Prove that function is bi-lipshitz given it'a differential is zero
$begingroup$
Let $f$ be a function $f:R^n -> R^n$ which is differentially continuous. Moreover, it is given that in the origin: $D_f(0)=Id$, which is the identity function.
I need to prove that around the origin, there exist $A,B>0$ that satisfy:
$$A||x-y|| le ||f(x)-f(y)|| le B||x-y||$$
I tried to maybe somehow use the Implicit Function Theorem, but no luck yet.
If someone could give me a hint or some direction, it will be very appreciated.
real-analysis calculus multivariable-calculus
asked Dec 5 '18 at 2:16
Gabi GGabi G
39819
$endgroup$
add a comment |
$begingroup$
Let $f$ be a function $f:R^n -> R^n$ which is differentially continuous. Moreover, it is given that in the origin: $D_f(0)=Id$, which is the identity function.
I need to prove that around the origin, there exist $A,B>0$ that satisfy:
$$A||x-y|| le ||f(x)-f(y)|| le B||x-y||$$
I tried to maybe somehow use the Implicit Function Theorem, but no luck yet.
If someone could give me a hint or some direction, it will be very appreciated.
real-analysis calculus multivariable-calculus
asked Dec 5 '18 at 2:16
Gabi GGabi G
39819
$endgroup$
add a comment |
$begingroup$
Let $f$ be a function $f:R^n -> R^n$ which is differentially continuous. Moreover, it is given that in the origin: $D_f(0)=Id$, which is the identity function.
I need to prove that around the origin, there exist $A,B>0$ that satisfy:
$$A||x-y|| le ||f(x)-f(y)|| le B||x-y||$$
I tried to maybe somehow use the Implicit Function Theorem, but no luck yet.
If someone could give me a hint or some direction, it will be very appreciated.
real-analysis calculus multivariable-calculus
asked Dec 5 '18 at 2:16
Gabi GGabi G
39819
$endgroup$
Let $f$ be a function $f:R^n -> R^n$ which is differentially continuous. Moreover, it is given that in the origin: $D_f(0)=Id$, which is the identity function.
I need to prove that around the origin, there exist $A,B>0$ that satisfy:
$$A||x-y|| le ||f(x)-f(y)|| le B||x-y||$$
I tried to maybe somehow use the Implicit Function Theorem, but no luck yet.
If someone could give me a hint or some direction, it will be very appreciated.
real-analysis calculus multivariable-calculus
real-analysis calculus multivariable-calculus
asked Dec 5 '18 at 2:16
Gabi GGabi G
39819
asked Dec 5 '18 at 2:16
Gabi GGabi G
39819
asked Dec 5 '18 at 2:16
Gabi GGabi G
39819
asked Dec 5 '18 at 2:16
Gabi GGabi G
39819
asked Dec 5 '18 at 2:16
Gabi GGabi G
39819
39819
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
By the Inverse Function Theorem, there exists two open sets $U$ and $V$ such that $0in U$, $f(U)=V$ and the restriction $f:U to V$ is a diffeomorphism. Let $varepsilon$ be a positive real number such that the closed ball $overline{B_varepsilon(0)}$ is included in $U$. Recall that the norm of $||cdot||$ of $mathbb{R}^n$ induces a norm on the vector space $mathbb{R}^{ntimes n}$ of real matrices of size $ntimes n$ defined by
$$
||A||=sup{||Ax||:||x||=1},
$$
for every $A$ in $mathbb{R}^{ntimes n}$. The main property of this induced norm is the inequality $||Ax||leq ||A||.||x||$, which holds for every matrix $A$ and every vector $x$. Now, the functions $Phi:xiinoverline{B_varepsilon(0)}mapsto Df(xi)in mathbb{R}^{ntimes n}$ and $Psi:xiin overline{B_varepsilon(0)}mapsto Df^{-1}(f(xi))in mathbb{R}^{ntimes n}$ are continuous and they are defined on a compact space. So there exists positive real numbers $A$ and $C$ such that
$$
||Phi(xi)||=||Df(xi)||leq A quadtext{ and }quad ||Psi(xi)||=||Df^{-1}(f(xi))||leq C,
$$
for every $xi$ in $overline{B_varepsilon(0)}$. Now, let $x$ and $y$ be different points in $B_varepsilon(0)$. By the Mean Value Theorem for Several Variables (if you do not know this version let me know and I will add the details), there exists $xi$ in the segment between $x$ and $y$ such that
$$
f(x)-f(y)=Df(xi).(x-y)
$$
And we know that
$$
||Df(xi).(x-y)||leq ||Df(xi)||.||x-y||leq A||x-y||
$$
and
$$
||x-y||=||Df^{-1}(f(xi)).Df(xi).(x-y)||leq ||Df^{-1}(f(xi))||.||Df(xi).(x-y)||leq C||Df(xi).(x-y)||,
$$
so $frac{1}{C}||x-y||leq ||Df(xi).(x-y)||$. If we set $B=frac{1}{C}$, it follows that
$$
B||x-y||leq ||f(x)-f(y)|| leq A||x-y||,
$$
for every $x$ and $y$ in $B_epsilon(0)$.
