How do I show that all continuous periodic functions are bounded and uniform continuous?












17












$begingroup$



A function $f:mathbb{R}to mathbb{R}$ is periodic if there exits $p>0$ such that $f(x+P)=f(x)$ for all $xin mathbb{R}$. Show that every continuous periodic function is bounded and uniformly continuous.




For boundedness, I first tried to show that since the a periodic function is continuous, it is continuous for the closed interval $[x_0,x_0+P]$. I know that there is a theorem saying that if it is continuous on a closed interval, then it is bounded. However, I'm not allowed to state that theorem directly. Should I just aim for a contradiction by supposing f is not bounded on the interval stated above?






real-analysis continuity uniform-continuity periodic-functions







share|cite|improve this question











$endgroup$

















    17












    $begingroup$



    A function $f:mathbb{R}to mathbb{R}$ is periodic if there exits $p>0$ such that $f(x+P)=f(x)$ for all $xin mathbb{R}$. Show that every continuous periodic function is bounded and uniformly continuous.




    For boundedness, I first tried to show that since the a periodic function is continuous, it is continuous for the closed interval $[x_0,x_0+P]$. I know that there is a theorem saying that if it is continuous on a closed interval, then it is bounded. However, I'm not allowed to state that theorem directly. Should I just aim for a contradiction by supposing f is not bounded on the interval stated above?






    real-analysis continuity uniform-continuity periodic-functions







    share|cite|improve this question











    $endgroup$















      17












      17








      17


      6



      $begingroup$



      A function $f:mathbb{R}to mathbb{R}$ is periodic if there exits $p>0$ such that $f(x+P)=f(x)$ for all $xin mathbb{R}$. Show that every continuous periodic function is bounded and uniformly continuous.




      For boundedness, I first tried to show that since the a periodic function is continuous, it is continuous for the closed interval $[x_0,x_0+P]$. I know that there is a theorem saying that if it is continuous on a closed interval, then it is bounded. However, I'm not allowed to state that theorem directly. Should I just aim for a contradiction by supposing f is not bounded on the interval stated above?






      real-analysis continuity uniform-continuity periodic-functions







      share|cite|improve this question











      $endgroup$





      A function $f:mathbb{R}to mathbb{R}$ is periodic if there exits $p>0$ such that $f(x+P)=f(x)$ for all $xin mathbb{R}$. Show that every continuous periodic function is bounded and uniformly continuous.




      For boundedness, I first tried to show that since the a periodic function is continuous, it is continuous for the closed interval $[x_0,x_0+P]$. I know that there is a theorem saying that if it is continuous on a closed interval, then it is bounded. However, I'm not allowed to state that theorem directly. Should I just aim for a contradiction by supposing f is not bounded on the interval stated above?






      real-analysis continuity uniform-continuity periodic-functions




      real-analysis continuity uniform-continuity periodic-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 14 at 13:07









      Martin Sleziak

      44.8k9118272




      44.8k9118272










      asked Apr 30 '14 at 1:13









      Oscar FloresOscar Flores

      623618




      623618






















          3 Answers
          3






          active

          oldest

          votes


















          16












          $begingroup$

          Suppose that $f$ has period one. Since $f:[0,2]toBbb R$ is continuous, it is bounded, so $f$ is bounded all over $Bbb R$ (why?). Also, $f:[0,2]toBbb R$ is uniformly continuous, being continuous on a compact set. Thus, given $varepsilon >0$ there exists $delta>0$ such that, whenever $|x-y|<delta,x,yin[0,2]$, then $|f(x)-f(y)|<varepsilon$. Pick arbitrary $x,y$, and assume $x<y$ with $|x-y|<delta$. We may take $delta <1$. I claim there is an integer $n$ such that $x-n,y-nin [0,2]$. Then $|x-y|=|x-n-(y-n)|$ and $f(x-n)=f(x)$, $f(y-n)=f(y)$. Can you continue now?



          Drawing a picture would prove useful. Essentially, you're translating the problem to $[0,2]$ where we already solved the issue.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ah, I was trying to deal with the case where, say, we have $x= 0.99$ and $y = 1.01$. Considering $f$ on $[0,2]$ instead of $[0,1]$ is a smart way to deal with that!
            $endgroup$
            – MCT
            Mar 4 '15 at 22:06



















          4












          $begingroup$

          A periodic continuous function is simply a function defined on a circle,
          $$f:S^1 longrightarrow R.$$
          Circle is a compact space. A continuous function on a compact space is both bounded and uniformly continuous.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But $f$ is defined on $mathbb R$, not on the circle.
            $endgroup$
            – Al Jebr
            Jun 24 '18 at 18:54










          • $begingroup$
            AI Jebr: We can identify a periodic function with one defined on a circle. You could also argue by noting that the image of a periodic function is the same as its image when restricted to one cycle.
            $endgroup$
            – Behnam Esmayli
            Jun 24 '18 at 22:38



















          1












          $begingroup$

          Here's proof for boundedness part by contradiction :



          Fix $a in Bbb R$. Consider the domain $I=[a,a+p]$. Assume $f$ is not bounded. Then there exists a sequence $x_n in I$ such that $|f(x_n)| gt n$.



