroup$
Parts (a) and (c) look good. Re part (b), the question should probably say to prove that $Df$ is invertible.
$endgroup$
– Jordan Green
Jan 23 at 21:33
1
$begingroup$
For part (b), what if there is a point $x$ so that $Df(x)$ is not invertible?
$endgroup$
– Jordan Green
Jan 23 at 21:34
$begingroup$
If $Df$ is not invertible for some $x$, there is a unity vector $v$ such that $D_{x}f(v) = 0$. But, $D_{x}f(v) = lim_{t to 0}frac{Vert f(x+tv) - f(x)Vert}{|t|} = 0$, that is, for $epsilon >0$ there is $delta > 0$ such that $|t|<delta$ implies $Vert f(x+tv) - f(x)Vert < k|t| = kVert tv Vert = kVert x+tv - x$, a contradiction.
$endgroup$
– Lucas Corrêa
Jan 23 at 22:03
2
2
$begingroup$
Parts (a) and (c) look good. Re part (b), the question should probably say to prove that $Df$ is invertible.
$endgroup$
– Jordan Green
Jan 23 at 21:33
$begingroup$
Parts (a) and (c) look good. Re part (b), the question should probably say to prove that $Df$ is invertible.
$endgroup$
– Jordan Green
Jan 23 at 21:33
1
1
$begingroup$
For part (b), what if there is a point $x$ so that $Df(x)$ is not invertible?
$endgroup$
– Jordan Green
Jan 23 at 21:34
$begingroup$
For part (b), what if there is a point $x$ so that $Df(x)$ is not invertible?
$endgroup$
– Jordan Green
Jan 23 at 21:34
$begingroup$
If $Df$ is not invertible for some $x$, there is a unity vector $v$ such that $D_{x}f(v) = 0$. But, $D_{x}f(v) = lim_{t to 0}frac{Vert f(x+tv) - f(x)Vert}{|t|} = 0$, that is, for $epsilon >0$ there is $delta > 0$ such that $|t|<delta$ implies $Vert f(x+tv) - f(x)Vert < k|t| = kVert tv Vert = kVert x+tv - x$, a contradiction.
$endgroup$
– Lucas Corrêa
Jan 23 at 22:03
$begingroup$
If $Df$ is not invertible for some $x$, there is a unity vector $v$ such that $D_{x}f(v) = 0$. But, $D_{x}f(v) = lim_{t to 0}frac{Vert f(x+tv) - f(x)Vert}{|t|} = 0$, that is, for $epsilon >0$ there is $delta > 0$ such that $|t|<delta$ implies $Vert f(x+tv) - f(x)Vert < k|t| = kVert tv Vert = kVert x+tv - x$, a contradiction.
$endgroup$
– Lucas Corrêa
Jan 23 at 22:03
add a comment |
1 Answer
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$begingroup$
Let $yin mathbb{R}^{n}setminus{0 }$, $tinmathbb{R}$. Then by definition of $Df(x_0)$
$$lim_{trightarrow0}frac{Vert{f(x_0+ty)-f(x_0)-Df(x_0)cdot ty}Vert}{Vert tyVert}=0$$
Then:
$$frac{Vert{f(x_0+ty)-f(x_0)-Dfcdot ty}Vert}{Vert tyVert}geqfrac{Vert{f(x_0+ty)-f(x_0)Vert-Vert Df(x_0)cdot ty}Vert}{Vert tyVert}geq frac{kVert ty Vert-Vert Df(x_0)cdot tyVert}{Vert ty Vert}=k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}$$
This leads to:
$$0=lim_{trightarrow0}frac{Vert{f(x_0+ty)-f(x_0)-Df(x_0)cdot ty}Vert}{Vert tyVert}geqlim_{trightarrow0}k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}=k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}$$
$$Rightarrow Vert Df(x_0) cdot y Vertgeq k Vert y Vert >0$$
So $Df(x_0)$ is invertible.
(b) and (c) are correct.
answered Jan 23 at 22:00
Martin ErhardtMartin Erhardt
21019
$endgroup$
add a comment |
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$begingroup$
Let $yin mathbb{R}^{n}setminus{0 }$, $tinmathbb{R}$. Then by definition of $Df(x_0)$
$$lim_{trightarrow0}frac{Vert{f(x_0+ty)-f(x_0)-Df(x_0)cdot ty}Vert}{Vert tyVert}=0$$
Then:
$$frac{Vert{f(x_0+ty)-f(x_0)-Dfcdot ty}Vert}{Vert tyVert}geqfrac{Vert{f(x_0+ty)-f(x_0)Vert-Vert Df(x_0)cdot ty}Vert}{Vert tyVert}geq frac{kVert ty Vert-Vert Df(x_0)cdot tyVert}{Vert ty Vert}=k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}$$
This leads to:
$$0=lim_{trightarrow0}frac{Vert{f(x_0+ty)-f(x_0)-Df(x_0)cdot ty}Vert}{Vert tyVert}geqlim_{trightarrow0}k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}=k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}$$
$$Rightarrow Vert Df(x_0) cdot y Vertgeq k Vert y Vert >0$$
So $Df(x_0)$ is invertible.
