Prove upper bound on derivative












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$begingroup$


Let $f$ be twice differentiable function on interval $I$. Let $M_0=sup_{xin I}|f(x)|, M_1=sup_{xin I}|f'(x)|, M_2=sup_{xin I}|f''(x)|$. Show that



a) if $I=[-a, a]$, then
$$|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2$$
b) 1. if length of $I$ is not less than $2sqrt{M_0/M_2}$, then $M_1leqslant 2sqrt{M_0M_2}$




  1. if $I=mathbb{R}$, then $M_1leqslant sqrt{2M_0M_2}$


c) Constants $2$ and $sqrt{2}$ from b) cannot be improved



d) if $f$ is $p$ times differentiable on $mathbb{R}$ and $M_0$ and $M_p=sup_{xinmathbb{R}}|f^{(p)}(x)|$ are finite, then for $1leqslant kleqslant p$ $M_k=sup_{xinmathbb{R}}|f^{(k)}(x)|$ are finite and



$$M_kleqslant2^{k(p-k)/2}M_0^{1-k/p}M_p^{k/p}$$



Solutions for b) and c) were given here and here.



d) was solved here.



Now I am trying to solve a):



Using Taylor's expansion we have



$$f(a)=f(x)+f'(x)(a-x)+frac{1}{2}f''(xi_1)(a-x)^2\
f(-a)=f(x)+f'(x)(-a-x)+frac{1}{2}f''(xi_2)(-a-x)^2,$$
for some $xi_1, xi_2$. From this



$$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



Using triangle inequality gives $|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2+|x|cdot M_2$



How to eliminate that $|x|cdot M_2$?






real-analysis derivatives taylor-expansion







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $f$ be twice differentiable function on interval $I$. Let $M_0=sup_{xin I}|f(x)|, M_1=sup_{xin I}|f'(x)|, M_2=sup_{xin I}|f''(x)|$. Show that



    a) if $I=[-a, a]$, then
    $$|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2$$
    b) 1. if length of $I$ is not less than $2sqrt{M_0/M_2}$, then $M_1leqslant 2sqrt{M_0M_2}$




    1. if $I=mathbb{R}$, then $M_1leqslant sqrt{2M_0M_2}$


    c) Constants $2$ and $sqrt{2}$ from b) cannot be improved



    d) if $f$ is $p$ times differentiable on $mathbb{R}$ and $M_0$ and $M_p=sup_{xinmathbb{R}}|f^{(p)}(x)|$ are finite, then for $1leqslant kleqslant p$ $M_k=sup_{xinmathbb{R}}|f^{(k)}(x)|$ are finite and



    $$M_kleqslant2^{k(p-k)/2}M_0^{1-k/p}M_p^{k/p}$$



    Solutions for b) and c) were given here and here.



    d) was solved here.



    Now I am trying to solve a):



    Using Taylor's expansion we have



    $$f(a)=f(x)+f'(x)(a-x)+frac{1}{2}f''(xi_1)(a-x)^2\
    f(-a)=f(x)+f'(x)(-a-x)+frac{1}{2}f''(xi_2)(-a-x)^2,$$
    for some $xi_1, xi_2$. From this



    $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



    Using triangle inequality gives $|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2+|x|cdot M_2$



    How to eliminate that $|x|cdot M_2$?






    real-analysis derivatives taylor-expansion







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      0



      $begingroup$


      Let $f$ be twice differentiable function on interval $I$. Let $M_0=sup_{xin I}|f(x)|, M_1=sup_{xin I}|f'(x)|, M_2=sup_{xin I}|f''(x)|$. Show that



      a) if $I=[-a, a]$, then
      $$|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2$$
      b) 1. if length of $I$ is not less than $2sqrt{M_0/M_2}$, then $M_1leqslant 2sqrt{M_0M_2}$




      1. if $I=mathbb{R}$, then $M_1leqslant sqrt{2M_0M_2}$


      c) Constants $2$ and $sqrt{2}$ from b) cannot be improved



      d) if $f$ is $p$ times differentiable on $mathbb{R}$ and $M_0$ and $M_p=sup_{xinmathbb{R}}|f^{(p)}(x)|$ are finite, then for $1leqslant kleqslant p$ $M_k=sup_{xinmathbb{R}}|f^{(k)}(x)|$ are finite and



      $$M_kleqslant2^{k(p-k)/2}M_0^{1-k/p}M_p^{k/p}$$



      Solutions for b) and c) were given here and here.



      d) was solved here.



