Show $T(x) := x+sf(x)$ is a bijection with $f$ Lipschitz and $vert s vert lt frac{1}{L}$
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1
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Let $f: mathbb{R}^n rightarrow mathbb{R}^n$ be a function and $L gt 0$ such that
$Vert f(x)-f(y)Vert leq L Vert x-yVert $for all $x,y in mathbb{R}^n$
- Show that $T(x) := x+sf(x)$ defines a bijection on $mathbb{R}^n$ if $vert s vert lt frac{1}{L}$.
- Prove that there is a constant $tilde{L} gt 0$, such that
$Vert T^{-1}(x)-T^{-1}(y)Vert leq tilde{L} Vert x-y Vert $ for all $x,y in mathbb{R}^n$
I can prove that $T$ is injective but I am currently struggling with showing that it is also surjective. I don't really now how to approach 2. - I mean I know that the inverse of $T$ exists I thought about using the surjectivity in order to rewrite $x,y$ as $T(tilde{x}),T(tilde{y})$.
real-analysis analysis numerical-methods normed-spaces lipschitz-functions
edited Nov 23 at 22:23
Scientifica
6,13141332
asked Nov 23 at 21:32
thehardyreader
19819
add a comment |
up vote
1
down vote
favorite
Let $f: mathbb{R}^n rightarrow mathbb{R}^n$ be a function and $L gt 0$ such that
$Vert f(x)-f(y)Vert leq L Vert x-yVert $for all $x,y in mathbb{R}^n$
- Show that $T(x) := x+sf(x)$ defines a bijection on $mathbb{R}^n$ if $vert s vert lt frac{1}{L}$.
- Prove that there is a constant $tilde{L} gt 0$, such that
$Vert T^{-1}(x)-T^{-1}(y)Vert leq tilde{L} Vert x-y Vert $ for all $x,y in mathbb{R}^n$
I can prove that $T$ is injective but I am currently struggling with showing that it is also surjective. I don't really now how to approach 2. - I mean I know that the inverse of $T$ exists I thought about using the surjectivity in order to rewrite $x,y$ as $T(tilde{x}),T(tilde{y})$.
real-analysis analysis numerical-methods normed-spaces lipschitz-functions
edited Nov 23 at 22:23
Scientifica
6,13141332
asked Nov 23 at 21:32
thehardyreader
19819
What have you tried?
– Ben W
Nov 23 at 21:35
For the proof of the surjectivity I have tried to show that $F_y(x):=T(x)-y$ is a contraction. Then it would follow that it has a unique fixed point which would give me the surjectivity. But I didn't manage to estimate a lipschitz constant which is smaller than $1$.
– thehardyreader
Nov 23 at 22:14
@thehardyreader trying to prove the existence of a unique fixed point is a nice idea, but that's not the right function to consider, because if $F_y(x)$ has a fixed point $x_0$, that would mean that $T(x_0)-y=x_0$, i.e, $T(x_0)=y+x_0$, which doesn't necessarily show surjectivity.
– Scientifica
Nov 23 at 22:16
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f: mathbb{R}^n rightarrow mathbb{R}^n$ be a function and $L gt 0$ such that
$Vert f(x)-f(y)Vert leq L Vert x-yVert $for all $x,y in mathbb{R}^n$
- Show that $T(x) := x+sf(x)$ defines a bijection on $mathbb{R}^n$ if $vert s vert lt frac{1}{L}$.
- Prove that there is a constant $tilde{L} gt 0$, such that
$Vert T^{-1}(x)-T^{-1}(y)Vert leq tilde{L} Vert x-y Vert $ for all $x,y in mathbb{R}^n$
I can prove that $T$ is injective but I am currently struggling with showing that it is also surjective. I don't really now how to approach 2. - I mean I know that the inverse of $T$ exists I thought about using the surjectivity in order to rewrite $x,y$ as $T(tilde{x}),T(tilde{y})$.
real-analysis analysis numerical-methods normed-spaces lipschitz-functions
edited Nov 23 at 22:23
Scientifica
6,13141332
asked Nov 23 at 21:32
thehardyreader
19819
Let $f: mathbb{R}^n rightarrow mathbb{R}^n$ be a function and $L gt 0$ such that
$Vert f(x)-f(y)Vert leq L Vert x-yVert $for all $x,y in mathbb{R}^n$
- Show that $T(x) := x+sf(x)$ defines a bijection on $mathbb{R}^n$ if $vert s vert lt frac{1}{L}$.
