Question related to Darboux's theorem












2












$begingroup$


Darboux's theorem says that $f'$ has intermediate value property. More precisely,




( Darboux's theorem) If $f$ is differentiable on $[a,b]$, and if r is any number for which $f'(a)<r<f'(b)$ then $exists$ c in (a,b) such that $f'(r)=c$.




Thus Darboux's theorem implies that $f'$ cannot have any simple discontinuities on $[a,b].$



I have a question as follows:




Is there a function $f$ satisfying $displaystyle lim_{xto d} f'(x)=infty$ for some $d in (a,b)$ under the assumptions in Darboux's theorem?




If so, I can conclude that for each $x in (a,b),$ either $(i)$ $f'$ is continuous at $x$ or $(ii)$ $f'$ oscillates near $x.$



It seems that there is no differentiable function $f$ on $[a,b]$ satisfying function $displaystyle lim_{xto d} f'(x)=infty$ and $exists f'(d)$ for some $d in (a,b)$.



Please let me know if you have any idea or comment for my question. Thanks in advance!






real-analysis calculus analysis







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$endgroup$








  • 1




    $begingroup$
    It is possible that $f'(d)$ exists but $lim_{x to d} f'(x)$ fails to exist (oscillating). It can even be unbounded. An example is $f(x) =x^2 sin x^{-2} $ for $x neq 0$ and $f(0) = 0$. You just can't have the derivative tend to $infty$.
    $endgroup$
    – RRL
    Dec 30 '18 at 3:35












  • $begingroup$
    @RRL Thanks for your comment. Now I get to know the information a lot from Darboux theorem.
    $endgroup$
    – 0706
    Dec 30 '18 at 3:43
















2












$begingroup$


Darboux's theorem says that $f'$ has intermediate value property. More precisely,




( Darboux's theorem) If $f$ is differentiable on $[a,b]$, and if r is any number for which $f'(a)<r<f'(b)$ then $exists$ c in (a,b) such that $f'(r)=c$.




Thus Darboux's theorem implies that $f'$ cannot have any simple discontinuities on $[a,b].$



I have a question as follows:




Is there a function $f$ satisfying $displaystyle lim_{xto d} f'(x)=infty$ for some $d in (a,b)$ under the assumptions in Darboux's theorem?




If so, I can conclude that for each $x in (a,b),$ either $(i)$ $f'$ is continuous at $x$ or $(ii)$ $f'$ oscillates near $x.$



It seems that there is no differentiable function $f$ on $[a,b]$ satisfying function $displaystyle lim_{xto d} f'(x)=infty$ and $exists f'(d)$ for some $d in (a,b)$.



Please let me know if you have any idea or comment for my question. Thanks in advance!






real-analysis calculus analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It is possible that $f'(d)$ exists but $lim_{x to d} f'(x)$ fails to exist (oscillating). It can even be unbounded. An example is $f(x) =x^2 sin x^{-2} $ for $x neq 0$ and $f(0) = 0$. You just can't have the derivative tend to $infty$.
    $endgroup$
    – RRL
    Dec 30 '18 at 3:35












  • $begingroup$
    @RRL Thanks for your comment. Now I get to know the information a lot from Darboux theorem.
    $endgroup$
    – 0706
    Dec 30 '18 at 3:43














2












2








2


1



$begingroup$


Darboux's theorem says that $f'$ has intermediate value property. More precisely,




( Darboux's theorem) If $f$ is differentiable on $[a,b]$, and if r is any number for which $f'(a)<r<f'(b)$ then $exists$ c in (a,b) such that $f'(r)=c$.




Thus Darboux's theorem implies that $f'$ cannot have any simple discontinuities on $[a,b].$



I have a question as follows:




Is there a function $f$ satisfying $displaystyle lim_{xto d} f'(x)=infty$ for some $d in (a,b)$ under the assumptions in Darboux's theorem?




If so, I can conclude that for each $x in (a,b),$ either $(i)$ $f'$ is continuous at $x$ or $(ii)$ $f'$ oscillates near $x.$



It seems that there is no differentiable function $f$ on $[a,b]$ satisfying function $displaystyle lim_{xto d} f'(x)=infty$ and $exists f'(d)$ for some $d in (a,b)$.



