Prove a Continuous Function Bounded












2












$begingroup$


Let $f:mathbb{R}rightarrowmathbb{R}$ be a continuous function which satisfies
$$sup_{x,yinmathbb{R}}|f(x+y)-f(x)-f(y)|<infty$$
If we have $displaystyle{lim_{nrightarrowinfty,ninmathbb{N}}frac{f(n)}{n}=2014}$, prove $displaystyle{sup_{xinmathbb{R}}|f(x)-2014x|}<infty$.



(Problem 1 in S.-T. Yau College Student Mathematics Contests 2014, Analysis and Differential Equations Individual)



Firstly we can replace $2014$ by $0$ so we actually need to prove $f(x)$ to be bounded over $mathbb{R}$. And it's not difficult to prove that $displaystyle{lim_{xrightarrow+infty}frac{f(x)}{x}=0}$, but idk how to go on with it. Thanks for your help!!!






real-analysis







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $f:mathbb{R}rightarrowmathbb{R}$ be a continuous function which satisfies
    $$sup_{x,yinmathbb{R}}|f(x+y)-f(x)-f(y)|<infty$$
    If we have $displaystyle{lim_{nrightarrowinfty,ninmathbb{N}}frac{f(n)}{n}=2014}$, prove $displaystyle{sup_{xinmathbb{R}}|f(x)-2014x|}<infty$.



    (Problem 1 in S.-T. Yau College Student Mathematics Contests 2014, Analysis and Differential Equations Individual)



    Firstly we can replace $2014$ by $0$ so we actually need to prove $f(x)$ to be bounded over $mathbb{R}$. And it's not difficult to prove that $displaystyle{lim_{xrightarrow+infty}frac{f(x)}{x}=0}$, but idk how to go on with it. Thanks for your help!!!






    real-analysis







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Let $f:mathbb{R}rightarrowmathbb{R}$ be a continuous function which satisfies
      $$sup_{x,yinmathbb{R}}|f(x+y)-f(x)-f(y)|<infty$$
      If we have $displaystyle{lim_{nrightarrowinfty,ninmathbb{N}}frac{f(n)}{n}=2014}$, prove $displaystyle{sup_{xinmathbb{R}}|f(x)-2014x|}<infty$.



      (Problem 1 in S.-T. Yau College Student Mathematics Contests 2014, Analysis and Differential Equations Individual)



      Firstly we can replace $2014$ by $0$ so we actually need to prove $f(x)$ to be bounded over $mathbb{R}$. And it's not difficult to prove that $displaystyle{lim_{xrightarrow+infty}frac{f(x)}{x}=0}$, but idk how to go on with it. Thanks for your help!!!






      real-analysis







      share|cite|improve this question









      $endgroup$




      Let $f:mathbb{R}rightarrowmathbb{R}$ be a continuous function which satisfies
      $$sup_{x,yinmathbb{R}}|f(x+y)-f(x)-f(y)|<infty$$
      If we have $displaystyle{lim_{nrightarrowinfty,ninmathbb{N}}frac{f(n)}{n}=2014}$, prove $displaystyle{sup_{xinmathbb{R}}|f(x)-2014x|}<infty$.



      (Problem 1 in S.-T. Yau College Student Mathematics Contests 2014, Analysis and Differential Equations Individual)



      Firstly we can replace $2014$ by $0$ so we actually need to prove $f(x)$ to be bounded over $mathbb{R}$. And it's not difficult to prove that $displaystyle{lim_{xrightarrow+infty}frac{f(x)}{x}=0}$, but idk how to go on with it. Thanks for your help!!!






      real-analysis




      real-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 10 '17 at 8:19









      AbnerYeAbnerYe

      468211




      468211






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          WLOG, consider a continuous function $g:mathbb{R}tomathbb{R}$ such that





          • $g(0)=0$,


          • $g([0,+infty)) supset [0,+infty)$,


          • $|g(x+y)-g(x)-g(y)| le 1$ for any $x,yinmathbb{R}$.


          Since $g$ is continuous, we can define
          $$ a_k := inf { x>0 : g(x)=2^k }, quad k=1,2,cdots. $$
          It is clear that $a_k$ increases to $+infty$ as $ktoinfty$. Also, we will find that
          $$ g(a_k)/a_k ge 1/a_1 >0, $$
          which provides a contradiction.



          Note that $a_1 in (0,+infty)$ and $g(a_1)=2$. If $a>0$ and $g(a)=n>0$, then
          $$ |g(a+a_1)-g(a)-g(a_1)| le 1 implies g(a+a_1) ge n+1, $$
          and
          $$ |g(2a+a_1)-g(a)-g(a+a_1)| le 1 implies g(2a+a_1) ge 2n. $$
          Therefore, we have $a_{k+1} le 2 a_k + a_1$, and thus
          $$ a_{k+1}+a_1 le 2(a_k+a_1) implies a_k+a_1 le 2^{k-1}(a_1+a_1). $$
          It follows that $a_k le 2^k a_1$ and the proof is complete.






          share|cite|improve this answer









          $endgroup$














            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',
            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%2f2091487%2fprove-a-continuous-function-bounded%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            WLOG, consider a continuous function $g:mathbb{R}tomathbb{R}$ such that





            • $g(0)=0$,


            • $g([0,+infty)) supset [0,+infty)$,


            • $|g(x+y)-g(x)-g(y)| le 1$ for any $x,yinmathbb{R}$.