Note that we only need $Df(0)$ to be invertible.
answered Dec 5 '18 at 22:48
Dante GrevinoDante Grevino
96319
$endgroup$
$begingroup$
Nice answer! Makes me wonder why in the question it was given that the differential at zero is the identity function, and not merely invertible.
$endgroup$
– Gabi G
Dec 5 '18 at 23:59
$begingroup$
Thanks! Well, you can use the precise value of $||Df(0)||$ and $||Df^{−1}(f(0))||$ to control the constants $A$ and $B$ in the following way: For every $delta >0$ there exists an $varepsilon >0$ such that the constansts $A$ and $B$ in the answer above can be chosen $||Df(0)||leq A<||Df(0)||+δ$ and $1/||Df^{−1}(f(0))||geq C>1/(||Df^{−1}(f(0))||+δ)$.
$endgroup$
– Dante Grevino
Dec 6 '18 at 1:04
add a comment |
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$begingroup$
By the Inverse Function Theorem, there exists two open sets $U$ and $V$ such that $0in U$, $f(U)=V$ and the restriction $f:U to V$ is a diffeomorphism. Let $varepsilon$ be a positive real number such that the closed ball $overline{B_varepsilon(0)}$ is included in $U$. Recall that the norm of $||cdot||$ of $mathbb{R}^n$ induces a norm on the vector space $mathbb{R}^{ntimes n}$ of real matrices of size $ntimes n$ defined by
$$
||A||=sup{||Ax||:||x||=1},
$$
for every $A$ in $mathbb{R}^{ntimes n}$. The main property of this induced norm is the inequality $||Ax||leq ||A||.||x||$, which holds for every matrix $A$ and every vector $x$. Now, the functions $Phi:xiinoverline{B_varepsilon(0)}mapsto Df(xi)in mathbb{R}^{ntimes n}$ and $Psi:xiin overline{B_varepsilon(0)}mapsto Df^{-1}(f(xi))in mathbb{R}^{ntimes n}$ are continuous and they are defined on a compact space. So there exists positive real numbers $A$ and $C$ such that
$$
||Phi(xi)||=||Df(xi)||leq A quadtext{ and }quad ||Psi(xi)||=||Df^{-1}(f(xi))||leq C,
$$
for every $xi$ in $overline{B_varepsilon(0)}$. Now, let $x$ and $y$ be different points in $B_varepsilon(0)$. By the Mean Value Theorem for Several Variables (if you do not know this version let me know and I will add the details), there exists $xi$ in the segment between $x$ and $y$ such that
$$
f(x)-f(y)=Df(xi).(x-y)
$$
And we know that
$$
||Df(xi).(x-y)||leq ||Df(xi)||.||x-y||leq A||x-y||
$$
and
$$
||x-y||=||Df^{-1}(f(xi)).Df(xi).(x-y)||leq ||Df^{-1}(f(xi))||.||Df(xi).(x-y)||leq C||Df(xi).(x-y)||,
$$
so $frac{1}{C}||x-y||leq ||Df(xi).(x-y)||$. If we set $B=frac{1}{C}$, it follows that
$$
B||x-y||leq ||f(x)-f(y)|| leq A||x-y||,
$$
for every $x$ and $y$ in $B_epsilon(0)$.
