Is $f:(-1,1)to R$ defined as $f(x)= frac{x}{1-lvert x rvert}$ and $f^{-1}(x)$ uniformly continuos?












1












$begingroup$


I think this fuction is uniformly continuos, but the inverse is not since the limit in the proximities to $lvert x rvert = 1$ the function has a vertical asymptote.



However I tried a lot of algebra to prove the inverse function is uniformly continuos, but with no luck. Any Hints?






calculus limits continuity uniform-continuity







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I think this fuction is uniformly continuos, but the inverse is not since the limit in the proximities to $lvert x rvert = 1$ the function has a vertical asymptote.



    However I tried a lot of algebra to prove the inverse function is uniformly continuos, but with no luck. Any Hints?






    calculus limits continuity uniform-continuity







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I think this fuction is uniformly continuos, but the inverse is not since the limit in the proximities to $lvert x rvert = 1$ the function has a vertical asymptote.



      However I tried a lot of algebra to prove the inverse function is uniformly continuos, but with no luck. Any Hints?






      calculus limits continuity uniform-continuity







      share|cite|improve this question











      $endgroup$




      I think this fuction is uniformly continuos, but the inverse is not since the limit in the proximities to $lvert x rvert = 1$ the function has a vertical asymptote.



      However I tried a lot of algebra to prove the inverse function is uniformly continuos, but with no luck. Any Hints?






      calculus limits continuity uniform-continuity




      calculus limits continuity uniform-continuity






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 14 '18 at 17:20







      Maria Guthier

















      asked Dec 14 '18 at 17:03









      Maria GuthierMaria Guthier

      887




      887






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          This function cannot be uniformly continuous on the interval $(-1,1)$ due to the vertical asymptotes; you would have to cut back to $(-1+epsilon, 1-delta)$ in order to recover uniform continuity. See this post for some more tricks.



          EDIT: To address the flip side of this problem, note that $f^{-1}:mathbb{R}rightarrow (-1,1)$ can be written out explictly:
          $$
          f^{-1}(x) = left{
          begin{array}{ll}
          frac{x}{1-x} & quad x < 0 \
          0 & quad x=0 \
          frac{x}{1+x} & quad x > 0
          end{array}
          right.
          $$



          On the intervals $(-infty,-2]$ and $[2,infty)$ note that $f^{-1}$ has a bounded derivative. Thus we know that $f^{-1}$ is uniformly continuous there. The interval $[-2,2]$ is compact and the function $f^{-1}$ is continuous on that interval, and thus is uniformly continuous on that interval. Finally, this post shows us that a function which is uniformly continuous on finitely many adjacent intervals is also uniformly continuous on their union.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yeah, its the case with the inverse where I'm struggling.
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:08










          • $begingroup$
            @MariaGuthier Okay, I will attempt to address your other question as well, but you should still edit your post to indicate that you are not trying to prove the uniform continuity of the function $f$, since it isn't uniformly continuous.
            $endgroup$
            – GenericMathGuy
            Dec 14 '18 at 17:18










          • $begingroup$
            But I think it can be useful for people reading Mathexchange. But I edited so its less confusing!
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:19












          • $begingroup$
            @MariaGuthier I have added text addressing the 2nd part of your question.
            $endgroup$
            – GenericMathGuy
            Dec 14 '18 at 18:12



















          4












          $begingroup$

          Your function $f$ is not uniformly continuous. To see why, recall that a uniformly continuous function preserves Cauchy sequences but you can take a sequence $a_nin(-1,1)$ such that $a_nto 1$ but $f(a_n)toinfty$, henece $f$ cannot be unifromly continuous.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I know is the other case when I don't know how to prove uniformly continuity.
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:10










          • $begingroup$
            @MariaGuthier I don't understand your question then. Could you edit it to make it more understandable?
            $endgroup$
            – BigbearZzz
            Dec 14 '18 at 17:11











          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%2f3039640%2fis-f-1-1-to-r-defined-as-fx-fracx1-lvert-x-rvert-and-f-1x%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          This function cannot be uniformly continuous on the interval $(-1,1)$ due to the vertical asymptotes; you would have to cut back to $(-1+epsilon, 1-delta)$ in order to recover uniform continuity. See this post for some more tricks.



