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







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












$begingroup$


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

460311




460311








  • 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











$endgroup$














    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056433%2fquestion-related-to-darbouxs-theorem%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    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


















    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

    244k17312556




    244k17312556












    • $begingroup$
      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

        53.8k52675




        53.8k52675






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056433%2fquestion-related-to-darbouxs-theorem%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
            Read more