Is $ f colon (1, +infty) to mathbb{R} , f(x) = sin frac{1}{x} $ uniformly continuous? [duplicate]











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  • $sin frac{1}{x}$ is not uniformly continuous on $[1, infty]$

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Is $$ f colon (1, +infty) to mathbb{R} ,
f(x) = sin frac{1}{x} $$ uniformly continuous? I think it could be but can't prove it. Would appreciate the help.
Edit: I'm looking for an answer in which mean value theorem and boundedness of the derivative isn't used.
So, I'm left with the definition and Lipschitz






uniform-continuity







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marked as duplicate by Chinnapparaj R, Lord Shark the Unknown, Brahadeesh, max_zorn, KReiser 2 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • You are asking different questions in title and main body
    – ramanujan
    21 hours ago










  • sorry I fixed it now
    – user15269
    21 hours ago










  • What is $f$ here, Be specific!
    – Chinnapparaj R
    21 hours ago










  • yes but I'm looking at $ (1, +infty)$ not $mathbb{R}$ so it's not the same question?
    – user15269
    21 hours ago










  • Note that $f$ has bounded derivative. The result follows from mean value theorem.
    – Danny Pak-Keung Chan
    21 hours ago















up vote
1
down vote

favorite













This question already has an answer here:




  • $sin frac{1}{x}$ is not uniformly continuous on $[1, infty]$

    1 answer




Is $$ f colon (1, +infty) to mathbb{R} ,
f(x) = sin frac{1}{x} $$ uniformly continuous? I think it could be but can't prove it. Would appreciate the help.
Edit: I'm looking for an answer in which mean value theorem and boundedness of the derivative isn't used.
So, I'm left with the definition and Lipschitz






uniform-continuity







share|cite|improve this question















marked as duplicate by Chinnapparaj R, Lord Shark the Unknown, Brahadeesh, max_zorn, KReiser 2 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • You are asking different questions in title and main body
    – ramanujan
    21 hours ago










  • sorry I fixed it now
    – user15269
    21 hours ago










  • What is $f$ here, Be specific!
    – Chinnapparaj R
    21 hours ago










  • yes but I'm looking at $ (1, +infty)$ not $mathbb{R}$ so it's not the same question?
    – user15269
    21 hours ago










  • Note that $f$ has bounded derivative. The result follows from mean value theorem.
    – Danny Pak-Keung Chan
    21 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite












This question already has an answer here:




  • $sin frac{1}{x}$ is not uniformly continuous on $[1, infty]$

    1 answer




Is $$ f colon (1, +infty) to mathbb{R} ,
f(x) = sin frac{1}{x} $$ uniformly continuous? I think it could be but can't prove it. Would appreciate the help.
Edit: I'm looking for an answer in which mean value theorem and boundedness of the derivative isn't used.
So, I'm left with the definition and Lipschitz






uniform-continuity







share|cite|improve this question
















This question already has an answer here:




  • $sin frac{1}{x}$ is not uniformly continuous on $[1, infty]$

    1 answer




Is $$ f colon (1, +infty) to mathbb{R} ,
f(x) = sin frac{1}{x} $$ uniformly continuous? I think it could be but can't prove it. Would appreciate the help.
Edit: I'm looking for an answer in which mean value theorem and boundedness of the derivative isn't used.
So, I'm left with the definition and Lipschitz





This question already has an answer here:




  • $sin frac{1}{x}$ is not uniformly continuous on $[1, infty]$

    1 answer







uniform-continuity




uniform-continuity






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 21 hours ago

























asked 21 hours ago









user15269

1448




1448




marked as duplicate by Chinnapparaj R, Lord Shark the Unknown, Brahadeesh, max_zorn, KReiser 2 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Chinnapparaj R, Lord Shark the Unknown, Brahadeesh, max_zorn, KReiser 2 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • You are asking different questions in title and main body
    – ramanujan
    21 hours ago










  • sorry I fixed it now
    – user15269
    21 hours ago










  • What is $f$ here, Be specific!
    – Chinnapparaj R
    21 hours ago










  • yes but I'm looking at $ (1, +infty)$ not $mathbb{R}$ so it's not the same question?
    – user15269
    21 hours ago










  • Note that $f$ has bounded derivative. The result follows from mean value theorem.
    – Danny Pak-Keung Chan
    21 hours ago


















  • You are asking different questions in title and main body
    – ramanujan
    21 hours ago










  • sorry I fixed it now
    – user15269
    21 hours ago










  • What is $f$ here, Be specific!
    – Chinnapparaj R
    21 hours ago










  • yes but I'm looking at $ (1, +infty)$ not $mathbb{R}$ so it's not the same question?
    – user15269
    21 hours ago










  • Note that $f$ has bounded derivative. The result follows from mean value theorem.
    – Danny Pak-Keung Chan
    21 hours ago
















