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











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













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






share|cite|improve this question















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






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













    $f^{prime}(x)= (-1/ x^2)(cos (1/x))$, 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















                      Popular posts from this blog

                      'app-layout' is not a known element: how to share Component with different Modules

                      Image
                      0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
                      Read more

                      android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
                      Read more

                      WPF add header to Image with URL pettitions [duplicate]

                      Image
                      up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
                      Read more