Please let me know if you have any idea or comment for my question. Thanks in advance!






real-analysis calculus analysis







share|cite|improve this question









$endgroup$




Darboux's theorem says that $f'$ has intermediate value property. More precisely,




( Darboux's theorem) If $f$ is differentiable on $[a,b]$, and if r is any number for which $f'(a)<r<f'(b)$ then $exists$ c in (a,b) such that $f'(r)=c$.




Thus Darboux's theorem implies that $f'$ cannot have any simple discontinuities on $[a,b].$



I have a question as follows:




Is there a function $f$ satisfying $displaystyle lim_{xto d} f'(x)=infty$ for some $d in (a,b)$ under the assumptions in Darboux's theorem?




If so, I can conclude that for each $x in (a,b),$ either $(i)$ $f'$ is continuous at $x$ or $(ii)$ $f'$ oscillates near $x.$



It seems that there is no differentiable function $f$ on $[a,b]$ satisfying function $displaystyle lim_{xto d} f'(x)=infty$ and $exists f'(d)$ for some $d in (a,b)$.



Please let me know if you have any idea or comment for my question. Thanks in advance!






real-analysis calculus analysis




real-analysis calculus analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 30 '18 at 1:36









07060706

441311




441311








  • 1




    $begingroup$
    It is possible that $f'(d)$ exists but $lim_{x to d} f'(x)$ fails to exist (oscillating). It can even be unbounded. An example is $f(x) =x^2 sin x^{-2} $ for $x neq 0$ and $f(0) = 0$. You just can't have the derivative tend to $infty$.
    $endgroup$
    – RRL
    Dec 30 '18 at 3:35












  • $begingroup$
    @RRL Thanks for your comment. Now I get to know the information a lot from Darboux theorem.
    $endgroup$
    – 0706
    Dec 30 '18 at 3:43














  • 1




    $begingroup$
    It is possible that $f'(d)$ exists but $lim_{x to d} f'(x)$ fails to exist (oscillating). It can even be unbounded. An example is $f(x) =x^2 sin x^{-2} $ for $x neq 0$ and $f(0) = 0$. You just can't have the derivative tend to $infty$.
    $endgroup$
    – RRL
    Dec 30 '18 at 3:35












  • $begingroup$
    @RRL Thanks for your comment. Now I get to know the information a lot from Darboux theorem.
    $endgroup$
    – 0706
    Dec 30 '18 at 3:43








1




1




$begingroup$
It is possible that $f'(d)$ exists but $lim_{x to d} f'(x)$ fails to exist (oscillating). It can even be unbounded. An example is $f(x) =x^2 sin x^{-2} $ for $x neq 0$ and $f(0) = 0$. You just can't have the derivative tend to $infty$.
$endgroup$
– RRL
Dec 30 '18 at 3:35






$begingroup$
It is possible that $f'(d)$ exists but $lim_{x to d} f'(x)$ fails to exist (oscillating). It can even be unbounded. An example is $f(x) =x^2 sin x^{-2} $ for $x neq 0$ and $f(0) = 0$. You just can't have the derivative tend to $infty$.
$endgroup$
– RRL
Dec 30 '18 at 3:35














$begingroup$
@RRL Thanks for your comment. Now I get to know the information a lot from Darboux theorem.
$endgroup$
– 0706
Dec 30 '18 at 3:43




$begingroup$
@RRL Thanks for your comment. Now I get to know the information a lot from Darboux theorem.
$endgroup$
– 0706
Dec 30 '18 at 3:43










2 Answers
2






active

oldest

votes


















2












$begingroup$

No, there can't be.



If $f'(x)to +infty$ for $xto d$, then there will be $delta$ such that $f'(x) > f'(d)+2$ on $(d, d+delta)$. And this contradicts Darboux's theorem on $[d,d+delta/2]$ -- there won't be any place with $f'(x) = f'(d)+1$.



