Given that derivative of a function is bounded. Prove surjectivity












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$begingroup$



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







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    3












    $begingroup$



    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







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$



      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







      share|cite|improve this question











      $endgroup$





      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 '18 at 5:09









      Will M.

      2,495315




      2,495315










      asked Nov 27 '18 at 4:28









      AntonAnton

      163




      163






















          1 Answer
          1






          active

          oldest

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          1












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          1












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          1












          1








          1





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 27 '18 at 6:28

























          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


















          • $begingroup$
            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


















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