Note that we only need $Df(0)$ to be invertible.
answered Dec 5 '18 at 22:48
Dante GrevinoDante Grevino
96319
$endgroup$
$begingroup$
Nice answer! Makes me wonder why in the question it was given that the differential at zero is the identity function, and not merely invertible.
$endgroup$
– Gabi G
Dec 5 '18 at 23:59
$begingroup$
Thanks! Well, you can use the precise value of $||Df(0)||$ and $||Df^{−1}(f(0))||$ to control the constants $A$ and $B$ in the following way: For every $delta >0$ there exists an $varepsilon >0$ such that the constansts $A$ and $B$ in the answer above can be chosen $||Df(0)||leq A<||Df(0)||+δ$ and $1/||Df^{−1}(f(0))||geq C>1/(||Df^{−1}(f(0))||+δ)$.
$endgroup$
– Dante Grevino
Dec 6 '18 at 1:04
add a comment |
$begingroup$
By the Inverse Function Theorem, there exists two open sets $U$ and $V$ such that $0in U$, $f(U)=V$ and the restriction $f:U to V$ is a diffeomorphism. Let $varepsilon$ be a positive real number such that the closed ball $overline{B_varepsilon(0)}$ is included in $U$. Recall that the norm of $||cdot||$ of $mathbb{R}^n$ induces a norm on the vector space $mathbb{R}^{ntimes n}$ of real matrices of size $ntimes n$ defined by
$$
||A||=sup{||Ax||:||x||=1},
$$
for every $A$ in $mathbb{R}^{ntimes n}$. The main property of this induced norm is the inequality $||Ax||leq ||A||.||x||$, which holds for every matrix $A$ and every vector $x$. Now, the functions $Phi:xiinoverline{B_varepsilon(0)}mapsto Df(xi)in mathbb{R}^{ntimes n}$ and $Psi:xiin overline{B_varepsilon(0)}mapsto Df^{-1}(f(xi))in mathbb{R}^{ntimes n}$ are continuous and they are defined on a compact space. So there exists positive real numbers $A$ and $C$ such that
$$
||Phi(xi)||=||Df(xi)||leq A quadtext{ and }quad ||Psi(xi)||=||Df^{-1}(f(xi))||leq C,
$$
for every $xi$ in $overline{B_varepsilon(0)}$. Now, let $x$ and $y$ be different points in $B_varepsilon(0)$. By the Mean Value Theorem for Several Variables (if you do not know this version let me know and I will add the details), there exists $xi$ in the segment between $x$ and $y$ such that
$$
f(x)-f(y)=Df(xi).(x-y)
$$
And we know that
$$
||Df(xi).(x-y)||leq ||Df(xi)||.||x-y||leq A||x-y||
$$
and
$$
||x-y||=||Df^{-1}(f(xi)).Df(xi).(x-y)||leq ||Df^{-1}(f(xi))||.||Df(xi).(x-y)||leq C||Df(xi).(x-y)||,
$$
so $frac{1}{C}||x-y||leq ||Df(xi).(x-y)||$. If we set $B=frac{1}{C}$, it follows that
$$
B||x-y||leq ||f(x)-f(y)|| leq A||x-y||,
$$
for every $x$ and $y$ in $B_epsilon(0)$.
Note that we only need $Df(0)$ to be invertible.
answered Dec 5 '18 at 22:48
Dante GrevinoDante Grevino
96319
$endgroup$
$begingroup$
Nice answer! Makes me wonder why in the question it was given that the differential at zero is the identity function, and not merely invertible.
$endgroup$
– Gabi G
Dec 5 '18 at 23:59
$begingroup$
Thanks! Well, you can use the precise value of $||Df(0)||$ and $||Df^{−1}(f(0))||$ to control the constants $A$ and $B$ in the following way: For every $delta >0$ there exists an $varepsilon >0$ such that the constansts $A$ and $B$ in the answer above can be chosen $||Df(0)||leq A<||Df(0)||+δ$ and $1/||Df^{−1}(f(0))||geq C>1/(||Df^{−1}(f(0))||+δ)$.