Be a bit careful about the precise meaning of "oscillates" in your conclusion (ii), though. It is possible for $f'$ to have a strict global minimum at $d$, for example.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for your answer. In the last paragraph in your answer, you are right. The definition of oscillation is not clear. By the way, is it possible for $f’$ not to be oscillate? I think, at least one side, $f’$ should oscillate. Is it wrong?
    $endgroup$
    – 0706
    Dec 30 '18 at 2:50












  • $begingroup$
    @0706: That depends on exactly what "oscillate" means here. (I'm not sure if there is a standard definition and if so what it is).
    $endgroup$
    – Henning Makholm
    Dec 30 '18 at 2:52





















3












$begingroup$

If $f$ is differentiable on $(a,b)$ and $x_n to d in (a,b)$ from the right, we have



$$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d}$$



By the MVT there is a sequence $xi_n in (d,x_n)$ such that



$$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d} = lim_{n to infty} f'(xi_n)$$



However, if $f'(x) to infty$ as $x to d$, then we must have $f'(xi_n) to infty$ since $xi_n to d$, a contradiction.



Hence, your function does not exist.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

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    2












    $begingroup$

    No, there can't be.



    If $f'(x)to +infty$ for $xto d$, then there will be $delta$ such that $f'(x) > f'(d)+2$ on $(d, d+delta)$. And this contradicts Darboux's theorem on $[d,d+delta/2]$ -- there won't be any place with $f'(x) = f'(d)+1$.



    Be a bit careful about the precise meaning of "oscillates" in your conclusion (ii), though. It is possible for $f'$ to have a strict global minimum at $d$, for example.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for your answer. In the last paragraph in your answer, you are right. The definition of oscillation is not clear. By the way, is it possible for $f’$ not to be oscillate? I think, at least one side, $f’$ should oscillate. Is it wrong?
      $endgroup$
      – 0706
      Dec 30 '18 at 2:50












    • $begingroup$
      @0706: That depends on exactly what "oscillate" means here. (I'm not sure if there is a standard definition and if so what it is).
      $endgroup$
      – Henning Makholm
      Dec 30 '18 at 2:52


















    2












    $begingroup$

    No, there can't be.



    If $f'(x)to +infty$ for $xto d$, then there will be $delta$ such that $f'(x) > f'(d)+2$ on $(d, d+delta)$. And this contradicts Darboux's theorem on $[d,d+delta/2]$ -- there won't be any place with $f'(x) = f'(d)+1$.



    Be a bit careful about the precise meaning of "oscillates" in your conclusion (ii), though. It is possible for $f'$ to have a strict global minimum at $d$, for example.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for your answer. In the last paragraph in your answer, you are right. The definition of oscillation is not clear. By the way, is it possible for $f’$ not to be oscillate? I think, at least one side, $f’$ should oscillate. Is it wrong?
      $endgroup$
      – 0706
      Dec 30 '18 at 2:50












    • $begingroup$
      @0706: That depends on exactly what "oscillate" means here. (I'm not sure if there is a standard definition and if so what it is).
      $endgroup$
      – Henning Makholm
      Dec 30 '18 at 2:52
















    2












    2








    2





    $begingroup$

    No, there can't be.



    If $f'(x)to +infty$ for $xto d$, then there will be $delta$ such that $f'(x) > f'(d)+2$ on $(d, d+delta)$. And this contradicts Darboux's theorem on $[d,d+delta/2]$ -- there won't be any place with $f'(x) = f'(d)+1$.



    Be a bit careful about the precise meaning of "oscillates" in your conclusion (ii), though. It is possible for $f'$ to have a strict global minimum at $d$, for example.






    share|cite|improve this answer











    $endgroup$



    No, there can't be.



    If $f'(x)to +infty$ for $xto d$, then there will be $delta$ such that $f'(x) > f'(d)+2$ on $(d, d+delta)$. And this contradicts Darboux's theorem on $[d,d+delta/2]$ -- there won't be any place with $f'(x) = f'(d)+1$.



