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

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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







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    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







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      17












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      17


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      $begingroup$



      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






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      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
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          $begingroup$

          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









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            3 Answers
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            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









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