You are asking different questions in title and main body
– ramanujan
21 hours ago




You are asking different questions in title and main body
– ramanujan
21 hours ago












sorry I fixed it now
– user15269
21 hours ago




sorry I fixed it now
– user15269
21 hours ago












What is $f$ here, Be specific!
– Chinnapparaj R
21 hours ago




What is $f$ here, Be specific!
– Chinnapparaj R
21 hours ago












yes but I'm looking at $ (1, +infty)$ not $mathbb{R}$ so it's not the same question?
– user15269
21 hours ago




yes but I'm looking at $ (1, +infty)$ not $mathbb{R}$ so it's not the same question?
– user15269
21 hours ago












Note that $f$ has bounded derivative. The result follows from mean value theorem.
– Danny Pak-Keung Chan
21 hours ago




Note that $f$ has bounded derivative. The result follows from mean value theorem.
– Danny Pak-Keung Chan
21 hours ago










2 Answers
2






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0
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Let $x$ and $y$ be different points in $(1,+infty)$. By the Mean Value Theorem, there exists a point $z$ between $x$ and $y$ such that



$$
f(x)-f(y)=cos(frac{1}{z})(frac{1}{x}-frac{1}{y})=cos(frac{1}{z})(frac{y-x}{xy})
$$
Then
$$
|f(x)-f(y)|leq|x-y|
$$
So $f$ is $1$-Lipschitz and, in particular, uniformly continuous.






share|cite|improve this answer








New contributor




Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

























    up vote
    0
    down vote













    $f^{prime}(x)= (-1/ x^2)(cos (1/x))$, which is bounded. So, given function is uniformly continuous.






    share|cite|improve this answer




























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      0
      down vote













      Let $x$ and $y$ be different points in $(1,+infty)$. By the Mean Value Theorem, there exists a point $z$ between $x$ and $y$ such that



      $$
      f(x)-f(y)=cos(frac{1}{z})(frac{1}{x}-frac{1}{y})=cos(frac{1}{z})(frac{y-x}{xy})
      $$
      Then
      $$
      |f(x)-f(y)|leq|x-y|
      $$
      So $f$ is $1$-Lipschitz and, in particular, uniformly continuous.






      share|cite|improve this answer








      New contributor




      Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















        up vote
        0
        down vote













        Let $x$ and $y$ be different points in $(1,+infty)$. By the Mean Value Theorem, there exists a point $z$ between $x$ and $y$ such that



        $$
        f(x)-f(y)=cos(frac{1}{z})(frac{1}{x}-frac{1}{y})=cos(frac{1}{z})(frac{y-x}{xy})
        $$
        Then
        $$
        |f(x)-f(y)|leq|x-y|
        $$
        So $f$ is $1$-Lipschitz and, in particular, uniformly continuous.






        share|cite|improve this answer








        New contributor




        Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.




















          up vote
          0
          down vote










          up vote
          0
          down vote









          Let $x$ and $y$ be different points in $(1,+infty)$. By the Mean Value Theorem, there exists a point $z$ between $x$ and $y$ such that



          $$
          f(x)-f(y)=cos(frac{1}{z})(frac{1}{x}-frac{1}{y})=cos(frac{1}{z})(frac{y-x}{xy})
          $$
          Then
          $$
          |f(x)-f(y)|leq|x-y|
          $$
          So $f$ is $1$-Lipschitz and, in particular, uniformly continuous.






          share|cite|improve this answer








          New contributor




          Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          Let $x$ and $y$ be different points in $(1,+infty)$. By the Mean Value Theorem, there exists a point $z$ between $x$ and $y$ such that



          $$
          f(x)-f(y)=cos(frac{1}{z})(frac{1}{x}-frac{1}{y})=cos(frac{1}{z})(frac{y-x}{xy})
          $$
          Then
          $$
          |f(x)-f(y)|leq|x-y|
          $$
          So $f$ is $1$-Lipschitz and, in particular, uniformly continuous.







          share|cite|improve this answer








          New contributor




          Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|cite|improve this answer



          share|cite|improve this answer






          New contributor




          Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 21 hours ago









          Dante Grevino

          1463




          1463




          New contributor




          Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






















              up vote
              0
              down vote













              $f^{prime}(x)= (-1/ x^2)(cos (1/x))$, which is bounded. So, given function is uniformly continuous.






              share|cite|improve this answer

























                up vote
                0
                down vote













                $f^{prime}(x)= (-1/ x^2)(cos (1/x))$, which is bounded. So, given function is uniformly continuous.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  $f^{prime}(x)= (-1/ x^2)(cos (1/x))$, which is bounded. So, given function is uniformly continuous.






                  share|cite|improve this answer












                  $f^{prime}(x)= (-1/ x^2)(cos (1/x))$, which is bounded. So, given function is uniformly continuous.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 21 hours ago









                  ramanujan

                  640713




                  640713















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