- Prove that there is a constant $tilde{L} gt 0$, such that
$Vert T^{-1}(x)-T^{-1}(y)Vert leq tilde{L} Vert x-y Vert $ for all $x,y in mathbb{R}^n$
I can prove that $T$ is injective but I am currently struggling with showing that it is also surjective. I don't really now how to approach 2. - I mean I know that the inverse of $T$ exists I thought about using the surjectivity in order to rewrite $x,y$ as $T(tilde{x}),T(tilde{y})$.
real-analysis analysis numerical-methods normed-spaces lipschitz-functions
real-analysis analysis numerical-methods normed-spaces lipschitz-functions
edited Nov 23 at 22:23
Scientifica
6,13141332
asked Nov 23 at 21:32
thehardyreader
19819
edited Nov 23 at 22:23
Scientifica
6,13141332
asked Nov 23 at 21:32
thehardyreader
19819
edited Nov 23 at 22:23
Scientifica
6,13141332
edited Nov 23 at 22:23
Scientifica
6,13141332
edited Nov 23 at 22:23
Scientifica
6,13141332
6,13141332
asked Nov 23 at 21:32
thehardyreader
19819
asked Nov 23 at 21:32
thehardyreader
19819
asked Nov 23 at 21:32
thehardyreader
19819
19819
What have you tried?
– Ben W
Nov 23 at 21:35
For the proof of the surjectivity I have tried to show that $F_y(x):=T(x)-y$ is a contraction. Then it would follow that it has a unique fixed point which would give me the surjectivity. But I didn't manage to estimate a lipschitz constant which is smaller than $1$.
– thehardyreader
Nov 23 at 22:14
@thehardyreader trying to prove the existence of a unique fixed point is a nice idea, but that's not the right function to consider, because if $F_y(x)$ has a fixed point $x_0$, that would mean that $T(x_0)-y=x_0$, i.e, $T(x_0)=y+x_0$, which doesn't necessarily show surjectivity.
– Scientifica
Nov 23 at 22:16
add a comment |
What have you tried?
– Ben W
Nov 23 at 21:35
For the proof of the surjectivity I have tried to show that $F_y(x):=T(x)-y$ is a contraction. Then it would follow that it has a unique fixed point which would give me the surjectivity. But I didn't manage to estimate a lipschitz constant which is smaller than $1$.
– thehardyreader
Nov 23 at 22:14
@thehardyreader trying to prove the existence of a unique fixed point is a nice idea, but that's not the right function to consider, because if $F_y(x)$ has a fixed point $x_0$, that would mean that $T(x_0)-y=x_0$, i.e, $T(x_0)=y+x_0$, which doesn't necessarily show surjectivity.
– Scientifica
Nov 23 at 22:16
What have you tried?
– Ben W
Nov 23 at 21:35
What have you tried?
– Ben W
Nov 23 at 21:35
For the proof of the surjectivity I have tried to show that $F_y(x):=T(x)-y$ is a contraction. Then it would follow that it has a unique fixed point which would give me the surjectivity. But I didn't manage to estimate a lipschitz constant which is smaller than $1$.
– thehardyreader
Nov 23 at 22:14
For the proof of the surjectivity I have tried to show that $F_y(x):=T(x)-y$ is a contraction. Then it would follow that it has a unique fixed point which would give me the surjectivity. But I didn't manage to estimate a lipschitz constant which is smaller than $1$.
– thehardyreader
Nov 23 at 22:14
@thehardyreader trying to prove the existence of a unique fixed point is a nice idea, but that's not the right function to consider, because if $F_y(x)$ has a fixed point $x_0$, that would mean that $T(x_0)-y=x_0$, i.e, $T(x_0)=y+x_0$, which doesn't necessarily show surjectivity.
– Scientifica
Nov 23 at 22:16
@thehardyreader trying to prove the existence of a unique fixed point is a nice idea, but that's not the right function to consider, because if $F_y(x)$ has a fixed point $x_0$, that would mean that $T(x_0)-y=x_0$, i.e, $T(x_0)=y+x_0$, which doesn't necessarily show surjectivity.
– Scientifica
Nov 23 at 22:16
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
You want to prove that given $yinmathbb R^n$, there exists $xinmathbb R^n$ such that $T(x)=y$, i.e, $y-sf(x)=x$, i.e, $x$ is a fixed point of $y-sf(x)$. So consider the function $g_y(x)=y-sf(x)$. Prove that $g_y$ is $|s|L$-Lipschitz. Since $(mathbb R^n,||$) is complete and $|s|L<1$, by the Banach fixed-point theorem, $g_y$ has a unique fixed point. (this also shows that $T$ is also injective).
