A function that is in $C^k(mathbb R)$ but not in $C^k+1(mathbb R)$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite












Is there an example of a family of functions, index by $k$, that is in $C_b^k(mathbb R)$ but not in $C_b^k+1(mathbb R)$ for arbitrary $k$?



$C_b^k(mathbb R)$ is the space of functions with bounded derivatives up to $k$.






functional-analysis special-functions







share|cite|improve this question























  • This should be a duplicate? But I did not find the previous one.
    – GEdgar
    Sep 5 at 10:45






  • 1




    Let $f(x) = 0$ for $x<0$ and $f(x) = arctan(x)^k-1$ for $x>0$ to get an example with bounded derivatives.
    – Winther
    Sep 5 at 10:51






  • 2




    You may want to know that $C^k(Bbb R)$ is not the space of functions with bounded derivatives up to $k$, but the space of functions with continuous derivatives up to $k$.
    – Saucy O'Path
    Sep 5 at 10:54










  • Doesn't $arctan$ have bounded derivatives for all $k$ (with the bound increasing with $k$)?
    – Tohiko
    Sep 5 at 12:45














up vote
2
down vote

favorite












Is there an example of a family of functions, index by $k$, that is in $C_b^k(mathbb R)$ but not in $C_b^k+1(mathbb R)$ for arbitrary $k$?



$C_b^k(mathbb R)$ is the space of functions with bounded derivatives up to $k$.






functional-analysis special-functions







share|cite|improve this question























  • This should be a duplicate? But I did not find the previous one.
    – GEdgar
    Sep 5 at 10:45






  • 1




    Let $f(x) = 0$ for $x<0$ and $f(x) = arctan(x)^k-1$ for $x>0$ to get an example with bounded derivatives.
    – Winther
    Sep 5 at 10:51






  • 2




    You may want to know that $C^k(Bbb R)$ is not the space of functions with bounded derivatives up to $k$, but the space of functions with continuous derivatives up to $k$.
    – Saucy O'Path
    Sep 5 at 10:54










  • Doesn't $arctan$ have bounded derivatives for all $k$ (with the bound increasing with $k$)?
    – Tohiko
    Sep 5 at 12:45












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Is there an example of a family of functions, index by $k$, that is in $C_b^k(mathbb R)$ but not in $C_b^k+1(mathbb R)$ for arbitrary $k$?



$C_b^k(mathbb R)$ is the space of functions with bounded derivatives up to $k$.






functional-analysis special-functions







share|cite|improve this question















Is there an example of a family of functions, index by $k$, that is in $C_b^k(mathbb R)$ but not in $C_b^k+1(mathbb R)$ for arbitrary $k$?



$C_b^k(mathbb R)$ is the space of functions with bounded derivatives up to $k$.






functional-analysis special-functions




functional-analysis special-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 5 at 18:41

























asked Sep 5 at 10:16









Tohiko

1365




1365











  • This should be a duplicate? But I did not find the previous one.
    – GEdgar
    Sep 5 at 10:45






  • 1




    Let $f(x) = 0$ for $x<0$ and $f(x) = arctan(x)^k-1$ for $x>0$ to get an example with bounded derivatives.
    – Winther
    Sep 5 at 10:51






  • 2




    You may want to know that $C^k(Bbb R)$ is not the space of functions with bounded derivatives up to $k$, but the space of functions with continuous derivatives up to $k$.
    – Saucy O'Path
    Sep 5 at 10:54










  • Doesn't $arctan$ have bounded derivatives for all $k$ (with the bound increasing with $k$)?
    – Tohiko
    Sep 5 at 12:45
















  • This should be a duplicate? But I did not find the previous one.
    – GEdgar
    Sep 5 at 10:45






  • 1




    Let $f(x) = 0$ for $x<0$ and $f(x) = arctan(x)^k-1$ for $x>0$ to get an example with bounded derivatives.
    – Winther
    Sep 5 at 10:51