(b) and (c) are correct.
answered Jan 23 at 22:00
Martin ErhardtMartin Erhardt
21019
$endgroup$
add a comment |
$begingroup$
Let $yin mathbb{R}^{n}setminus{0 }$, $tinmathbb{R}$. Then by definition of $Df(x_0)$
$$lim_{trightarrow0}frac{Vert{f(x_0+ty)-f(x_0)-Df(x_0)cdot ty}Vert}{Vert tyVert}=0$$
Then:
$$frac{Vert{f(x_0+ty)-f(x_0)-Dfcdot ty}Vert}{Vert tyVert}geqfrac{Vert{f(x_0+ty)-f(x_0)Vert-Vert Df(x_0)cdot ty}Vert}{Vert tyVert}geq frac{kVert ty Vert-Vert Df(x_0)cdot tyVert}{Vert ty Vert}=k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}$$
This leads to:
$$0=lim_{trightarrow0}frac{Vert{f(x_0+ty)-f(x_0)-Df(x_0)cdot ty}Vert}{Vert tyVert}geqlim_{trightarrow0}k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}=k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}$$
$$Rightarrow Vert Df(x_0) cdot y Vertgeq k Vert y Vert >0$$
So $Df(x_0)$ is invertible.
(b) and (c) are correct.
answered Jan 23 at 22:00
Martin ErhardtMartin Erhardt
21019
$endgroup$
add a comment |
$begingroup$
Let $yin mathbb{R}^{n}setminus{0 }$, $tinmathbb{R}$. Then by definition of $Df(x_0)$
$$lim_{trightarrow0}frac{Vert{f(x_0+ty)-f(x_0)-Df(x_0)cdot ty}Vert}{Vert tyVert}=0$$
Then:
$$frac{Vert{f(x_0+ty)-f(x_0)-Dfcdot ty}Vert}{Vert tyVert}geqfrac{Vert{f(x_0+ty)-f(x_0)Vert-Vert Df(x_0)cdot ty}Vert}{Vert tyVert}geq frac{kVert ty Vert-Vert Df(x_0)cdot tyVert}{Vert ty Vert}=k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}$$
This leads to:
$$0=lim_{trightarrow0}frac{Vert{f(x_0+ty)-f(x_0)-Df(x_0)cdot ty}Vert}{Vert tyVert}geqlim_{trightarrow0}k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}=k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}$$
$$Rightarrow Vert Df(x_0) cdot y Vertgeq k Vert y Vert >0$$
So $Df(x_0)$ is invertible.
(b) and (c) are correct.
answered Jan 23 at 22:00
Martin ErhardtMartin Erhardt
21019
$endgroup$
Let $yin mathbb{R}^{n}setminus{0 }$, $tinmathbb{R}$. Then by definition of $Df(x_0)$
$$lim_{trightarrow0}frac{Vert{f(x_0+ty)-f(x_0)-Df(x_0)cdot ty}Vert}{Vert tyVert}=0$$
Then:
$$frac{Vert{f(x_0+ty)-f(x_0)-Dfcdot ty}Vert}{Vert tyVert}geqfrac{Vert{f(x_0+ty)-f(x_0)Vert-Vert Df(x_0)cdot ty}Vert}{Vert tyVert}geq frac{kVert ty Vert-Vert Df(x_0)cdot tyVert}{Vert ty Vert}=k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}$$
This leads to:
$$0=lim_{trightarrow0}frac{Vert{f(x_0+ty)-f(x_0)-Df(x_0)cdot ty}Vert}{Vert tyVert}geqlim_{trightarrow0}k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}=k-frac{Vert Df(x_0) cdot y Vert}{Vert y Vert}$$
$$Rightarrow Vert Df(x_0) cdot y Vertgeq k Vert y Vert >0$$
So $Df(x_0)$ is invertible.
(b) and (c) are correct.
answered Jan 23 at 22:00
Martin ErhardtMartin Erhardt
21019
answered Jan 23 at 22:00
Martin ErhardtMartin Erhardt
21019
answered Jan 23 at 22:00
Martin ErhardtMartin Erhardt
21019
answered Jan 23 at 22:00
Martin ErhardtMartin Erhardt
21019
21019
add a comment |
add a comment |
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$begingroup$
Parts (a) and (c) look good. Re part (b), the question should probably say to prove that $Df$ is invertible.
$endgroup$
– Jordan Green
Jan 23 at 21:33
1
$begingroup$
For part (b), what if there is a point $x$ so that $Df(x)$ is not invertible?
$endgroup$
– Jordan Green
Jan 23 at 21:34
$begingroup$
If $Df$ is not invertible for some $x$, there is a unity vector $v$ such that $D_{x}f(v) = 0$. But, $D_{x}f(v) = lim_{t to 0}frac{Vert f(x+tv) - f(x)Vert}{|t|} = 0$, that is, for $epsilon >0$ there is $delta > 0$ such that $|t|<delta$ implies $Vert f(x+tv) - f(x)Vert < k|t| = kVert tv Vert = kVert x+tv - x$, a contradiction.
$endgroup$
– Lucas Corrêa
Jan 23 at 22:03