Notice that, as shown in 1., for $xinmathbb R^n$, $T^{-1}(x)$ is the unique fixed point of $g_x$. So given $x,yinmathbb R^n$, let $x_0=T^{-1}(x)$ and $y_0=T^{-1}(y)$. So you know that $x_0=g_x(x_0)$ and $y_0=g_y(y_0)$. So $|T^{-1}(x)-T^{-1}(y)|=|g_x(x_0)-g_y(y_0)|cdots$
answered Nov 23 at 22:13
Scientifica
6,13141332
Thank you! I have understood the first part, and working on the second: $...=Vert x - s f(x_{0}) - y + f(y_0) Vert leq Vert x-y Vert + vert s vert L Vert x_0 - y_0Vert = Vert x-y Vert + vert s vert L Vert T^{-1}(x) - T^{-1}(y) Vert $ Subtracting the last term from both sides and dividing by $1-vert s vert L$
– thehardyreader
Nov 24 at 9:22
@thehardyreader It's a pleasure :) That's right!! Well done!
– Scientifica
Nov 24 at 9:39
@thehardyreader By the way, out of curiosity, is this exercise from a textbook (probably numerical analysis, looking at your tag). If so, what's the name of the book?
– Scientifica
Nov 24 at 9:40
I unfortunately don't know the original source of this exercise - I got this exercise during my numerical analysis course. I don't know if the professor made it up or took it from a textbook. Sorry, that I can't help - I ask him next week, in order to return the favor.
– thehardyreader
Nov 24 at 10:06
No problem :) And people don't help expecting a return of favor ;) I would have asked the question even if I wasn't the one who answered yours ;) Good luck !!
– Scientifica
Nov 24 at 10:09
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You want to prove that given $yinmathbb R^n$, there exists $xinmathbb R^n$ such that $T(x)=y$, i.e, $y-sf(x)=x$, i.e, $x$ is a fixed point of $y-sf(x)$. So consider the function $g_y(x)=y-sf(x)$. Prove that $g_y$ is $|s|L$-Lipschitz. Since $(mathbb R^n,||$) is complete and $|s|L<1$, by the Banach fixed-point theorem, $g_y$ has a unique fixed point. (this also shows that $T$ is also injective).
Notice that, as shown in 1., for $xinmathbb R^n$, $T^{-1}(x)$ is the unique fixed point of $g_x$. So given $x,yinmathbb R^n$, let $x_0=T^{-1}(x)$ and $y_0=T^{-1}(y)$. So you know that $x_0=g_x(x_0)$ and $y_0=g_y(y_0)$. So $|T^{-1}(x)-T^{-1}(y)|=|g_x(x_0)-g_y(y_0)|cdots$
answered Nov 23 at 22:13
Scientifica
6,13141332
Thank you! I have understood the first part, and working on the second: $...=Vert x - s f(x_{0}) - y + f(y_0) Vert leq Vert x-y Vert + vert s vert L Vert x_0 - y_0Vert = Vert x-y Vert + vert s vert L Vert T^{-1}(x) - T^{-1}(y) Vert $ Subtracting the last term from both sides and dividing by $1-vert s vert L$
– thehardyreader
Nov 24 at 9:22
@thehardyreader It's a pleasure :) That's right!! Well done!
– Scientifica
Nov 24 at 9:39
@thehardyreader By the way, out of curiosity, is this exercise from a textbook (probably numerical analysis, looking at your tag). If so, what's the name of the book?
– Scientifica
Nov 24 at 9:40
I unfortunately don't know the original source of this exercise - I got this exercise during my numerical analysis course. I don't know if the professor made it up or took it from a textbook. Sorry, that I can't help - I ask him next week, in order to return the favor.
– thehardyreader
Nov 24 at 10:06
No problem :) And people don't help expecting a return of favor ;) I would have asked the question even if I wasn't the one who answered yours ;) Good luck !!
– Scientifica
Nov 24 at 10:09
add a comment |
up vote
2
down vote
accepted
You want to prove that given $yinmathbb R^n$, there exists $xinmathbb R^n$ such that $T(x)=y$, i.e, $y-sf(x)=x$, i.e, $x$ is a fixed point of $y-sf(x)$. So consider the function $g_y(x)=y-sf(x)$. Prove that $g_y$ is $|s|L$-Lipschitz. Since $(mathbb R^n,||$) is complete and $|s|L<1$, by the Banach fixed-point theorem, $g_y$ has a unique fixed point. (this also shows that $T$ is also injective).