          Since $x_n$ is bounded and consists of real numbers, it has a convergent subsequence $x_{n_k}$. But $I$ is closed hence $x_{n_k}$ converges inside $I$. i.e. $exists x in I$ such that $x_{n_k} to x$.



          What we have got is a sequence $x_{n_k} to x in I$ and $|f(x_{n_k})| to infty ; (because |f(x_{n_k})| gt n_k)$. This is contradiction for $f$ being continuous on $I$ and hence at $x$.






          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%2f775045%2fhow-do-i-show-that-all-continuous-periodic-functions-are-bounded-and-uniform-con%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            16












            $begingroup$

            Suppose that $f$ has period one. Since $f:[0,2]toBbb R$ is continuous, it is bounded, so $f$ is bounded all over $Bbb R$ (why?). Also, $f:[0,2]toBbb R$ is uniformly continuous, being continuous on a compact set. Thus, given $varepsilon >0$ there exists $delta>0$ such that, whenever $|x-y|<delta,x,yin[0,2]$, then $|f(x)-f(y)|<varepsilon$. Pick arbitrary $x,y$, and assume $x<y$ with $|x-y|<delta$. We may take $delta <1$. I claim there is an integer $n$ such that $x-n,y-nin [0,2]$. Then $|x-y|=|x-n-(y-n)|$ and $f(x-n)=f(x)$, $f(y-n)=f(y)$. Can you continue now?



            Drawing a picture would prove useful. Essentially, you're translating the problem to $[0,2]$ where we already solved the issue.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              ah, I was trying to deal with the case where, say, we have $x= 0.99$ and $y = 1.01$. Considering $f$ on $[0,2]$ instead of $[0,1]$ is a smart way to deal with that!
              $endgroup$
              – MCT
              Mar 4 '15 at 22:06
















            16












            $begingroup$

            Suppose that $f$ has period one. Since $f:[0,2]toBbb R$ is continuous, it is bounded, so $f$ is bounded all over $Bbb R$ (why?). Also, $f:[0,2]toBbb R$ is uniformly continuous, being continuous on a compact set. Thus, given $varepsilon >0$ there exists $delta>0$ such that, whenever $|x-y|<delta,x,yin[0,2]$, then $|f(x)-f(y)|<varepsilon$. Pick arbitrary $x,y$, and assume $x<y$ with $|x-y|<delta$. We may take $delta <1$. I claim there is an integer $n$ such that $x-n,y-nin [0,2]$. Then $|x-y|=|x-n-(y-n)|$ and $f(x-n)=f(x)$, $f(y-n)=f(y)$. Can you continue now?



            Drawing a picture would prove useful. Essentially, you're translating the problem to $[0,2]$ where we already solved the issue.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              ah, I was trying to deal with the case where, say, we have $x= 0.99$ and $y = 1.01$. Considering $f$ on $[0,2]$ instead of $[0,1]$ is a smart way to deal with that!
              $endgroup$
              – MCT
              Mar 4 '15 at 22:06














            16












            16








            16





            $begingroup$

            Suppose that $f$ has period one. Since $f:[0,2]toBbb R$ is continuous, it is bounded, so $f$ is bounded all over $Bbb R$ (why?). Also, $f:[0,2]toBbb R$ is uniformly continuous, being continuous on a compact set. Thus, given $varepsilon >0$ there exists $delta>0$ such that, whenever $|x-y|<delta,x,yin[0,2]$, then $|f(x)-f(y)|<varepsilon$. Pick arbitrary $x,y$, and assume $x<y$ with $|x-y|<delta$. We may take $delta <1$. I claim there is an integer $n$ such that $x-n,y-nin [0,2]$. Then $|x-y|=|x-n-(y-n)|$ and $f(x-n)=f(x)$, $f(y-n)=f(y)$. Can you continue now?



