Fixed point of cotinuous differentiable function whose derivative is bounded?












0














Let $f$ be a real continously differentiable function such that $mid f'(x) mid leq 4/5$. Is it true that $f(x) = x$ for some $x$.



My approach:
I took $g(x) = f(x) - x$ and showed that $-9/5 leq g'(x) leq -1/5 $ i.e $g(x)$ is a monotonous decreasing function. Now I plan to show that $g(x_1) geq 0$ for some $x_1$ and $g(x_2) leq 0$ for some $x_2$ and then use the intermediate value property to conclude that $g(x) = 0$ for some $x$ but I am unable to find such $x_1$ and $x_2$.






calculus







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  • en.m.wikipedia.org/wiki/Banach_fixed-point_theorem
    – Peter Szilas
    Nov 27 at 8:38
















0














Let $f$ be a real continously differentiable function such that $mid f'(x) mid leq 4/5$. Is it true that $f(x) = x$ for some $x$.



My approach:
I took $g(x) = f(x) - x$ and showed that $-9/5 leq g'(x) leq -1/5 $ i.e $g(x)$ is a monotonous decreasing function. Now I plan to show that $g(x_1) geq 0$ for some $x_1$ and $g(x_2) leq 0$ for some $x_2$ and then use the intermediate value property to conclude that $g(x) = 0$ for some $x$ but I am unable to find such $x_1$ and $x_2$.






calculus







share|cite|improve this question






















  • en.m.wikipedia.org/wiki/Banach_fixed-point_theorem
    – Peter Szilas
    Nov 27 at 8:38














0












0








0







Let $f$ be a real continously differentiable function such that $mid f'(x) mid leq 4/5$. Is it true that $f(x) = x$ for some $x$.



My approach:
I took $g(x) = f(x) - x$ and showed that $-9/5 leq g'(x) leq -1/5 $ i.e $g(x)$ is a monotonous decreasing function. Now I plan to show that $g(x_1) geq 0$ for some $x_1$ and $g(x_2) leq 0$ for some $x_2$ and then use the intermediate value property to conclude that $g(x) = 0$ for some $x$ but I am unable to find such $x_1$ and $x_2$.






calculus







share|cite|improve this question













Let $f$ be a real continously differentiable function such that $mid f'(x) mid leq 4/5$. Is it true that $f(x) = x$ for some $x$.



My approach:
I took $g(x) = f(x) - x$ and showed that $-9/5 leq g'(x) leq -1/5 $ i.e $g(x)$ is a monotonous decreasing function. Now I plan to show that $g(x_1) geq 0$ for some $x_1$ and $g(x_2) leq 0$ for some $x_2$ and then use the intermediate value property to conclude that $g(x) = 0$ for some $x$ but I am unable to find such $x_1$ and $x_2$.






calculus




calculus






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asked Nov 27 at 6:31









henceproved

1358




1358












  • en.m.wikipedia.org/wiki/Banach_fixed-point_theorem
    – Peter Szilas
    Nov 27 at 8:38


















  • en.m.wikipedia.org/wiki/Banach_fixed-point_theorem
    – Peter Szilas
    Nov 27 at 8:38
















en.m.wikipedia.org/wiki/Banach_fixed-point_theorem
– Peter Szilas
Nov 27 at 8:38




en.m.wikipedia.org/wiki/Banach_fixed-point_theorem
– Peter Szilas
Nov 27 at 8:38










2 Answers
2






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2














Do you know "Banach" Fixed Point Theorem?



Theorem

Let $f : mathbb{R} to mathbb{R}$ and $exists delta, 0< delta <1$ such that $forall x, y in mathbb{R}, |f(x) - f(y)| leq delta |x-y|$, then $exists! x_{0} in mathbb{R}$ such that $f(x_{0}) = x_{0}$.



Now, we also need one more theorem.



Theorem (Mean value Theorem)

Let $f : mathbb{R} to mathbb{R}$ be a once differentiable function. then $forall x,y in mathbb{R}, exists c in mathbb{R},, x < c < y$ such that $frac{f(x)-f(y)}{x-y} = f'(c)$.



To show your function has a fixed point. Fix $x,y in mathbb{R}$. Then, by mean value theorem we have
$$bigg|frac{f(x)-f(y)}{x-y}bigg| = bigg|f'(c)bigg| leq frac{4}{5}$$
This implies
$$|f(x)-f(y)| leq frac{4}{5}|x-y|$$
Thus, $f$ satisfies the condition of Banach Fixed Point Theorem and $exists! x_{0} in mathbb{R}$ such that $f(x_{0}) = x_{0}$.






share|cite|improve this answer





























    3














    You may proceed as follows:




    • Choose any $x_0 in mathbb{R}$ as a starting value and consider the iteration

    • $x_{n+1} =f(x_n)$


    It follows
    $$left| x_{n+1} - x_n right| = left| f(x_{n}) - f(x_{n-1}) right| = |f'(xi_n)|left| x_{n} - x_{n-1} right| leq frac{4}{5}left| x_{n} - x_{n-1} right|$$
    $$Rightarrow left| x_{n+1} - x_n right| leq left(frac{4}{5} right)^n left| x_{1} - x_0 right|$$
    Now, it follows that this sequence converges as
    $$x_{n} = x_0 + sum_{k=1}^n(x_k - x_{k-1}) mbox{ and } left|sum_{k=1}^n(x_k - x_{k-1})right| < left| x_{1} - x_0 right|sum_{k=1}^{infty}left(frac{4}{5} right)^k < infty$$
    Because of continuity of $f$ you have for the limit $x^{star} = lim_{n to infty} x_n$
    $$boxed{f(x^{star}) = x^{star}}$$






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      Do you know "Banach" Fixed Point Theorem?