      Now I am trying to solve a):



      Using Taylor's expansion we have



      $$f(a)=f(x)+f'(x)(a-x)+frac{1}{2}f''(xi_1)(a-x)^2\
      f(-a)=f(x)+f'(x)(-a-x)+frac{1}{2}f''(xi_2)(-a-x)^2,$$
      for some $xi_1, xi_2$. From this



      $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



      Using triangle inequality gives $|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2+|x|cdot M_2$



      How to eliminate that $|x|cdot M_2$?






      real-analysis derivatives taylor-expansion







      share|cite|improve this question









      $endgroup$




      Let $f$ be twice differentiable function on interval $I$. Let $M_0=sup_{xin I}|f(x)|, M_1=sup_{xin I}|f'(x)|, M_2=sup_{xin I}|f''(x)|$. Show that



      a) if $I=[-a, a]$, then
      $$|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2$$
      b) 1. if length of $I$ is not less than $2sqrt{M_0/M_2}$, then $M_1leqslant 2sqrt{M_0M_2}$




      1. if $I=mathbb{R}$, then $M_1leqslant sqrt{2M_0M_2}$


      c) Constants $2$ and $sqrt{2}$ from b) cannot be improved



      d) if $f$ is $p$ times differentiable on $mathbb{R}$ and $M_0$ and $M_p=sup_{xinmathbb{R}}|f^{(p)}(x)|$ are finite, then for $1leqslant kleqslant p$ $M_k=sup_{xinmathbb{R}}|f^{(k)}(x)|$ are finite and



      $$M_kleqslant2^{k(p-k)/2}M_0^{1-k/p}M_p^{k/p}$$



      Solutions for b) and c) were given here and here.



      d) was solved here.



      Now I am trying to solve a):



      Using Taylor's expansion we have



      $$f(a)=f(x)+f'(x)(a-x)+frac{1}{2}f''(xi_1)(a-x)^2\
      f(-a)=f(x)+f'(x)(-a-x)+frac{1}{2}f''(xi_2)(-a-x)^2,$$
      for some $xi_1, xi_2$. From this



      $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



      Using triangle inequality gives $|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2+|x|cdot M_2$



      How to eliminate that $|x|cdot M_2$?






      real-analysis derivatives taylor-expansion




      real-analysis derivatives taylor-expansion






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 11 '18 at 6:37









      B. BekmaganbetovB. Bekmaganbetov

      214




      214






















          2 Answers
          2






          active

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          1












          $begingroup$

          Don't write $f'(x)$ as:



          $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



          but simply as:



          $$f'(x)=frac{f(a)-f(-a)}{2a}-frac{1}{4a}f''(xi_1)(a-x_0)^2+frac{1}{4a}f''(xi_2)(a+x_0)^2$$



          and so it follows that:



          $$|f'(x)|leq frac{M_0}{a}+frac{M_2}{4a}(a-x_0)^2+frac{M_2}{4a}(a+x_0)^2=frac{M_0}{a}+frac{M_2}{2a}(a^2+x_0^2)$$



