How to arrive at $(f'(x))^2<2f(x)$?












3












$begingroup$


Let we see a problem ahead of I asking what I want to ask.



Define $f:mathbb{R}to(0,+infty)$ differentiable satisfying $|f'(x)-f'(y)|le|x-y|,, forall x,yinmathbb{R}$.



Prove: $(f'(x))^2<2f(x)$



I came up with a proof which is as follows:




In order to make the proof more rigorous, we pointed out in advance that $f'$ is (uniformly) continuous.



1. $,$ If $f'(x)=0$.

Then through $f>0$ we arrive at the conclusion.



2. $,$ If $f'(x)>0$.$,,,$ Let $x_0=x-f'(x)$.



so $f(x)=int_{x_0}^{x}f'(t),dt+f(x_0)>int_{x_0}^{x}f'(t),dtgeint_{x_0}^{x}(f'(x)+t-x),dt=frac{1}{2}(f'(x))^2$



3. $,$ If $f'(x)<0$.$,,,$ Let $x_0=x-f'(x)$.



so $f(x)=f(x_0)-int_{x}^{x_0}f'(t),dt>int_{x}^{x_0}-f'(t),dtgeint_{x}^{x_0}(-f'(x)+x-t),dt=frac{1}{2}(f'(x))^2$




Actually I got stuck on how to arrive at $(f'(x))^2<2f(x)$.



I thought $f(x)$ is influenced by $f'(t)$ where $t$ is around $x$, so I came up with the proof as mentioned above.



But at the same time I thought that for the similar questions such as $f(x)>f'(x)$ we can construct the auxiliary function $g(x)=frac{f(x)}{e^x}$.



So what I really want to ask is how to arrive at $(f'(x))^2<2f(x)$?



I thought this maybe relates to the solution of $(f'(x))^2=2f(x)$.



What's more, any new ideas for the above problem are welcomed. Thank you in advance!






real-analysis ordinary-differential-equations







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Let we see a problem ahead of I asking what I want to ask.



    Define $f:mathbb{R}to(0,+infty)$ differentiable satisfying $|f'(x)-f'(y)|le|x-y|,, forall x,yinmathbb{R}$.



    Prove: $(f'(x))^2<2f(x)$



    I came up with a proof which is as follows:




    In order to make the proof more rigorous, we pointed out in advance that $f'$ is (uniformly) continuous.



    1. $,$ If $f'(x)=0$.

    Then through $f>0$ we arrive at the conclusion.



    2. $,$ If $f'(x)>0$.$,,,$ Let $x_0=x-f'(x)$.



    so $f(x)=int_{x_0}^{x}f'(t),dt+f(x_0)>int_{x_0}^{x}f'(t),dtgeint_{x_0}^{x}(f'(x)+t-x),dt=frac{1}{2}(f'(x))^2$



    3. $,$ If $f'(x)<0$.$,,,$ Let $x_0=x-f'(x)$.



    so $f(x)=f(x_0)-int_{x}^{x_0}f'(t),dt>int_{x}^{x_0}-f'(t),dtgeint_{x}^{x_0}(-f'(x)+x-t),dt=frac{1}{2}(f'(x))^2$




    Actually I got stuck on how to arrive at $(f'(x))^2<2f(x)$.



    I thought $f(x)$ is influenced by $f'(t)$ where $t$ is around $x$, so I came up with the proof as mentioned above.



    But at the same time I thought that for the similar questions such as $f(x)>f'(x)$ we can construct the auxiliary function $g(x)=frac{f(x)}{e^x}$.



    So what I really want to ask is how to arrive at $(f'(x))^2<2f(x)$?



    I thought this maybe relates to the solution of $(f'(x))^2=2f(x)$.



    What's more, any new ideas for the above problem are welcomed. Thank you in advance!






    real-analysis ordinary-differential-equations







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      2



      $begingroup$


      Let we see a problem ahead of I asking what I want to ask.



      Define $f:mathbb{R}to(0,+infty)$ differentiable satisfying $|f'(x)-f'(y)|le|x-y|,, forall x,yinmathbb{R}$.



      Prove: $(f'(x))^2<2f(x)$



      I came up with a proof which is as follows:




      In order to make the proof more rigorous, we pointed out in advance that $f'$ is (uniformly) continuous.