      Theorem

      Let $f : mathbb{R} to mathbb{R}$ and $exists delta, 0< delta <1$ such that $forall x, y in mathbb{R}, |f(x) - f(y)| leq delta |x-y|$, then $exists! x_{0} in mathbb{R}$ such that $f(x_{0}) = x_{0}$.



      Now, we also need one more theorem.



      Theorem (Mean value Theorem)

      Let $f : mathbb{R} to mathbb{R}$ be a once differentiable function. then $forall x,y in mathbb{R}, exists c in mathbb{R},, x < c < y$ such that $frac{f(x)-f(y)}{x-y} = f'(c)$.



      To show your function has a fixed point. Fix $x,y in mathbb{R}$. Then, by mean value theorem we have
      $$bigg|frac{f(x)-f(y)}{x-y}bigg| = bigg|f'(c)bigg| leq frac{4}{5}$$
      This implies
      $$|f(x)-f(y)| leq frac{4}{5}|x-y|$$
      Thus, $f$ satisfies the condition of Banach Fixed Point Theorem and $exists! x_{0} in mathbb{R}$ such that $f(x_{0}) = x_{0}$.






      share|cite|improve this answer


























        2














        Do you know "Banach" Fixed Point Theorem?



        Theorem

        Let $f : mathbb{R} to mathbb{R}$ and $exists delta, 0< delta <1$ such that $forall x, y in mathbb{R}, |f(x) - f(y)| leq delta |x-y|$, then $exists! x_{0} in mathbb{R}$ such that $f(x_{0}) = x_{0}$.



        Now, we also need one more theorem.



        Theorem (Mean value Theorem)

        Let $f : mathbb{R} to mathbb{R}$ be a once differentiable function. then $forall x,y in mathbb{R}, exists c in mathbb{R},, x < c < y$ such that $frac{f(x)-f(y)}{x-y} = f'(c)$.



        To show your function has a fixed point. Fix $x,y in mathbb{R}$. Then, by mean value theorem we have
        $$bigg|frac{f(x)-f(y)}{x-y}bigg| = bigg|f'(c)bigg| leq frac{4}{5}$$
        This implies
        $$|f(x)-f(y)| leq frac{4}{5}|x-y|$$
        Thus, $f$ satisfies the condition of Banach Fixed Point Theorem and $exists! x_{0} in mathbb{R}$ such that $f(x_{0}) = x_{0}$.






        share|cite|improve this answer
























          2












          2








          2






          Do you know "Banach" Fixed Point Theorem?



          Theorem

          Let $f : mathbb{R} to mathbb{R}$ and $exists delta, 0< delta <1$ such that $forall x, y in mathbb{R}, |f(x) - f(y)| leq delta |x-y|$, then $exists! x_{0} in mathbb{R}$ such that $f(x_{0}) = x_{0}$.



          Now, we also need one more theorem.



          Theorem (Mean value Theorem)

          Let $f : mathbb{R} to mathbb{R}$ be a once differentiable function. then $forall x,y in mathbb{R}, exists c in mathbb{R},, x < c < y$ such that $frac{f(x)-f(y)}{x-y} = f'(c)$.



          To show your function has a fixed point. Fix $x,y in mathbb{R}$. Then, by mean value theorem we have
          $$bigg|frac{f(x)-f(y)}{x-y}bigg| = bigg|f'(c)bigg| leq frac{4}{5}$$
          This implies
          $$|f(x)-f(y)| leq frac{4}{5}|x-y|$$
          Thus, $f$ satisfies the condition of Banach Fixed Point Theorem and $exists! x_{0} in mathbb{R}$ such that $f(x_{0}) = x_{0}$.






          share|cite|improve this answer












          Do you know "Banach" Fixed Point Theorem?



          Theorem

          Let $f : mathbb{R} to mathbb{R}$ and $exists delta, 0< delta <1$ such that $forall x, y in mathbb{R}, |f(x) - f(y)| leq delta |x-y|$, then $exists! x_{0} in mathbb{R}$ such that $f(x_{0}) = x_{0}$.



          Now, we also need one more theorem.



          Theorem (Mean value Theorem)

          Let $f : mathbb{R} to mathbb{R}$ be a once differentiable function. then $forall x,y in mathbb{R}, exists c in mathbb{R},, x < c < y$ such that $frac{f(x)-f(y)}{x-y} = f'(c)$.



