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












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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







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    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
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          $begingroup$

          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





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          2 Answers
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          2 Answers
          2






          active

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          active

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          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


















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