      1. $,$ If $f'(x)=0$.

      Then through $f>0$ we arrive at the conclusion.



      2. $,$ If $f'(x)>0$.$,,,$ Let $x_0=x-f'(x)$.



      so $f(x)=int_{x_0}^{x}f'(t),dt+f(x_0)>int_{x_0}^{x}f'(t),dtgeint_{x_0}^{x}(f'(x)+t-x),dt=frac{1}{2}(f'(x))^2$



      3. $,$ If $f'(x)<0$.$,,,$ Let $x_0=x-f'(x)$.



      so $f(x)=f(x_0)-int_{x}^{x_0}f'(t),dt>int_{x}^{x_0}-f'(t),dtgeint_{x}^{x_0}(-f'(x)+x-t),dt=frac{1}{2}(f'(x))^2$




      Actually I got stuck on how to arrive at $(f'(x))^2<2f(x)$.



      I thought $f(x)$ is influenced by $f'(t)$ where $t$ is around $x$, so I came up with the proof as mentioned above.



      But at the same time I thought that for the similar questions such as $f(x)>f'(x)$ we can construct the auxiliary function $g(x)=frac{f(x)}{e^x}$.



      So what I really want to ask is how to arrive at $(f'(x))^2<2f(x)$?



      I thought this maybe relates to the solution of $(f'(x))^2=2f(x)$.



      What's more, any new ideas for the above problem are welcomed. Thank you in advance!






      real-analysis ordinary-differential-equations







      share|cite|improve this question









      $endgroup$




      Let we see a problem ahead of I asking what I want to ask.



      Define $f:mathbb{R}to(0,+infty)$ differentiable satisfying $|f'(x)-f'(y)|le|x-y|,, forall x,yinmathbb{R}$.



      Prove: $(f'(x))^2<2f(x)$



      I came up with a proof which is as follows:




      In order to make the proof more rigorous, we pointed out in advance that $f'$ is (uniformly) continuous.



      1. $,$ If $f'(x)=0$.

      Then through $f>0$ we arrive at the conclusion.



      2. $,$ If $f'(x)>0$.$,,,$ Let $x_0=x-f'(x)$.



      so $f(x)=int_{x_0}^{x}f'(t),dt+f(x_0)>int_{x_0}^{x}f'(t),dtgeint_{x_0}^{x}(f'(x)+t-x),dt=frac{1}{2}(f'(x))^2$



      3. $,$ If $f'(x)<0$.$,,,$ Let $x_0=x-f'(x)$.



      so $f(x)=f(x_0)-int_{x}^{x_0}f'(t),dt>int_{x}^{x_0}-f'(t),dtgeint_{x}^{x_0}(-f'(x)+x-t),dt=frac{1}{2}(f'(x))^2$




      Actually I got stuck on how to arrive at $(f'(x))^2<2f(x)$.



      I thought $f(x)$ is influenced by $f'(t)$ where $t$ is around $x$, so I came up with the proof as mentioned above.



      But at the same time I thought that for the similar questions such as $f(x)>f'(x)$ we can construct the auxiliary function $g(x)=frac{f(x)}{e^x}$.



      So what I really want to ask is how to arrive at $(f'(x))^2<2f(x)$?



      I thought this maybe relates to the solution of $(f'(x))^2=2f(x)$.



      What's more, any new ideas for the above problem are welcomed. Thank you in advance!






      real-analysis ordinary-differential-equations




      real-analysis ordinary-differential-equations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 18 '18 at 14:24









      ZeroZero

      35310




      35310






















          1 Answer
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          $begingroup$

          When exploring the values between $x$ and $x-f'(x)$, it usually makes sense to parametrize that segment as $x-sf'(x)$, $sin[0,1]$. This avoids discuss the different cases of the order between these points.



