Given that derivative of a function is bounded. Prove surjectivity
$begingroup$
Given a differentiable function $f:mathbf{R} to mathbf{R},$ such that $|f'(x)| < c < 1$. Consider a function $g:mathbf{R}^2 to mathbf{R}^2$, such that $g(x,y) = (x+f(y),y+f(x))$. Prove that g is surjective.
My attempt:
determinant of derivative of g is always non-zero. Consider a point $(a,b)$ which we want to show will have a preimage, by implicit function theorem, there's a neighbourhood around the point and also a neighbourhood around the point $(a+f(b),b+f(a))$, such that function is a bijection. Now I wanted to make sure that the distance between $(a,b)$ and $(a+f(b),b+f(a))$ is small so that $(a,b)$ is attainable.
multivariable-calculus inverse-function-theorem diffeomorphism
edited Nov 27 '18 at 5:09
Will M.
2,495315
asked Nov 27 '18 at 4:28
AntonAnton
163
$endgroup$
add a comment |
$begingroup$
Given a differentiable function $f:mathbf{R} to mathbf{R},$ such that $|f'(x)| < c < 1$. Consider a function $g:mathbf{R}^2 to mathbf{R}^2$, such that $g(x,y) = (x+f(y),y+f(x))$. Prove that g is surjective.
My attempt:
determinant of derivative of g is always non-zero. Consider a point $(a,b)$ which we want to show will have a preimage, by implicit function theorem, there's a neighbourhood around the point and also a neighbourhood around the point $(a+f(b),b+f(a))$, such that function is a bijection. Now I wanted to make sure that the distance between $(a,b)$ and $(a+f(b),b+f(a))$ is small so that $(a,b)$ is attainable.
multivariable-calculus inverse-function-theorem diffeomorphism
edited Nov 27 '18 at 5:09
Will M.
2,495315
asked Nov 27 '18 at 4:28
AntonAnton
163
$endgroup$
add a comment |
$begingroup$
Given a differentiable function $f:mathbf{R} to mathbf{R},$ such that $|f'(x)| < c < 1$. Consider a function $g:mathbf{R}^2 to mathbf{R}^2$, such that $g(x,y) = (x+f(y),y+f(x))$. Prove that g is surjective.
My attempt:
determinant of derivative of g is always non-zero. Consider a point $(a,b)$ which we want to show will have a preimage, by implicit function theorem, there's a neighbourhood around the point and also a neighbourhood around the point $(a+f(b),b+f(a))$, such that function is a bijection. Now I wanted to make sure that the distance between $(a,b)$ and $(a+f(b),b+f(a))$ is small so that $(a,b)$ is attainable.
multivariable-calculus inverse-function-theorem diffeomorphism
edited Nov 27 '18 at 5:09
Will M.
2,495315
asked Nov 27 '18 at 4:28
AntonAnton
163
$endgroup$
Given a differentiable function $f:mathbf{R} to mathbf{R},$ such that $|f'(x)| < c < 1$. Consider a function $g:mathbf{R}^2 to mathbf{R}^2$, such that $g(x,y) = (x+f(y),y+f(x))$. Prove that g is surjective.
My attempt:
determinant of derivative of g is always non-zero. Consider a point $(a,b)$ which we want to show will have a preimage, by implicit function theorem, there's a neighbourhood around the point and also a neighbourhood around the point $(a+f(b),b+f(a))$, such that function is a bijection. Now I wanted to make sure that the distance between $(a,b)$ and $(a+f(b),b+f(a))$ is small so that $(a,b)$ is attainable.
multivariable-calculus inverse-function-theorem diffeomorphism
multivariable-calculus inverse-function-theorem diffeomorphism
edited Nov 27 '18 at 5:09
Will M.
2,495315
asked Nov 27 '18 at 4:28
AntonAnton
163
edited Nov 27 '18 at 5:09
Will M.
2,495315
asked Nov 27 '18 at 4:28
AntonAnton
163
edited Nov 27 '18 at 5:09
Will M.
2,495315
edited Nov 27 '18 at 5:09
Will M.
2,495315
edited Nov 27 '18 at 5:09
Will M.
