Prove that function is bi-lipshitz given it'a differential is zero












1












$begingroup$


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







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 5 '18 at 2:16









      Gabi GGabi G

      39819




      39819






















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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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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













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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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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


















          1












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          1












          1








          1





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 6 '18 at 0:31

























          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




















          • $begingroup$
            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




















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          Mont Emei

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          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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