          Let $L$ be the Lipschitz constant of $f'$, then we know from the fundamental theorem that
          begin{align}
          |f(x+u)-f(x)-f'(x)u|&leint_0^1|f'(x+su)-f'(x)|,ds,|u|
          \
          &le int_us,ds,L|u|^2
          \
          &=frac12L|u|^2
          end{align}



          Resolving the absolute value to one side it follows that
          $$
          f(x)+f'(x)uge f(x+u)-frac12L|u|^2.
          $$
          Now insert $u=-frac{f'(x)}{L}$ to find
          $$
          2f(x)ge 2fleft(x-frac{f'(x)}{L}right)+frac{f'(x)^2}{L}
          $$
          To conclude now use that $f>0$.






          share|cite|improve this answer











          $endgroup$













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            1












            $begingroup$

            When exploring the values between $x$ and $x-f'(x)$, it usually makes sense to parametrize that segment as $x-sf'(x)$, $sin[0,1]$. This avoids discuss the different cases of the order between these points.



            Let $L$ be the Lipschitz constant of $f'$, then we know from the fundamental theorem that
            begin{align}
            |f(x+u)-f(x)-f'(x)u|&leint_0^1|f'(x+su)-f'(x)|,ds,|u|
            \
            &le int_us,ds,L|u|^2
            \
            &=frac12L|u|^2
            end{align}



            Resolving the absolute value to one side it follows that
            $$
            f(x)+f'(x)uge f(x+u)-frac12L|u|^2.
            $$
            Now insert $u=-frac{f'(x)}{L}$ to find
            $$
            2f(x)ge 2fleft(x-frac{f'(x)}{L}right)+frac{f'(x)^2}{L}
            $$
            To conclude now use that $f>0$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              When exploring the values between $x$ and $x-f'(x)$, it usually makes sense to parametrize that segment as $x-sf'(x)$, $sin[0,1]$. This avoids discuss the different cases of the order between these points.



              Let $L$ be the Lipschitz constant of $f'$, then we know from the fundamental theorem that
              begin{align}
              |f(x+u)-f(x)-f'(x)u|&leint_0^1|f'(x+su)-f'(x)|,ds,|u|
              \
              &le int_us,ds,L|u|^2
              \
              &=frac12L|u|^2
              end{align}



              Resolving the absolute value to one side it follows that
              $$
              f(x)+f'(x)uge f(x+u)-frac12L|u|^2.
              $$
              Now insert $u=-frac{f'(x)}{L}$ to find
              $$
              2f(x)ge 2fleft(x-frac{f'(x)}{L}right)+frac{f'(x)^2}{L}
              $$
              To conclude now use that $f>0$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                When exploring the values between $x$ and $x-f'(x)$, it usually makes sense to parametrize that segment as $x-sf'(x)$, $sin[0,1]$. This avoids discuss the different cases of the order between these points.



                Let $L$ be the Lipschitz constant of $f'$, then we know from the fundamental theorem that
                begin{align}
                |f(x+u)-f(x)-f'(x)u|&leint_0^1|f'(x+su)-f'(x)|,ds,|u|
                \
                &le int_us,ds,L|u|^2
                \
                &=frac12L|u|^2
                end{align}



                Resolving the absolute value to one side it follows that
                $$
                f(x)+f'(x)uge f(x+u)-frac12L|u|^2.
                $$
                Now insert $u=-frac{f'(x)}{L}$ to find
                $$
                2f(x)ge 2fleft(x-frac{f'(x)}{L}right)+frac{f'(x)^2}{L}
                $$
                To conclude now use that $f>0$.






                share|cite|improve this answer











                $endgroup$



                When exploring the values between $x$ and $x-f'(x)$, it usually makes sense to parametrize that segment as $x-sf'(x)$, $sin[0,1]$. This avoids discuss the different cases of the order between these points.



                Let $L$ be the Lipschitz constant of $f'$, then we know from the fundamental theorem that
                begin{align}
                |f(x+u)-f(x)-f'(x)u|&leint_0^1|f'(x+su)-f'(x)|,ds,|u|
                \
                &le int_us,ds,L|u|^2
                \
                &=frac12L|u|^2
                end{align}



                Resolving the absolute value to one side it follows that
                $$
                f(x)+f'(x)uge f(x+u)-frac12L|u|^2.
                $$
                Now insert $u=-frac{f'(x)}{L}$ to find
                $$
                2f(x)ge 2fleft(x-frac{f'(x)}{L}right)+frac{f'(x)^2}{L}
                $$
                To conclude now use that $f>0$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 18 '18 at 17:31

























                answered Dec 18 '18 at 16:16









                LutzLLutzL

                58.8k42056




                58.8k42056






























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