2,495315
2,495315
asked Nov 27 '18 at 4:28
AntonAnton
163
asked Nov 27 '18 at 4:28
AntonAnton
163
asked Nov 27 '18 at 4:28
AntonAnton
163
163
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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I think there is another possible approach. Let $(a,b)$ a point in the plane $mathbb{R}^2$ and we want to find another point $(x,y)$ such that $a=x+f(y)$ and $b=y+f(x)$. So from the last equation we get $y=b-f(x)$. And inserting this into the first equation we get $a=x+f(b-f(x))$. So we reduce the problem to find such an $x$. Let us define $h:mathbb{R}tomathbb{R}$ by the formula
$$
h(x)=x+f(b-f(x))
$$
The derivative is $h'(x)=1-f'(b-f(x))f'(x)$ for every $x$. By hypothesis $0<1-c<h'(x) <1+c$. So integrating the inequality we get
$$
h(0)+(1-c)xleq h(x)leq h(0) + (1+c)x
$$
Thus checking the limits $xto-infty$ and $xto +infty$ we see that $h$ is surjective.
answered Nov 27 '18 at 6:05
Dante GrevinoDante Grevino
96319
$endgroup$
$begingroup$
very noice. This is very simple
$endgroup$
– Anton
Nov 27 '18 at 6:25
$begingroup$
I think this solution is simple and it avoid uses of strong theorems!
$endgroup$
– Dante Grevino
Nov 27 '18 at 6:31
$begingroup$
Alternatively, you can use Banach's Fixed Point Theorem to the function $r(x) = a-f(b-f(x))$.
$endgroup$
– Dante Grevino
Nov 27 '18 at 7:01
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
I think there is another possible approach. Let $(a,b)$ a point in the plane $mathbb{R}^2$ and we want to find another point $(x,y)$ such that $a=x+f(y)$ and $b=y+f(x)$. So from the last equation we get $y=b-f(x)$. And inserting this into the first equation we get $a=x+f(b-f(x))$. So we reduce the problem to find such an $x$. Let us define $h:mathbb{R}tomathbb{R}$ by the formula
$$
h(x)=x+f(b-f(x))
$$
The derivative is $h'(x)=1-f'(b-f(x))f'(x)$ for every $x$. By hypothesis $0<1-c<h'(x) <1+c$. So integrating the inequality we get
$$
h(0)+(1-c)xleq h(x)leq h(0) + (1+c)x
$$
Thus checking the limits $xto-infty$ and $xto +infty$ we see that $h$ is surjective.
answered Nov 27 '18 at 6:05
Dante GrevinoDante Grevino
96319
$endgroup$
$begingroup$
very noice. This is very simple
$endgroup$
– Anton
Nov 27 '18 at 6:25
$begingroup$
I think this solution is simple and it avoid uses of strong theorems!
$endgroup$
– Dante Grevino
Nov 27 '18 at 6:31
$begingroup$
Alternatively, you can use Banach's Fixed Point Theorem to the function $r(x) = a-f(b-f(x))$.
$endgroup$
– Dante Grevino
Nov 27 '18 at 7:01
add a comment |
$begingroup$
I think there is another possible approach. Let $(a,b)$ a point in the plane $mathbb{R}^2$ and we want to find another point $(x,y)$ such that $a=x+f(y)$ and $b=y+f(x)$. So from the last equation we get $y=b-f(x)$. And inserting this into the first equation we get $a=x+f(b-f(x))$. So we reduce the problem to find such an $x$. Let us define $h:mathbb{R}tomathbb{R}$ by the formula
$$
h(x)=x+f(b-f(x))
$$
The derivative is $h'(x)=1-f'(b-f(x))f'(x)$ for every $x$. By hypothesis $0<1-c<h'(x) <1+c$. So integrating the inequality we get
$$
h(0)+(1-c)xleq h(x)leq h(0) + (1+c)x
$$
Thus checking the limits $xto-infty$ and $xto +infty$ we see that $h$ is surjective.
answered Nov 27 '18 at 6:05
Dante GrevinoDante Grevino
96319
$endgroup$
$begingroup$
very noice. This is very simple
$endgroup$
– Anton
Nov 27 '18 at 6:25
$begingroup$
I think this solution is simple and it avoid uses of strong theorems!
$endgroup$
– Dante Grevino
Nov 27 '18 at 6:31
$begingroup$
Alternatively, you can use Banach's Fixed Point Theorem to the function $r(x) = a-f(b-f(x))$.