          EDIT: To address the flip side of this problem, note that $f^{-1}:mathbb{R}rightarrow (-1,1)$ can be written out explictly:
          $$
          f^{-1}(x) = left{
          begin{array}{ll}
          frac{x}{1-x} & quad x < 0 \
          0 & quad x=0 \
          frac{x}{1+x} & quad x > 0
          end{array}
          right.
          $$



          On the intervals $(-infty,-2]$ and $[2,infty)$ note that $f^{-1}$ has a bounded derivative. Thus we know that $f^{-1}$ is uniformly continuous there. The interval $[-2,2]$ is compact and the function $f^{-1}$ is continuous on that interval, and thus is uniformly continuous on that interval. Finally, this post shows us that a function which is uniformly continuous on finitely many adjacent intervals is also uniformly continuous on their union.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yeah, its the case with the inverse where I'm struggling.
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:08










          • $begingroup$
            @MariaGuthier Okay, I will attempt to address your other question as well, but you should still edit your post to indicate that you are not trying to prove the uniform continuity of the function $f$, since it isn't uniformly continuous.
            $endgroup$
            – GenericMathGuy
            Dec 14 '18 at 17:18










          • $begingroup$
            But I think it can be useful for people reading Mathexchange. But I edited so its less confusing!
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:19












          • $begingroup$
            @MariaGuthier I have added text addressing the 2nd part of your question.
            $endgroup$
            – GenericMathGuy
            Dec 14 '18 at 18:12
















          2












          $begingroup$

          This function cannot be uniformly continuous on the interval $(-1,1)$ due to the vertical asymptotes; you would have to cut back to $(-1+epsilon, 1-delta)$ in order to recover uniform continuity. See this post for some more tricks.



          EDIT: To address the flip side of this problem, note that $f^{-1}:mathbb{R}rightarrow (-1,1)$ can be written out explictly:
          $$
          f^{-1}(x) = left{
          begin{array}{ll}
          frac{x}{1-x} & quad x < 0 \
          0 & quad x=0 \
          frac{x}{1+x} & quad x > 0
          end{array}
          right.
          $$



          On the intervals $(-infty,-2]$ and $[2,infty)$ note that $f^{-1}$ has a bounded derivative. Thus we know that $f^{-1}$ is uniformly continuous there. The interval $[-2,2]$ is compact and the function $f^{-1}$ is continuous on that interval, and thus is uniformly continuous on that interval. Finally, this post shows us that a function which is uniformly continuous on finitely many adjacent intervals is also uniformly continuous on their union.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yeah, its the case with the inverse where I'm struggling.
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:08










          • $begingroup$
            @MariaGuthier Okay, I will attempt to address your other question as well, but you should still edit your post to indicate that you are not trying to prove the uniform continuity of the function $f$, since it isn't uniformly continuous.
            $endgroup$
            – GenericMathGuy
            Dec 14 '18 at 17:18










          • $begingroup$
            But I think it can be useful for people reading Mathexchange. But I edited so its less confusing!
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:19












          • $begingroup$
            @MariaGuthier I have added text addressing the 2nd part of your question.
            $endgroup$
            – GenericMathGuy
            Dec 14 '18 at 18:12














          2












          2








          2





          $begingroup$

          This function cannot be uniformly continuous on the interval $(-1,1)$ due to the vertical asymptotes; you would have to cut back to $(-1+epsilon, 1-delta)$ in order to recover uniform continuity. See this post for some more tricks.