            Drawing a picture would prove useful. Essentially, you're translating the problem to $[0,2]$ where we already solved the issue.






            share|cite|improve this answer











            $endgroup$



            Suppose that $f$ has period one. Since $f:[0,2]toBbb R$ is continuous, it is bounded, so $f$ is bounded all over $Bbb R$ (why?). Also, $f:[0,2]toBbb R$ is uniformly continuous, being continuous on a compact set. Thus, given $varepsilon >0$ there exists $delta>0$ such that, whenever $|x-y|<delta,x,yin[0,2]$, then $|f(x)-f(y)|<varepsilon$. Pick arbitrary $x,y$, and assume $x<y$ with $|x-y|<delta$. We may take $delta <1$. I claim there is an integer $n$ such that $x-n,y-nin [0,2]$. Then $|x-y|=|x-n-(y-n)|$ and $f(x-n)=f(x)$, $f(y-n)=f(y)$. Can you continue now?



            Drawing a picture would prove useful. Essentially, you're translating the problem to $[0,2]$ where we already solved the issue.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 4 '15 at 22:13

























            answered Apr 30 '14 at 1:19









            Pedro Tamaroff♦Pedro Tamaroff

            96.7k10153297




            96.7k10153297












            • $begingroup$
              ah, I was trying to deal with the case where, say, we have $x= 0.99$ and $y = 1.01$. Considering $f$ on $[0,2]$ instead of $[0,1]$ is a smart way to deal with that!
              $endgroup$
              – MCT
              Mar 4 '15 at 22:06


















            • $begingroup$
              ah, I was trying to deal with the case where, say, we have $x= 0.99$ and $y = 1.01$. Considering $f$ on $[0,2]$ instead of $[0,1]$ is a smart way to deal with that!
              $endgroup$
              – MCT
              Mar 4 '15 at 22:06
















            $begingroup$
            ah, I was trying to deal with the case where, say, we have $x= 0.99$ and $y = 1.01$. Considering $f$ on $[0,2]$ instead of $[0,1]$ is a smart way to deal with that!
            $endgroup$
            – MCT
            Mar 4 '15 at 22:06




            $begingroup$
            ah, I was trying to deal with the case where, say, we have $x= 0.99$ and $y = 1.01$. Considering $f$ on $[0,2]$ instead of $[0,1]$ is a smart way to deal with that!
            $endgroup$
            – MCT
            Mar 4 '15 at 22:06











            4












            $begingroup$

            A periodic continuous function is simply a function defined on a circle,
            $$f:S^1 longrightarrow R.$$
            Circle is a compact space. A continuous function on a compact space is both bounded and uniformly continuous.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But $f$ is defined on $mathbb R$, not on the circle.
              $endgroup$
              – Al Jebr
              Jun 24 '18 at 18:54










            • $begingroup$
              AI Jebr: We can identify a periodic function with one defined on a circle. You could also argue by noting that the image of a periodic function is the same as its image when restricted to one cycle.
              $endgroup$
              – Behnam Esmayli
              Jun 24 '18 at 22:38
















            4












            $begingroup$

            A periodic continuous function is simply a function defined on a circle,
            $$f:S^1 longrightarrow R.$$
            Circle is a compact space. A continuous function on a compact space is both bounded and uniformly continuous.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But $f$ is defined on $mathbb R$, not on the circle.
              $endgroup$
              – Al Jebr
              Jun 24 '18 at 18:54










            • $begingroup$
              AI Jebr: We can identify a periodic function with one defined on a circle. You could also argue by noting that the image of a periodic function is the same as its image when restricted to one cycle.
              $endgroup$
              – Behnam Esmayli
              Jun 24 '18 at 22:38














            4












            4








            4





            $begingroup$

            A periodic continuous function is simply a function defined on a circle,
            $$f:S^1 longrightarrow R.$$
            Circle is a compact space. A continuous function on a compact space is both bounded and uniformly continuous.






            share|cite|improve this answer









            $endgroup$



            A periodic continuous function is simply a function defined on a circle,
            $$f:S^1 longrightarrow R.$$
            Circle is a compact space. A continuous function on a compact space is both bounded and uniformly continuous.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 9 '17 at 6:12









            Behnam EsmayliBehnam Esmayli

            1,956515




            1,956515












            • $begingroup$
              But $f$ is defined on $mathbb R$, not on the circle.
              $endgroup$
              – Al Jebr
              Jun 24 '18 at 18:54










            • $begingroup$
              AI Jebr: We can identify a periodic function with one defined on a circle. You could also argue by noting that the image of a periodic function is the same as its image when restricted to one cycle.
              $endgroup$
              – Behnam Esmayli
              Jun 24 '18 at 22:38


















            • $begingroup$
              But $f$ is defined on $mathbb R$, not on the circle.
              $endgroup$
              – Al Jebr
              Jun 24 '18 at 18:54