$endgroup$
– Dante Grevino
Dec 6 '18 at 1:04
add a comment |
$begingroup$
By the Inverse Function Theorem, there exists two open sets $U$ and $V$ such that $0in U$, $f(U)=V$ and the restriction $f:U to V$ is a diffeomorphism. Let $varepsilon$ be a positive real number such that the closed ball $overline{B_varepsilon(0)}$ is included in $U$. Recall that the norm of $||cdot||$ of $mathbb{R}^n$ induces a norm on the vector space $mathbb{R}^{ntimes n}$ of real matrices of size $ntimes n$ defined by
$$
||A||=sup{||Ax||:||x||=1},
$$
for every $A$ in $mathbb{R}^{ntimes n}$. The main property of this induced norm is the inequality $||Ax||leq ||A||.||x||$, which holds for every matrix $A$ and every vector $x$. Now, the functions $Phi:xiinoverline{B_varepsilon(0)}mapsto Df(xi)in mathbb{R}^{ntimes n}$ and $Psi:xiin overline{B_varepsilon(0)}mapsto Df^{-1}(f(xi))in mathbb{R}^{ntimes n}$ are continuous and they are defined on a compact space. So there exists positive real numbers $A$ and $C$ such that
$$
||Phi(xi)||=||Df(xi)||leq A quadtext{ and }quad ||Psi(xi)||=||Df^{-1}(f(xi))||leq C,
$$
for every $xi$ in $overline{B_varepsilon(0)}$. Now, let $x$ and $y$ be different points in $B_varepsilon(0)$. By the Mean Value Theorem for Several Variables (if you do not know this version let me know and I will add the details), there exists $xi$ in the segment between $x$ and $y$ such that
$$
f(x)-f(y)=Df(xi).(x-y)
$$
And we know that
$$
||Df(xi).(x-y)||leq ||Df(xi)||.||x-y||leq A||x-y||
$$
and
$$
||x-y||=||Df^{-1}(f(xi)).Df(xi).(x-y)||leq ||Df^{-1}(f(xi))||.||Df(xi).(x-y)||leq C||Df(xi).(x-y)||,
$$
so $frac{1}{C}||x-y||leq ||Df(xi).(x-y)||$. If we set $B=frac{1}{C}$, it follows that
$$
B||x-y||leq ||f(x)-f(y)|| leq A||x-y||,
$$
for every $x$ and $y$ in $B_epsilon(0)$.
Note that we only need $Df(0)$ to be invertible.
answered Dec 5 '18 at 22:48
Dante GrevinoDante Grevino
96319
$endgroup$
By the Inverse Function Theorem, there exists two open sets $U$ and $V$ such that $0in U$, $f(U)=V$ and the restriction $f:U to V$ is a diffeomorphism. Let $varepsilon$ be a positive real number such that the closed ball $overline{B_varepsilon(0)}$ is included in $U$. Recall that the norm of $||cdot||$ of $mathbb{R}^n$ induces a norm on the vector space $mathbb{R}^{ntimes n}$ of real matrices of size $ntimes n$ defined by
$$
||A||=sup{||Ax||:||x||=1},
$$
for every $A$ in $mathbb{R}^{ntimes n}$. The main property of this induced norm is the inequality $||Ax||leq ||A||.||x||$, which holds for every matrix $A$ and every vector $x$. Now, the functions $Phi:xiinoverline{B_varepsilon(0)}mapsto Df(xi)in mathbb{R}^{ntimes n}$ and $Psi:xiin overline{B_varepsilon(0)}mapsto Df^{-1}(f(xi))in mathbb{R}^{ntimes n}$ are continuous and they are defined on a compact space. So there exists positive real numbers $A$ and $C$ such that
$$
||Phi(xi)||=||Df(xi)||leq A quadtext{ and }quad ||Psi(xi)||=||Df^{-1}(f(xi))||leq C,
$$
for every $xi$ in $overline{B_varepsilon(0)}$. Now, let $x$ and $y$ be different points in $B_varepsilon(0)$. By the Mean Value Theorem for Several Variables (if you do not know this version let me know and I will add the details), there exists $xi$ in the segment between $x$ and $y$ such that
$$
f(x)-f(y)=Df(xi).(x-y)
$$
And we know that
$$
||Df(xi).(x-y)||leq ||Df(xi)||.||x-y||leq A||x-y||
$$
and
$$
||x-y||=||Df^{-1}(f(xi)).Df(xi).(x-y)||leq ||Df^{-1}(f(xi))||.||Df(xi).(x-y)||leq C||Df(xi).(x-y)||,
$$
so $frac{1}{C}||x-y||leq ||Df(xi).(x-y)||$. If we set $B=frac{1}{C}$, it follows that
$$
B||x-y||leq ||f(x)-f(y)|| leq A||x-y||,
$$
for every $x$ and $y$ in $B_epsilon(0)$.