          To show your function has a fixed point. Fix $x,y in mathbb{R}$. Then, by mean value theorem we have
          $$bigg|frac{f(x)-f(y)}{x-y}bigg| = bigg|f'(c)bigg| leq frac{4}{5}$$
          This implies
          $$|f(x)-f(y)| leq frac{4}{5}|x-y|$$
          Thus, $f$ satisfies the condition of Banach Fixed Point Theorem and $exists! x_{0} in mathbb{R}$ such that $f(x_{0}) = x_{0}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 at 7:02









          Evan William Chandra

          508313




          508313























              3














              You may proceed as follows:




              • Choose any $x_0 in mathbb{R}$ as a starting value and consider the iteration

              • $x_{n+1} =f(x_n)$


              It follows
              $$left| x_{n+1} - x_n right| = left| f(x_{n}) - f(x_{n-1}) right| = |f'(xi_n)|left| x_{n} - x_{n-1} right| leq frac{4}{5}left| x_{n} - x_{n-1} right|$$
              $$Rightarrow left| x_{n+1} - x_n right| leq left(frac{4}{5} right)^n left| x_{1} - x_0 right|$$
              Now, it follows that this sequence converges as
              $$x_{n} = x_0 + sum_{k=1}^n(x_k - x_{k-1}) mbox{ and } left|sum_{k=1}^n(x_k - x_{k-1})right| < left| x_{1} - x_0 right|sum_{k=1}^{infty}left(frac{4}{5} right)^k < infty$$
              Because of continuity of $f$ you have for the limit $x^{star} = lim_{n to infty} x_n$
              $$boxed{f(x^{star}) = x^{star}}$$






              share|cite|improve this answer


























                3














                You may proceed as follows:




                • Choose any $x_0 in mathbb{R}$ as a starting value and consider the iteration

                • $x_{n+1} =f(x_n)$


                It follows
                $$left| x_{n+1} - x_n right| = left| f(x_{n}) - f(x_{n-1}) right| = |f'(xi_n)|left| x_{n} - x_{n-1} right| leq frac{4}{5}left| x_{n} - x_{n-1} right|$$
                $$Rightarrow left| x_{n+1} - x_n right| leq left(frac{4}{5} right)^n left| x_{1} - x_0 right|$$
                Now, it follows that this sequence converges as
                $$x_{n} = x_0 + sum_{k=1}^n(x_k - x_{k-1}) mbox{ and } left|sum_{k=1}^n(x_k - x_{k-1})right| < left| x_{1} - x_0 right|sum_{k=1}^{infty}left(frac{4}{5} right)^k < infty$$
                Because of continuity of $f$ you have for the limit $x^{star} = lim_{n to infty} x_n$
                $$boxed{f(x^{star}) = x^{star}}$$






                share|cite|improve this answer
























                  3












                  3








                  3






                  You may proceed as follows:




                  • Choose any $x_0 in mathbb{R}$ as a starting value and consider the iteration

                  • $x_{n+1} =f(x_n)$


                  It follows
                  $$left| x_{n+1} - x_n right| = left| f(x_{n}) - f(x_{n-1}) right| = |f'(xi_n)|left| x_{n} - x_{n-1} right| leq frac{4}{5}left| x_{n} - x_{n-1} right|$$
                  $$Rightarrow left| x_{n+1} - x_n right| leq left(frac{4}{5} right)^n left| x_{1} - x_0 right|$$
                  Now, it follows that this sequence converges as
                  $$x_{n} = x_0 + sum_{k=1}^n(x_k - x_{k-1}) mbox{ and } left|sum_{k=1}^n(x_k - x_{k-1})right| < left| x_{1} - x_0 right|sum_{k=1}^{infty}left(frac{4}{5} right)^k < infty$$
                  Because of continuity of $f$ you have for the limit $x^{star} = lim_{n to infty} x_n$
                  $$boxed{f(x^{star}) = x^{star}}$$






                  share|cite|improve this answer












                  You may proceed as follows:




                  • Choose any $x_0 in mathbb{R}$ as a starting value and consider the iteration

                  • $x_{n+1} =f(x_n)$


                  It follows
                  $$left| x_{n+1} - x_n right| = left| f(x_{n}) - f(x_{n-1}) right| = |f'(xi_n)|left| x_{n} - x_{n-1} right| leq frac{4}{5}left| x_{n} - x_{n-1} right|$$
                  $$Rightarrow left| x_{n+1} - x_n right| leq left(frac{4}{5} right)^n left| x_{1} - x_0 right|$$
                  Now, it follows that this sequence converges as
                  $$x_{n} = x_0 + sum_{k=1}^n(x_k - x_{k-1}) mbox{ and } left|sum_{k=1}^n(x_k - x_{k-1})right| < left| x_{1} - x_0 right|sum_{k=1}^{infty}left(frac{4}{5} right)^k < infty$$
                  Because of continuity of $f$ you have for the limit $x^{star} = lim_{n to infty} x_n$
                  $$boxed{f(x^{star}) = x^{star}}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 27 at 6:53









                  trancelocation

                  9,1051521




                  9,1051521






























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