$endgroup$
– Dante Grevino
Nov 27 '18 at 7:01
add a comment |
$begingroup$
I think there is another possible approach. Let $(a,b)$ a point in the plane $mathbb{R}^2$ and we want to find another point $(x,y)$ such that $a=x+f(y)$ and $b=y+f(x)$. So from the last equation we get $y=b-f(x)$. And inserting this into the first equation we get $a=x+f(b-f(x))$. So we reduce the problem to find such an $x$. Let us define $h:mathbb{R}tomathbb{R}$ by the formula
$$
h(x)=x+f(b-f(x))
$$
The derivative is $h'(x)=1-f'(b-f(x))f'(x)$ for every $x$. By hypothesis $0<1-c<h'(x) <1+c$. So integrating the inequality we get
$$
h(0)+(1-c)xleq h(x)leq h(0) + (1+c)x
$$
Thus checking the limits $xto-infty$ and $xto +infty$ we see that $h$ is surjective.
answered Nov 27 '18 at 6:05
Dante GrevinoDante Grevino
96319
$endgroup$
I think there is another possible approach. Let $(a,b)$ a point in the plane $mathbb{R}^2$ and we want to find another point $(x,y)$ such that $a=x+f(y)$ and $b=y+f(x)$. So from the last equation we get $y=b-f(x)$. And inserting this into the first equation we get $a=x+f(b-f(x))$. So we reduce the problem to find such an $x$. Let us define $h:mathbb{R}tomathbb{R}$ by the formula
$$
h(x)=x+f(b-f(x))
$$
The derivative is $h'(x)=1-f'(b-f(x))f'(x)$ for every $x$. By hypothesis $0<1-c<h'(x) <1+c$. So integrating the inequality we get
$$
h(0)+(1-c)xleq h(x)leq h(0) + (1+c)x
$$
Thus checking the limits $xto-infty$ and $xto +infty$ we see that $h$ is surjective.
answered Nov 27 '18 at 6:05
Dante GrevinoDante Grevino
96319
edited Nov 27 '18 at 6:28
answered Nov 27 '18 at 6:05
Dante GrevinoDante Grevino
96319
answered Nov 27 '18 at 6:05
Dante GrevinoDante Grevino
96319
answered Nov 27 '18 at 6:05
Dante GrevinoDante Grevino
96319
96319
$begingroup$
very noice. This is very simple
$endgroup$
– Anton
Nov 27 '18 at 6:25
$begingroup$
I think this solution is simple and it avoid uses of strong theorems!
$endgroup$
– Dante Grevino
Nov 27 '18 at 6:31
$begingroup$
Alternatively, you can use Banach's Fixed Point Theorem to the function $r(x) = a-f(b-f(x))$.
$endgroup$
– Dante Grevino
Nov 27 '18 at 7:01
add a comment |
$begingroup$
very noice. This is very simple
$endgroup$
– Anton
Nov 27 '18 at 6:25
$begingroup$
I think this solution is simple and it avoid uses of strong theorems!
$endgroup$
– Dante Grevino
Nov 27 '18 at 6:31
$begingroup$
Alternatively, you can use Banach's Fixed Point Theorem to the function $r(x) = a-f(b-f(x))$.
$endgroup$
– Dante Grevino
Nov 27 '18 at 7:01
$begingroup$
very noice. This is very simple
$endgroup$
– Anton
Nov 27 '18 at 6:25
$begingroup$
very noice. This is very simple
$endgroup$
– Anton
Nov 27 '18 at 6:25
$begingroup$
I think this solution is simple and it avoid uses of strong theorems!
$endgroup$
– Dante Grevino
Nov 27 '18 at 6:31
$begingroup$
I think this solution is simple and it avoid uses of strong theorems!
$endgroup$
– Dante Grevino
Nov 27 '18 at 6:31
$begingroup$
Alternatively, you can use Banach's Fixed Point Theorem to the function $r(x) = a-f(b-f(x))$.
$endgroup$
– Dante Grevino
Nov 27 '18 at 7:01
$begingroup$
Alternatively, you can use Banach's Fixed Point Theorem to the function $r(x) = a-f(b-f(x))$.
$endgroup$
– Dante Grevino
Nov 27 '18 at 7:01
add a comment |
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