  • 2




    You may want to know that $C^k(Bbb R)$ is not the space of functions with bounded derivatives up to $k$, but the space of functions with continuous derivatives up to $k$.
    – Saucy O'Path
    Sep 5 at 10:54










  • Doesn't $arctan$ have bounded derivatives for all $k$ (with the bound increasing with $k$)?
    – Tohiko
    Sep 5 at 12:45















This should be a duplicate? But I did not find the previous one.
– GEdgar
Sep 5 at 10:45




This should be a duplicate? But I did not find the previous one.
– GEdgar
Sep 5 at 10:45




1




1




Let $f(x) = 0$ for $x<0$ and $f(x) = arctan(x)^k-1$ for $x>0$ to get an example with bounded derivatives.
– Winther
Sep 5 at 10:51




Let $f(x) = 0$ for $x<0$ and $f(x) = arctan(x)^k-1$ for $x>0$ to get an example with bounded derivatives.
– Winther
Sep 5 at 10:51




2




2




You may want to know that $C^k(Bbb R)$ is not the space of functions with bounded derivatives up to $k$, but the space of functions with continuous derivatives up to $k$.
– Saucy O'Path
Sep 5 at 10:54




You may want to know that $C^k(Bbb R)$ is not the space of functions with bounded derivatives up to $k$, but the space of functions with continuous derivatives up to $k$.
– Saucy O'Path
Sep 5 at 10:54












Doesn't $arctan$ have bounded derivatives for all $k$ (with the bound increasing with $k$)?
– Tohiko
Sep 5 at 12:45




Doesn't $arctan$ have bounded derivatives for all $k$ (with the bound increasing with $k$)?
– Tohiko
Sep 5 at 12:45










3 Answers
3






active

oldest

votes

















up vote
0
down vote



accepted










Define $f$ by
$$
f(x) = begincases (1-x^2)^k+1 & |x|<1\0 & |x|ge1
endcases
$$
Then all derivatives of $f$ up to order $k$ are continuous (they are zero at $|x|=1$).
But the $k+1$-st derivative does not exist at $|x|=1$.






share|cite|improve this answer



























    up vote
    0
    down vote













    Take a continuous function with compact support but not differentiable function $f(x)$ and integrate it $k$ times:



    Let $g_1(x) = int_-infty^x f(x) dx$ and define:
    $g_i(x) = int_-infty^x g_i-1(x) dx$



    Use : Lebesgue's Differentiation Theorem for Continuous Functions
    to prove that the derivatives exist.



    $g_k(x)$ is the required function.






    share|cite|improve this answer



























      up vote
      0
      down vote













      $F (x)=|x|^lambda , ~~k<lambda<k+1$






      share|cite|improve this answer




















        Your Answer




        StackExchange.ifUsing("editor", function ()
        return StackExchange.using("mathjaxEditing", function ()
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        );
        );
        , "mathjax-editing");

        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',
        convertImagesToLinks: true,
        noModals: false,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        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%2f2906101%2fa-function-that-is-in-ck-mathbb-r-but-not-in-ck1-mathbb-r%23new-answer', 'question_page');

        );

        Post as a guest

















        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        0
        down vote



        accepted










        Define $f$ by
        $$
        f(x) = begincases (1-x^2)^k+1 & |x|<1\0 & |x|ge1
        endcases
        $$
        Then all derivatives of $f$ up to order $k$ are continuous (they are zero at $|x|=1$).
        But the $k+1$-st derivative does not exist at $|x|=1$.






        share|cite|improve this answer
























          up vote
          0
          down vote



          accepted










          Define $f$ by
          $$
          f(x) = begincases (1-x^2)^k+1 & |x|<1\0 & |x|ge1
          endcases
          $$
          Then all derivatives of $f$ up to order $k$ are continuous (they are zero at $|x|=1$).
          But the $k+1$-st derivative does not exist at $|x|=1$.