            Since $g$ is continuous, we can define
            $$ a_k := inf { x>0 : g(x)=2^k }, quad k=1,2,cdots. $$
            It is clear that $a_k$ increases to $+infty$ as $ktoinfty$. Also, we will find that
            $$ g(a_k)/a_k ge 1/a_1 >0, $$
            which provides a contradiction.



            Note that $a_1 in (0,+infty)$ and $g(a_1)=2$. If $a>0$ and $g(a)=n>0$, then
            $$ |g(a+a_1)-g(a)-g(a_1)| le 1 implies g(a+a_1) ge n+1, $$
            and
            $$ |g(2a+a_1)-g(a)-g(a+a_1)| le 1 implies g(2a+a_1) ge 2n. $$
            Therefore, we have $a_{k+1} le 2 a_k + a_1$, and thus
            $$ a_{k+1}+a_1 le 2(a_k+a_1) implies a_k+a_1 le 2^{k-1}(a_1+a_1). $$
            It follows that $a_k le 2^k a_1$ and the proof is complete.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              WLOG, consider a continuous function $g:mathbb{R}tomathbb{R}$ such that





              • $g(0)=0$,


              • $g([0,+infty)) supset [0,+infty)$,


              • $|g(x+y)-g(x)-g(y)| le 1$ for any $x,yinmathbb{R}$.


              Since $g$ is continuous, we can define
              $$ a_k := inf { x>0 : g(x)=2^k }, quad k=1,2,cdots. $$
              It is clear that $a_k$ increases to $+infty$ as $ktoinfty$. Also, we will find that
              $$ g(a_k)/a_k ge 1/a_1 >0, $$
              which provides a contradiction.



              Note that $a_1 in (0,+infty)$ and $g(a_1)=2$. If $a>0$ and $g(a)=n>0$, then
              $$ |g(a+a_1)-g(a)-g(a_1)| le 1 implies g(a+a_1) ge n+1, $$
              and
              $$ |g(2a+a_1)-g(a)-g(a+a_1)| le 1 implies g(2a+a_1) ge 2n. $$
              Therefore, we have $a_{k+1} le 2 a_k + a_1$, and thus
              $$ a_{k+1}+a_1 le 2(a_k+a_1) implies a_k+a_1 le 2^{k-1}(a_1+a_1). $$
              It follows that $a_k le 2^k a_1$ and the proof is complete.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                WLOG, consider a continuous function $g:mathbb{R}tomathbb{R}$ such that





                • $g(0)=0$,


                • $g([0,+infty)) supset [0,+infty)$,


                • $|g(x+y)-g(x)-g(y)| le 1$ for any $x,yinmathbb{R}$.


                Since $g$ is continuous, we can define
                $$ a_k := inf { x>0 : g(x)=2^k }, quad k=1,2,cdots. $$
                It is clear that $a_k$ increases to $+infty$ as $ktoinfty$. Also, we will find that
                $$ g(a_k)/a_k ge 1/a_1 >0, $$
                which provides a contradiction.



                Note that $a_1 in (0,+infty)$ and $g(a_1)=2$. If $a>0$ and $g(a)=n>0$, then
                $$ |g(a+a_1)-g(a)-g(a_1)| le 1 implies g(a+a_1) ge n+1, $$
                and
                $$ |g(2a+a_1)-g(a)-g(a+a_1)| le 1 implies g(2a+a_1) ge 2n. $$
                Therefore, we have $a_{k+1} le 2 a_k + a_1$, and thus
                $$ a_{k+1}+a_1 le 2(a_k+a_1) implies a_k+a_1 le 2^{k-1}(a_1+a_1). $$
                It follows that $a_k le 2^k a_1$ and the proof is complete.






                share|cite|improve this answer









                $endgroup$



                WLOG, consider a continuous function $g:mathbb{R}tomathbb{R}$ such that





                • $g(0)=0$,


                • $g([0,+infty)) supset [0,+infty)$,


                • $|g(x+y)-g(x)-g(y)| le 1$ for any $x,yinmathbb{R}$.


                Since $g$ is continuous, we can define
                $$ a_k := inf { x>0 : g(x)=2^k }, quad k=1,2,cdots. $$
                It is clear that $a_k$ increases to $+infty$ as $ktoinfty$. Also, we will find that
                $$ g(a_k)/a_k ge 1/a_1 >0, $$
                which provides a contradiction.



                Note that $a_1 in (0,+infty)$ and $g(a_1)=2$. If $a>0$ and $g(a)=n>0$, then
                $$ |g(a+a_1)-g(a)-g(a_1)| le 1 implies g(a+a_1) ge n+1, $$
                and
                $$ |g(2a+a_1)-g(a)-g(a+a_1)| le 1 implies g(2a+a_1) ge 2n. $$
                Therefore, we have $a_{k+1} le 2 a_k + a_1$, and thus
                $$ a_{k+1}+a_1 le 2(a_k+a_1) implies a_k+a_1 le 2^{k-1}(a_1+a_1). $$
                It follows that $a_k le 2^k a_1$ and the proof is complete.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 31 at 13:10









                kelltykellty

                9417




                9417






























                    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%2f2091487%2fprove-a-continuous-function-bounded%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

                    MongoDB - Not Authorized To Execute Command

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
                    Read more

                    in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

                    Image
                    1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
                    Read more

                    How to fix TextFormField cause rebuild widget in Flutter

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
                    Read more