Notice that, as shown in 1., for $xinmathbb R^n$, $T^{-1}(x)$ is the unique fixed point of $g_x$. So given $x,yinmathbb R^n$, let $x_0=T^{-1}(x)$ and $y_0=T^{-1}(y)$. So you know that $x_0=g_x(x_0)$ and $y_0=g_y(y_0)$. So $|T^{-1}(x)-T^{-1}(y)|=|g_x(x_0)-g_y(y_0)|cdots$
answered Nov 23 at 22:13
Scientifica
6,13141332
Thank you! I have understood the first part, and working on the second: $...=Vert x - s f(x_{0}) - y + f(y_0) Vert leq Vert x-y Vert + vert s vert L Vert x_0 - y_0Vert = Vert x-y Vert + vert s vert L Vert T^{-1}(x) - T^{-1}(y) Vert $ Subtracting the last term from both sides and dividing by $1-vert s vert L$
– thehardyreader
Nov 24 at 9:22
@thehardyreader It's a pleasure :) That's right!! Well done!
– Scientifica
Nov 24 at 9:39
@thehardyreader By the way, out of curiosity, is this exercise from a textbook (probably numerical analysis, looking at your tag). If so, what's the name of the book?
– Scientifica
Nov 24 at 9:40
I unfortunately don't know the original source of this exercise - I got this exercise during my numerical analysis course. I don't know if the professor made it up or took it from a textbook. Sorry, that I can't help - I ask him next week, in order to return the favor.
– thehardyreader
Nov 24 at 10:06
No problem :) And people don't help expecting a return of favor ;) I would have asked the question even if I wasn't the one who answered yours ;) Good luck !!
– Scientifica
Nov 24 at 10:09
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You want to prove that given $yinmathbb R^n$, there exists $xinmathbb R^n$ such that $T(x)=y$, i.e, $y-sf(x)=x$, i.e, $x$ is a fixed point of $y-sf(x)$. So consider the function $g_y(x)=y-sf(x)$. Prove that $g_y$ is $|s|L$-Lipschitz. Since $(mathbb R^n,||$) is complete and $|s|L<1$, by the Banach fixed-point theorem, $g_y$ has a unique fixed point. (this also shows that $T$ is also injective).
Notice that, as shown in 1., for $xinmathbb R^n$, $T^{-1}(x)$ is the unique fixed point of $g_x$. So given $x,yinmathbb R^n$, let $x_0=T^{-1}(x)$ and $y_0=T^{-1}(y)$. So you know that $x_0=g_x(x_0)$ and $y_0=g_y(y_0)$. So $|T^{-1}(x)-T^{-1}(y)|=|g_x(x_0)-g_y(y_0)|cdots$
answered Nov 23 at 22:13
Scientifica
6,13141332
You want to prove that given $yinmathbb R^n$, there exists $xinmathbb R^n$ such that $T(x)=y$, i.e, $y-sf(x)=x$, i.e, $x$ is a fixed point of $y-sf(x)$. So consider the function $g_y(x)=y-sf(x)$. Prove that $g_y$ is $|s|L$-Lipschitz. Since $(mathbb R^n,||$) is complete and $|s|L<1$, by the Banach fixed-point theorem, $g_y$ has a unique fixed point. (this also shows that $T$ is also injective).
Notice that, as shown in 1., for $xinmathbb R^n$, $T^{-1}(x)$ is the unique fixed point of $g_x$. So given $x,yinmathbb R^n$, let $x_0=T^{-1}(x)$ and $y_0=T^{-1}(y)$. So you know that $x_0=g_x(x_0)$ and $y_0=g_y(y_0)$. So $|T^{-1}(x)-T^{-1}(y)|=|g_x(x_0)-g_y(y_0)|cdots$
answered Nov 23 at 22:13
Scientifica
6,13141332
edited Nov 23 at 22:22
answered Nov 23 at 22:13
Scientifica
6,13141332
answered Nov 23 at 22:13
Scientifica
6,13141332
answered Nov 23 at 22:13
Scientifica
6,13141332
6,13141332
Thank you! I have understood the first part, and working on the second: $...=Vert x - s f(x_{0}) - y + f(y_0) Vert leq Vert x-y Vert + vert s vert L Vert x_0 - y_0Vert = Vert x-y Vert + vert s vert L Vert T^{-1}(x) - T^{-1}(y) Vert $ Subtracting the last term from both sides and dividing by $1-vert s vert L$
– thehardyreader
Nov 24 at 9:22
@thehardyreader It's a pleasure :) That's right!! Well done!
– Scientifica
Nov 24 at 9:39
@thehardyreader By the way, out of curiosity, is this exercise from a textbook (probably numerical analysis, looking at your tag). If so, what's the name of the book?