          share|cite|improve this answer






















            up vote
            0
            down vote



            accepted







            up vote
            0
            down vote



            accepted






            Define $f$ by
            $$
            f(x) = begincases (1-x^2)^k+1 & |x|<1\0 & |x|ge1
            endcases
            $$
            Then all derivatives of $f$ up to order $k$ are continuous (they are zero at $|x|=1$).
            But the $k+1$-st derivative does not exist at $|x|=1$.






            share|cite|improve this answer












            Define $f$ by
            $$
            f(x) = begincases (1-x^2)^k+1 & |x|<1\0 & |x|ge1
            endcases
            $$
            Then all derivatives of $f$ up to order $k$ are continuous (they are zero at $|x|=1$).
            But the $k+1$-st derivative does not exist at $|x|=1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 5 at 17:42









            daw

            22.2k1542




            22.2k1542




















                up vote
                0
                down vote













                Take a continuous function with compact support but not differentiable function $f(x)$ and integrate it $k$ times:



                Let $g_1(x) = int_-infty^x f(x) dx$ and define:
                $g_i(x) = int_-infty^x g_i-1(x) dx$



                Use : Lebesgue's Differentiation Theorem for Continuous Functions
                to prove that the derivatives exist.



                $g_k(x)$ is the required function.






                share|cite|improve this answer
























                  up vote
                  0
                  down vote













                  Take a continuous function with compact support but not differentiable function $f(x)$ and integrate it $k$ times:



                  Let $g_1(x) = int_-infty^x f(x) dx$ and define:
                  $g_i(x) = int_-infty^x g_i-1(x) dx$



                  Use : Lebesgue's Differentiation Theorem for Continuous Functions
                  to prove that the derivatives exist.



                  $g_k(x)$ is the required function.






                  share|cite|improve this answer






















                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    Take a continuous function with compact support but not differentiable function $f(x)$ and integrate it $k$ times:



                    Let $g_1(x) = int_-infty^x f(x) dx$ and define:
                    $g_i(x) = int_-infty^x g_i-1(x) dx$



                    Use : Lebesgue's Differentiation Theorem for Continuous Functions
                    to prove that the derivatives exist.



                    $g_k(x)$ is the required function.






                    share|cite|improve this answer












                    Take a continuous function with compact support but not differentiable function $f(x)$ and integrate it $k$ times:



                    Let $g_1(x) = int_-infty^x f(x) dx$ and define:
                    $g_i(x) = int_-infty^x g_i-1(x) dx$



                    Use : Lebesgue's Differentiation Theorem for Continuous Functions
                    to prove that the derivatives exist.



                    $g_k(x)$ is the required function.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 5 at 10:48









                    Balaji sb

                    40325




                    40325




















                        up vote
                        0
                        down vote













                        $F (x)=|x|^lambda , ~~k<lambda<k+1$






                        share|cite|improve this answer
























                          up vote
                          0
                          down vote













                          $F (x)=|x|^lambda , ~~k<lambda<k+1$






                          share|cite|improve this answer






















                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            $F (x)=|x|^lambda , ~~k<lambda<k+1$






                            share|cite|improve this answer












                            $F (x)=|x|^lambda , ~~k<lambda<k+1$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Sep 5 at 19:14









                            Eduardo

                            413112




                            413112



























                                 

                                draft saved


                                draft discarded















































                                 


                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function ()
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2906101%2fa-function-that-is-in-ck-mathbb-r-but-not-in-ck1-mathbb-r%23new-answer', 'question_page');

                                );

                                Post as a guest
































































                                -

                                這個網誌中的熱門文章

                                How to combine Bézier curves to a surface?

                                圖片
                                Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
                                閱讀完整內容

                                Mutual Information Always Non-negative

                                圖片
                                Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
                                閱讀完整內容

                                Why am i infinitely getting the same tweet with the Twitter Search API?

                                圖片
                                up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia...
                                閱讀完整內容