          that is the claim.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Taylor expansion about $x=x_0$ is given by
            $$f(x)=f(x_0)+f'(x_0)(x-x_0)+frac{f''(x_0)}{2}(x-x_0)^2.$$
            So,
            $$f(a)=f(x_0)+f'(x_0)(a-x_0)+frac{f''(x_0)}{2}(a-x_0)^2,~~~~~~~~~~~~~~~(1)$$
            and
            $$f(-a)=f(x_0)+f'(x_0)(-a-x_0)+frac{f''(x_0)}{2}(-a-x_0)^2~~~~~~~~~~~~~~~(2)$$
            Subtracting $(2)$ from $(1)$, we get
            $$f(a)-f(-a)=2a~f'(x_0)+frac{1}{2}(-4ax_0)f''(x_0)$$
            $$implies f'(x_0)=frac{f(a)-f(-a)}{2a}+frac{1}{2a}(2ax_0)f''(x_0).$$
            Using triangle inequality, we have
            $$|f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(2ax_0)M_2$$
            $$implies |f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(x_0^2+a^2)M_2~~~(because (x_0-a)^2geq 0).$$
            Since $x_0$ was arbitrary, so
            $$|f'(x)|leqfrac{M_0}{a}+frac{1}{2a}(x^2+a^2)M_2.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              (1) and (2) are incorrect. See en.wikipedia.org/wiki/Taylor's_theorem. You cannot write $f''(x_0)$ because the point lies between $x_0$ and $a$ or $-a$, but it is not $x_0$. Consider $f(x)=x^3$ on $[-1, 1]$.
              $endgroup$
              – B. Bekmaganbetov
              Mar 11 '18 at 8:35











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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Don't write $f'(x)$ as:



            $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



            but simply as:



            $$f'(x)=frac{f(a)-f(-a)}{2a}-frac{1}{4a}f''(xi_1)(a-x_0)^2+frac{1}{4a}f''(xi_2)(a+x_0)^2$$



            and so it follows that:



            $$|f'(x)|leq frac{M_0}{a}+frac{M_2}{4a}(a-x_0)^2+frac{M_2}{4a}(a+x_0)^2=frac{M_0}{a}+frac{M_2}{2a}(a^2+x_0^2)$$



            that is the claim.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Don't write $f'(x)$ as:



              $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



              but simply as:



              $$f'(x)=frac{f(a)-f(-a)}{2a}-frac{1}{4a}f''(xi_1)(a-x_0)^2+frac{1}{4a}f''(xi_2)(a+x_0)^2$$



              and so it follows that:



              $$|f'(x)|leq frac{M_0}{a}+frac{M_2}{4a}(a-x_0)^2+frac{M_2}{4a}(a+x_0)^2=frac{M_0}{a}+frac{M_2}{2a}(a^2+x_0^2)$$



              that is the claim.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Don't write $f'(x)$ as:



                $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



                but simply as:



                $$f'(x)=frac{f(a)-f(-a)}{2a}-frac{1}{4a}f''(xi_1)(a-x_0)^2+frac{1}{4a}f''(xi_2)(a+x_0)^2$$



                and so it follows that:



                $$|f'(x)|leq frac{M_0}{a}+frac{M_2}{4a}(a-x_0)^2+frac{M_2}{4a}(a+x_0)^2=frac{M_0}{a}+frac{M_2}{2a}(a^2+x_0^2)$$



                that is the claim.






                share|cite|improve this answer









                $endgroup$



                Don't write $f'(x)$ as:



                $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



                but simply as:



                $$f'(x)=frac{f(a)-f(-a)}{2a}-frac{1}{4a}f''(xi_1)(a-x_0)^2+frac{1}{4a}f''(xi_2)(a+x_0)^2$$



                and so it follows that:



                $$|f'(x)|leq frac{M_0}{a}+frac{M_2}{4a}(a-x_0)^2+frac{M_2}{4a}(a+x_0)^2=frac{M_0}{a}+frac{M_2}{2a}(a^2+x_0^2)$$



                that is the claim.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 24 at 20:34