          EDIT: To address the flip side of this problem, note that $f^{-1}:mathbb{R}rightarrow (-1,1)$ can be written out explictly:
          $$
          f^{-1}(x) = left{
          begin{array}{ll}
          frac{x}{1-x} & quad x < 0 \
          0 & quad x=0 \
          frac{x}{1+x} & quad x > 0
          end{array}
          right.
          $$



          On the intervals $(-infty,-2]$ and $[2,infty)$ note that $f^{-1}$ has a bounded derivative. Thus we know that $f^{-1}$ is uniformly continuous there. The interval $[-2,2]$ is compact and the function $f^{-1}$ is continuous on that interval, and thus is uniformly continuous on that interval. Finally, this post shows us that a function which is uniformly continuous on finitely many adjacent intervals is also uniformly continuous on their union.






          share|cite|improve this answer











          $endgroup$



          This function cannot be uniformly continuous on the interval $(-1,1)$ due to the vertical asymptotes; you would have to cut back to $(-1+epsilon, 1-delta)$ in order to recover uniform continuity. See this post for some more tricks.



          EDIT: To address the flip side of this problem, note that $f^{-1}:mathbb{R}rightarrow (-1,1)$ can be written out explictly:
          $$
          f^{-1}(x) = left{
          begin{array}{ll}
          frac{x}{1-x} & quad x < 0 \
          0 & quad x=0 \
          frac{x}{1+x} & quad x > 0
          end{array}
          right.
          $$



          On the intervals $(-infty,-2]$ and $[2,infty)$ note that $f^{-1}$ has a bounded derivative. Thus we know that $f^{-1}$ is uniformly continuous there. The interval $[-2,2]$ is compact and the function $f^{-1}$ is continuous on that interval, and thus is uniformly continuous on that interval. Finally, this post shows us that a function which is uniformly continuous on finitely many adjacent intervals is also uniformly continuous on their union.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 18:11

























          answered Dec 14 '18 at 17:06









          GenericMathGuyGenericMathGuy

          1312




          1312












          • $begingroup$
            Yeah, its the case with the inverse where I'm struggling.
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:08










          • $begingroup$
            @MariaGuthier Okay, I will attempt to address your other question as well, but you should still edit your post to indicate that you are not trying to prove the uniform continuity of the function $f$, since it isn't uniformly continuous.
            $endgroup$
            – GenericMathGuy
            Dec 14 '18 at 17:18










          • $begingroup$
            But I think it can be useful for people reading Mathexchange. But I edited so its less confusing!
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:19












          • $begingroup$
            @MariaGuthier I have added text addressing the 2nd part of your question.
            $endgroup$
            – GenericMathGuy
            Dec 14 '18 at 18:12


















          • $begingroup$
            Yeah, its the case with the inverse where I'm struggling.
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:08










          • $begingroup$
            @MariaGuthier Okay, I will attempt to address your other question as well, but you should still edit your post to indicate that you are not trying to prove the uniform continuity of the function $f$, since it isn't uniformly continuous.
            $endgroup$
            – GenericMathGuy
            Dec 14 '18 at 17:18










          • $begingroup$
            But I think it can be useful for people reading Mathexchange. But I edited so its less confusing!
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:19












          • $begingroup$
            @MariaGuthier I have added text addressing the 2nd part of your question.
            $endgroup$
            – GenericMathGuy
            Dec 14 '18 at 18:12
















          $begingroup$
          Yeah, its the case with the inverse where I'm struggling.
          $endgroup$
          – Maria Guthier
          Dec 14 '18 at 17:08




          $begingroup$
          Yeah, its the case with the inverse where I'm struggling.
          $endgroup$
          – Maria Guthier
          Dec 14 '18 at 17:08












          $begingroup$
          @MariaGuthier Okay, I will attempt to address your other question as well, but you should still edit your post to indicate that you are not trying to prove the uniform continuity of the function $f$, since it isn't uniformly continuous.
          $endgroup$
          – GenericMathGuy
          Dec 14 '18 at 17:18




          $begingroup$
          @MariaGuthier Okay, I will attempt to address your other question as well, but you should still edit your post to indicate that you are not trying to prove the uniform continuity of the function $f$, since it isn't uniformly continuous.
          $endgroup$
          – GenericMathGuy
          Dec 14 '18 at 17:18












          $begingroup$
          But I think it can be useful for people reading Mathexchange. But I edited so its less confusing!
          $endgroup$
          – Maria Guthier
          Dec 14 '18 at 17:19






          $begingroup$
          But I think it can be useful for people reading Mathexchange. But I edited so its less confusing!
          $endgroup$
          – Maria Guthier
          Dec 14 '18 at 17:19