Note that we only need $Df(0)$ to be invertible.
answered Dec 5 '18 at 22:48
Dante GrevinoDante Grevino
96319
edited Dec 6 '18 at 0:31
answered Dec 5 '18 at 22:48
Dante GrevinoDante Grevino
96319
answered Dec 5 '18 at 22:48
Dante GrevinoDante Grevino
96319
answered Dec 5 '18 at 22:48
Dante GrevinoDante Grevino
96319
96319
$begingroup$
Nice answer! Makes me wonder why in the question it was given that the differential at zero is the identity function, and not merely invertible.
$endgroup$
– Gabi G
Dec 5 '18 at 23:59
$begingroup$
Thanks! Well, you can use the precise value of $||Df(0)||$ and $||Df^{−1}(f(0))||$ to control the constants $A$ and $B$ in the following way: For every $delta >0$ there exists an $varepsilon >0$ such that the constansts $A$ and $B$ in the answer above can be chosen $||Df(0)||leq A<||Df(0)||+δ$ and $1/||Df^{−1}(f(0))||geq C>1/(||Df^{−1}(f(0))||+δ)$.
$endgroup$
– Dante Grevino
Dec 6 '18 at 1:04
add a comment |
$begingroup$
Nice answer! Makes me wonder why in the question it was given that the differential at zero is the identity function, and not merely invertible.
$endgroup$
– Gabi G
Dec 5 '18 at 23:59
$begingroup$
Thanks! Well, you can use the precise value of $||Df(0)||$ and $||Df^{−1}(f(0))||$ to control the constants $A$ and $B$ in the following way: For every $delta >0$ there exists an $varepsilon >0$ such that the constansts $A$ and $B$ in the answer above can be chosen $||Df(0)||leq A<||Df(0)||+δ$ and $1/||Df^{−1}(f(0))||geq C>1/(||Df^{−1}(f(0))||+δ)$.
$endgroup$
– Dante Grevino
Dec 6 '18 at 1:04
$begingroup$
Nice answer! Makes me wonder why in the question it was given that the differential at zero is the identity function, and not merely invertible.
$endgroup$
– Gabi G
Dec 5 '18 at 23:59
$begingroup$
Nice answer! Makes me wonder why in the question it was given that the differential at zero is the identity function, and not merely invertible.
$endgroup$
– Gabi G
Dec 5 '18 at 23:59
$begingroup$
Thanks! Well, you can use the precise value of $||Df(0)||$ and $||Df^{−1}(f(0))||$ to control the constants $A$ and $B$ in the following way: For every $delta >0$ there exists an $varepsilon >0$ such that the constansts $A$ and $B$ in the answer above can be chosen $||Df(0)||leq A<||Df(0)||+δ$ and $1/||Df^{−1}(f(0))||geq C>1/(||Df^{−1}(f(0))||+δ)$.
$endgroup$
– Dante Grevino
Dec 6 '18 at 1:04
$begingroup$
Thanks! Well, you can use the precise value of $||Df(0)||$ and $||Df^{−1}(f(0))||$ to control the constants $A$ and $B$ in the following way: For every $delta >0$ there exists an $varepsilon >0$ such that the constansts $A$ and $B$ in the answer above can be chosen $||Df(0)||leq A<||Df(0)||+δ$ and $1/||Df^{−1}(f(0))||geq C>1/(||Df^{−1}(f(0))||+δ)$.
$endgroup$
– Dante Grevino
Dec 6 '18 at 1:04
add a comment |
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