                NamelessNameless

                788




                788























                    0












                    $begingroup$

                    Taylor expansion about $x=x_0$ is given by
                    $$f(x)=f(x_0)+f'(x_0)(x-x_0)+frac{f''(x_0)}{2}(x-x_0)^2.$$
                    So,
                    $$f(a)=f(x_0)+f'(x_0)(a-x_0)+frac{f''(x_0)}{2}(a-x_0)^2,~~~~~~~~~~~~~~~(1)$$
                    and
                    $$f(-a)=f(x_0)+f'(x_0)(-a-x_0)+frac{f''(x_0)}{2}(-a-x_0)^2~~~~~~~~~~~~~~~(2)$$
                    Subtracting $(2)$ from $(1)$, we get
                    $$f(a)-f(-a)=2a~f'(x_0)+frac{1}{2}(-4ax_0)f''(x_0)$$
                    $$implies f'(x_0)=frac{f(a)-f(-a)}{2a}+frac{1}{2a}(2ax_0)f''(x_0).$$
                    Using triangle inequality, we have
                    $$|f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(2ax_0)M_2$$
                    $$implies |f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(x_0^2+a^2)M_2~~~(because (x_0-a)^2geq 0).$$
                    Since $x_0$ was arbitrary, so
                    $$|f'(x)|leqfrac{M_0}{a}+frac{1}{2a}(x^2+a^2)M_2.$$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      (1) and (2) are incorrect. See en.wikipedia.org/wiki/Taylor's_theorem. You cannot write $f''(x_0)$ because the point lies between $x_0$ and $a$ or $-a$, but it is not $x_0$. Consider $f(x)=x^3$ on $[-1, 1]$.
                      $endgroup$
                      – B. Bekmaganbetov
                      Mar 11 '18 at 8:35
















                    0












                    $begingroup$

                    Taylor expansion about $x=x_0$ is given by
                    $$f(x)=f(x_0)+f'(x_0)(x-x_0)+frac{f''(x_0)}{2}(x-x_0)^2.$$
                    So,
                    $$f(a)=f(x_0)+f'(x_0)(a-x_0)+frac{f''(x_0)}{2}(a-x_0)^2,~~~~~~~~~~~~~~~(1)$$
                    and
                    $$f(-a)=f(x_0)+f'(x_0)(-a-x_0)+frac{f''(x_0)}{2}(-a-x_0)^2~~~~~~~~~~~~~~~(2)$$
                    Subtracting $(2)$ from $(1)$, we get
                    $$f(a)-f(-a)=2a~f'(x_0)+frac{1}{2}(-4ax_0)f''(x_0)$$
                    $$implies f'(x_0)=frac{f(a)-f(-a)}{2a}+frac{1}{2a}(2ax_0)f''(x_0).$$
                    Using triangle inequality, we have
                    $$|f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(2ax_0)M_2$$
                    $$implies |f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(x_0^2+a^2)M_2~~~(because (x_0-a)^2geq 0).$$
                    Since $x_0$ was arbitrary, so
                    $$|f'(x)|leqfrac{M_0}{a}+frac{1}{2a}(x^2+a^2)M_2.$$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      (1) and (2) are incorrect. See en.wikipedia.org/wiki/Taylor's_theorem. You cannot write $f''(x_0)$ because the point lies between $x_0$ and $a$ or $-a$, but it is not $x_0$. Consider $f(x)=x^3$ on $[-1, 1]$.
                      $endgroup$
                      – B. Bekmaganbetov
                      Mar 11 '18 at 8:35