          $begingroup$
          @MariaGuthier I have added text addressing the 2nd part of your question.
          $endgroup$
          – GenericMathGuy
          Dec 14 '18 at 18:12




          $begingroup$
          @MariaGuthier I have added text addressing the 2nd part of your question.
          $endgroup$
          – GenericMathGuy
          Dec 14 '18 at 18:12











          4












          $begingroup$

          Your function $f$ is not uniformly continuous. To see why, recall that a uniformly continuous function preserves Cauchy sequences but you can take a sequence $a_nin(-1,1)$ such that $a_nto 1$ but $f(a_n)toinfty$, henece $f$ cannot be unifromly continuous.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I know is the other case when I don't know how to prove uniformly continuity.
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:10










          • $begingroup$
            @MariaGuthier I don't understand your question then. Could you edit it to make it more understandable?
            $endgroup$
            – BigbearZzz
            Dec 14 '18 at 17:11
















          4












          $begingroup$

          Your function $f$ is not uniformly continuous. To see why, recall that a uniformly continuous function preserves Cauchy sequences but you can take a sequence $a_nin(-1,1)$ such that $a_nto 1$ but $f(a_n)toinfty$, henece $f$ cannot be unifromly continuous.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I know is the other case when I don't know how to prove uniformly continuity.
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:10










          • $begingroup$
            @MariaGuthier I don't understand your question then. Could you edit it to make it more understandable?
            $endgroup$
            – BigbearZzz
            Dec 14 '18 at 17:11














          4












          4








          4





          $begingroup$

          Your function $f$ is not uniformly continuous. To see why, recall that a uniformly continuous function preserves Cauchy sequences but you can take a sequence $a_nin(-1,1)$ such that $a_nto 1$ but $f(a_n)toinfty$, henece $f$ cannot be unifromly continuous.






          share|cite|improve this answer











          $endgroup$



          Your function $f$ is not uniformly continuous. To see why, recall that a uniformly continuous function preserves Cauchy sequences but you can take a sequence $a_nin(-1,1)$ such that $a_nto 1$ but $f(a_n)toinfty$, henece $f$ cannot be unifromly continuous.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 17 '18 at 9:10

























          answered Dec 14 '18 at 17:09









          BigbearZzzBigbearZzz

          8,75121652




          8,75121652












          • $begingroup$
            I know is the other case when I don't know how to prove uniformly continuity.
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:10










          • $begingroup$
            @MariaGuthier I don't understand your question then. Could you edit it to make it more understandable?
            $endgroup$
            – BigbearZzz
            Dec 14 '18 at 17:11


















          • $begingroup$
            I know is the other case when I don't know how to prove uniformly continuity.
            $endgroup$
            – Maria Guthier
            Dec 14 '18 at 17:10










          • $begingroup$
            @MariaGuthier I don't understand your question then. Could you edit it to make it more understandable?
            $endgroup$
            – BigbearZzz
            Dec 14 '18 at 17:11
















          $begingroup$
          I know is the other case when I don't know how to prove uniformly continuity.
          $endgroup$
          – Maria Guthier
          Dec 14 '18 at 17:10




          $begingroup$
          I know is the other case when I don't know how to prove uniformly continuity.
          $endgroup$
          – Maria Guthier
          Dec 14 '18 at 17:10












          $begingroup$
          @MariaGuthier I don't understand your question then. Could you edit it to make it more understandable?
          $endgroup$
          – BigbearZzz
          Dec 14 '18 at 17:11




          $begingroup$
          @MariaGuthier I don't understand your question then. Could you edit it to make it more understandable?
          $endgroup$
          – BigbearZzz
          Dec 14 '18 at 17:11


















          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%2f3039640%2fis-f-1-1-to-r-defined-as-fx-fracx1-lvert-x-rvert-and-f-1x%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

          How do I know what Microsoft account the skydrive app is syncing to?

          Image
          up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
          Read more

          Grease: Live!

          Image
          A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
          Read more

          When does type information flow backwards in C++?

          Image
          up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
          Read more