            • $begingroup$
              AI Jebr: We can identify a periodic function with one defined on a circle. You could also argue by noting that the image of a periodic function is the same as its image when restricted to one cycle.
              $endgroup$
              – Behnam Esmayli
              Jun 24 '18 at 22:38
















            $begingroup$
            But $f$ is defined on $mathbb R$, not on the circle.
            $endgroup$
            – Al Jebr
            Jun 24 '18 at 18:54




            $begingroup$
            But $f$ is defined on $mathbb R$, not on the circle.
            $endgroup$
            – Al Jebr
            Jun 24 '18 at 18:54












            $begingroup$
            AI Jebr: We can identify a periodic function with one defined on a circle. You could also argue by noting that the image of a periodic function is the same as its image when restricted to one cycle.
            $endgroup$
            – Behnam Esmayli
            Jun 24 '18 at 22:38




            $begingroup$
            AI Jebr: We can identify a periodic function with one defined on a circle. You could also argue by noting that the image of a periodic function is the same as its image when restricted to one cycle.
            $endgroup$
            – Behnam Esmayli
            Jun 24 '18 at 22:38











            1












            $begingroup$

            Here's proof for boundedness part by contradiction :



            Fix $a in Bbb R$. Consider the domain $I=[a,a+p]$. Assume $f$ is not bounded. Then there exists a sequence $x_n in I$ such that $|f(x_n)| gt n$.



            Since $x_n$ is bounded and consists of real numbers, it has a convergent subsequence $x_{n_k}$. But $I$ is closed hence $x_{n_k}$ converges inside $I$. i.e. $exists x in I$ such that $x_{n_k} to x$.



            What we have got is a sequence $x_{n_k} to x in I$ and $|f(x_{n_k})| to infty ; (because |f(x_{n_k})| gt n_k)$. This is contradiction for $f$ being continuous on $I$ and hence at $x$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Here's proof for boundedness part by contradiction :



              Fix $a in Bbb R$. Consider the domain $I=[a,a+p]$. Assume $f$ is not bounded. Then there exists a sequence $x_n in I$ such that $|f(x_n)| gt n$.



              Since $x_n$ is bounded and consists of real numbers, it has a convergent subsequence $x_{n_k}$. But $I$ is closed hence $x_{n_k}$ converges inside $I$. i.e. $exists x in I$ such that $x_{n_k} to x$.



              What we have got is a sequence $x_{n_k} to x in I$ and $|f(x_{n_k})| to infty ; (because |f(x_{n_k})| gt n_k)$. This is contradiction for $f$ being continuous on $I$ and hence at $x$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Here's proof for boundedness part by contradiction :



                Fix $a in Bbb R$. Consider the domain $I=[a,a+p]$. Assume $f$ is not bounded. Then there exists a sequence $x_n in I$ such that $|f(x_n)| gt n$.



                Since $x_n$ is bounded and consists of real numbers, it has a convergent subsequence $x_{n_k}$. But $I$ is closed hence $x_{n_k}$ converges inside $I$. i.e. $exists x in I$ such that $x_{n_k} to x$.



                What we have got is a sequence $x_{n_k} to x in I$ and $|f(x_{n_k})| to infty ; (because |f(x_{n_k})| gt n_k)$. This is contradiction for $f$ being continuous on $I$ and hence at $x$.






                share|cite|improve this answer









                $endgroup$



                Here's proof for boundedness part by contradiction :



                Fix $a in Bbb R$. Consider the domain $I=[a,a+p]$. Assume $f$ is not bounded. Then there exists a sequence $x_n in I$ such that $|f(x_n)| gt n$.



                Since $x_n$ is bounded and consists of real numbers, it has a convergent subsequence $x_{n_k}$. But $I$ is closed hence $x_{n_k}$ converges inside $I$. i.e. $exists x in I$ such that $x_{n_k} to x$.



                What we have got is a sequence $x_{n_k} to x in I$ and $|f(x_{n_k})| to infty ; (because |f(x_{n_k})| gt n_k)$. This is contradiction for $f$ being continuous on $I$ and hence at $x$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 9 '17 at 5:22









                Error 404Error 404

                3,84321336




                3,84321336






























                    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%2f775045%2fhow-do-i-show-that-all-continuous-periodic-functions-are-bounded-and-uniform-con%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

                    Mario Kart Wii

                    Image
                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
                    Read more

                    What does “Dominus providebit” mean?

                    Image
                    5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
                    Read more

                    Antonio Litta Visconti Arese

                    Image
                    Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
                    Read more