                    0












                    0








                    0





                    $begingroup$

                    Taylor expansion about $x=x_0$ is given by
                    $$f(x)=f(x_0)+f'(x_0)(x-x_0)+frac{f''(x_0)}{2}(x-x_0)^2.$$
                    So,
                    $$f(a)=f(x_0)+f'(x_0)(a-x_0)+frac{f''(x_0)}{2}(a-x_0)^2,~~~~~~~~~~~~~~~(1)$$
                    and
                    $$f(-a)=f(x_0)+f'(x_0)(-a-x_0)+frac{f''(x_0)}{2}(-a-x_0)^2~~~~~~~~~~~~~~~(2)$$
                    Subtracting $(2)$ from $(1)$, we get
                    $$f(a)-f(-a)=2a~f'(x_0)+frac{1}{2}(-4ax_0)f''(x_0)$$
                    $$implies f'(x_0)=frac{f(a)-f(-a)}{2a}+frac{1}{2a}(2ax_0)f''(x_0).$$
                    Using triangle inequality, we have
                    $$|f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(2ax_0)M_2$$
                    $$implies |f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(x_0^2+a^2)M_2~~~(because (x_0-a)^2geq 0).$$
                    Since $x_0$ was arbitrary, so
                    $$|f'(x)|leqfrac{M_0}{a}+frac{1}{2a}(x^2+a^2)M_2.$$






                    share|cite|improve this answer









                    $endgroup$



                    Taylor expansion about $x=x_0$ is given by
                    $$f(x)=f(x_0)+f'(x_0)(x-x_0)+frac{f''(x_0)}{2}(x-x_0)^2.$$
                    So,
                    $$f(a)=f(x_0)+f'(x_0)(a-x_0)+frac{f''(x_0)}{2}(a-x_0)^2,~~~~~~~~~~~~~~~(1)$$
                    and
                    $$f(-a)=f(x_0)+f'(x_0)(-a-x_0)+frac{f''(x_0)}{2}(-a-x_0)^2~~~~~~~~~~~~~~~(2)$$
                    Subtracting $(2)$ from $(1)$, we get
                    $$f(a)-f(-a)=2a~f'(x_0)+frac{1}{2}(-4ax_0)f''(x_0)$$
                    $$implies f'(x_0)=frac{f(a)-f(-a)}{2a}+frac{1}{2a}(2ax_0)f''(x_0).$$
                    Using triangle inequality, we have
                    $$|f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(2ax_0)M_2$$
                    $$implies |f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(x_0^2+a^2)M_2~~~(because (x_0-a)^2geq 0).$$
                    Since $x_0$ was arbitrary, so
                    $$|f'(x)|leqfrac{M_0}{a}+frac{1}{2a}(x^2+a^2)M_2.$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 11 '18 at 7:48









                    GoldyGoldy

                    43414




                    43414












                    • $begingroup$
                      (1) and (2) are incorrect. See en.wikipedia.org/wiki/Taylor's_theorem. You cannot write $f''(x_0)$ because the point lies between $x_0$ and $a$ or $-a$, but it is not $x_0$. Consider $f(x)=x^3$ on $[-1, 1]$.
                      $endgroup$
                      – B. Bekmaganbetov
                      Mar 11 '18 at 8:35


















                    • $begingroup$
                      (1) and (2) are incorrect. See en.wikipedia.org/wiki/Taylor's_theorem. You cannot write $f''(x_0)$ because the point lies between $x_0$ and $a$ or $-a$, but it is not $x_0$. Consider $f(x)=x^3$ on $[-1, 1]$.
                      $endgroup$
                      – B. Bekmaganbetov
                      Mar 11 '18 at 8:35
















                    $begingroup$
                    (1) and (2) are incorrect. See en.wikipedia.org/wiki/Taylor's_theorem. You cannot write $f''(x_0)$ because the point lies between $x_0$ and $a$ or $-a$, but it is not $x_0$. Consider $f(x)=x^3$ on $[-1, 1]$.
                    $endgroup$
                    – B. Bekmaganbetov
                    Mar 11 '18 at 8:35




                    $begingroup$
                    (1) and (2) are incorrect. See en.wikipedia.org/wiki/Taylor's_theorem. You cannot write $f''(x_0)$ because the point lies between $x_0$ and $a$ or $-a$, but it is not $x_0$. Consider $f(x)=x^3$ on $[-1, 1]$.
                    $endgroup$
                    – B. Bekmaganbetov
                    Mar 11 '18 at 8:35


















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