– Scientifica
Nov 24 at 9:40
I unfortunately don't know the original source of this exercise - I got this exercise during my numerical analysis course. I don't know if the professor made it up or took it from a textbook. Sorry, that I can't help - I ask him next week, in order to return the favor.
– thehardyreader
Nov 24 at 10:06
No problem :) And people don't help expecting a return of favor ;) I would have asked the question even if I wasn't the one who answered yours ;) Good luck !!
– Scientifica
Nov 24 at 10:09
add a comment |
Thank you! I have understood the first part, and working on the second: $...=Vert x - s f(x_{0}) - y + f(y_0) Vert leq Vert x-y Vert + vert s vert L Vert x_0 - y_0Vert = Vert x-y Vert + vert s vert L Vert T^{-1}(x) - T^{-1}(y) Vert $ Subtracting the last term from both sides and dividing by $1-vert s vert L$
– thehardyreader
Nov 24 at 9:22
@thehardyreader It's a pleasure :) That's right!! Well done!
– Scientifica
Nov 24 at 9:39
@thehardyreader By the way, out of curiosity, is this exercise from a textbook (probably numerical analysis, looking at your tag). If so, what's the name of the book?
– Scientifica
Nov 24 at 9:40
I unfortunately don't know the original source of this exercise - I got this exercise during my numerical analysis course. I don't know if the professor made it up or took it from a textbook. Sorry, that I can't help - I ask him next week, in order to return the favor.
– thehardyreader
Nov 24 at 10:06
No problem :) And people don't help expecting a return of favor ;) I would have asked the question even if I wasn't the one who answered yours ;) Good luck !!
– Scientifica
Nov 24 at 10:09
Thank you! I have understood the first part, and working on the second: $...=Vert x - s f(x_{0}) - y + f(y_0) Vert leq Vert x-y Vert + vert s vert L Vert x_0 - y_0Vert = Vert x-y Vert + vert s vert L Vert T^{-1}(x) - T^{-1}(y) Vert $ Subtracting the last term from both sides and dividing by $1-vert s vert L$
– thehardyreader
Nov 24 at 9:22
Thank you! I have understood the first part, and working on the second: $...=Vert x - s f(x_{0}) - y + f(y_0) Vert leq Vert x-y Vert + vert s vert L Vert x_0 - y_0Vert = Vert x-y Vert + vert s vert L Vert T^{-1}(x) - T^{-1}(y) Vert $ Subtracting the last term from both sides and dividing by $1-vert s vert L$
– thehardyreader
Nov 24 at 9:22
@thehardyreader It's a pleasure :) That's right!! Well done!
– Scientifica
Nov 24 at 9:39
@thehardyreader It's a pleasure :) That's right!! Well done!
– Scientifica
Nov 24 at 9:39
@thehardyreader By the way, out of curiosity, is this exercise from a textbook (probably numerical analysis, looking at your tag). If so, what's the name of the book?
– Scientifica
Nov 24 at 9:40
@thehardyreader By the way, out of curiosity, is this exercise from a textbook (probably numerical analysis, looking at your tag). If so, what's the name of the book?
– Scientifica
Nov 24 at 9:40
I unfortunately don't know the original source of this exercise - I got this exercise during my numerical analysis course. I don't know if the professor made it up or took it from a textbook. Sorry, that I can't help - I ask him next week, in order to return the favor.
– thehardyreader
Nov 24 at 10:06
I unfortunately don't know the original source of this exercise - I got this exercise during my numerical analysis course. I don't know if the professor made it up or took it from a textbook. Sorry, that I can't help - I ask him next week, in order to return the favor.
– thehardyreader
Nov 24 at 10:06
No problem :) And people don't help expecting a return of favor ;) I would have asked the question even if I wasn't the one who answered yours ;) Good luck !!
– Scientifica
Nov 24 at 10:09
No problem :) And people don't help expecting a return of favor ;) I would have asked the question even if I wasn't the one who answered yours ;) Good luck !!
– Scientifica
Nov 24 at 10:09
add a comment |
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What have you tried?
– Ben W
Nov 23 at 21:35
For the proof of the surjectivity I have tried to show that $F_y(x):=T(x)-y$ is a contraction. Then it would follow that it has a unique fixed point which would give me the surjectivity. But I didn't manage to estimate a lipschitz constant which is smaller than $1$.
– thehardyreader
Nov 23 at 22:14
@thehardyreader trying to prove the existence of a unique fixed point is a nice idea, but that's not the right function to consider, because if $F_y(x)$ has a fixed point $x_0$, that would mean that $T(x_0)-y=x_0$, i.e, $T(x_0)=y+x_0$, which doesn't necessarily show surjectivity.
– Scientifica
Nov 23 at 22:16