    Be a bit careful about the precise meaning of "oscillates" in your conclusion (ii), though. It is possible for $f'$ to have a strict global minimum at $d$, for example.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 30 '18 at 2:33

























    answered Dec 30 '18 at 2:26









    Henning MakholmHenning Makholm

    242k17308550




    242k17308550












    • $begingroup$
      Thanks for your answer. In the last paragraph in your answer, you are right. The definition of oscillation is not clear. By the way, is it possible for $f’$ not to be oscillate? I think, at least one side, $f’$ should oscillate. Is it wrong?
      $endgroup$
      – 0706
      Dec 30 '18 at 2:50












    • $begingroup$
      @0706: That depends on exactly what "oscillate" means here. (I'm not sure if there is a standard definition and if so what it is).
      $endgroup$
      – Henning Makholm
      Dec 30 '18 at 2:52




















    • $begingroup$
      Thanks for your answer. In the last paragraph in your answer, you are right. The definition of oscillation is not clear. By the way, is it possible for $f’$ not to be oscillate? I think, at least one side, $f’$ should oscillate. Is it wrong?
      $endgroup$
      – 0706
      Dec 30 '18 at 2:50












    • $begingroup$
      @0706: That depends on exactly what "oscillate" means here. (I'm not sure if there is a standard definition and if so what it is).
      $endgroup$
      – Henning Makholm
      Dec 30 '18 at 2:52


















    $begingroup$
    Thanks for your answer. In the last paragraph in your answer, you are right. The definition of oscillation is not clear. By the way, is it possible for $f’$ not to be oscillate? I think, at least one side, $f’$ should oscillate. Is it wrong?
    $endgroup$
    – 0706
    Dec 30 '18 at 2:50






    $begingroup$
    Thanks for your answer. In the last paragraph in your answer, you are right. The definition of oscillation is not clear. By the way, is it possible for $f’$ not to be oscillate? I think, at least one side, $f’$ should oscillate. Is it wrong?
    $endgroup$
    – 0706
    Dec 30 '18 at 2:50














    $begingroup$
    @0706: That depends on exactly what "oscillate" means here. (I'm not sure if there is a standard definition and if so what it is).
    $endgroup$
    – Henning Makholm
    Dec 30 '18 at 2:52






    $begingroup$
    @0706: That depends on exactly what "oscillate" means here. (I'm not sure if there is a standard definition and if so what it is).
    $endgroup$
    – Henning Makholm
    Dec 30 '18 at 2:52













    3












    $begingroup$

    If $f$ is differentiable on $(a,b)$ and $x_n to d in (a,b)$ from the right, we have



    $$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d}$$



    By the MVT there is a sequence $xi_n in (d,x_n)$ such that



    $$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d} = lim_{n to infty} f'(xi_n)$$



    However, if $f'(x) to infty$ as $x to d$, then we must have $f'(xi_n) to infty$ since $xi_n to d$, a contradiction.



    Hence, your function does not exist.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      If $f$ is differentiable on $(a,b)$ and $x_n to d in (a,b)$ from the right, we have



      $$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d}$$



      By the MVT there is a sequence $xi_n in (d,x_n)$ such that



      $$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d} = lim_{n to infty} f'(xi_n)$$



      However, if $f'(x) to infty$ as $x to d$, then we must have $f'(xi_n) to infty$ since $xi_n to d$, a contradiction.



      Hence, your function does not exist.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        If $f$ is differentiable on $(a,b)$ and $x_n to d in (a,b)$ from the right, we have



        $$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d}$$



        By the MVT there is a sequence $xi_n in (d,x_n)$ such that



        $$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d} = lim_{n to infty} f'(xi_n)$$



        However, if $f'(x) to infty$ as $x to d$, then we must have $f'(xi_n) to infty$ since $xi_n to d$, a contradiction.



        Hence, your function does not exist.






        share|cite|improve this answer











        $endgroup$



        If $f$ is differentiable on $(a,b)$ and $x_n to d in (a,b)$ from the right, we have



        $$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d}$$



        By the MVT there is a sequence $xi_n in (d,x_n)$ such that



        $$f'(d) = lim_{n to infty} frac{f(x_n) -f(d)}{x_n - d} = lim_{n to infty} f'(xi_n)$$



        However, if $f'(x) to infty$ as $x to d$, then we must have $f'(xi_n) to infty$ since $xi_n to d$, a contradiction.



        Hence, your function does not exist.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 30 '18 at 2:38

























        answered Dec 30 '18 at 2:15









        RRLRRL

        52.9k42573




        52.9k42573






























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