Are $F$ and $F^{-1}$ continuous in this given metric space?












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The full question is below:



Let $S _ { infty }$ be the set of continuous functions $f : [ 0,1 ] rightarrow [0,1]$, with the distance $d _ { infty } ( f , g ) = sup | f ( x ) - g ( x ) |$, and let $S_1$ be the same set of continuous functions on $[0,1]$ but with the distance $d _ { 1 } ( f , g ) = int _ { 0 } ^ { 1 } | f ( x ) - g ( x ) | d x$ . Now let $F : S _ { infty } rightarrow S _ { 1 }$ be the identity map taking any function $f$ to itself. Are $F$ and $F^{-1}$ continuous in this given metric space?



I believe that $F^{-1}$ will not be continuous due to a counter example of $f_n(x) = x^n$. Unless I am mistaken, this is because $f_n$ will converge to $0$ in $ (S_1,d_1),$ but won't converge in $(S_{infty},d_{infty})$. This would be enough proof that it does not converge, right? I am not sure how to prove that $F$ does or does not converge though. I would assume that it does converge, but I am not certain.



By the way, this is from a class on Real Analysis so my knowledge of topology is fairly limited.






real-analysis convergence metric-spaces







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    0












    $begingroup$


    The full question is below:



    Let $S _ { infty }$ be the set of continuous functions $f : [ 0,1 ] rightarrow [0,1]$, with the distance $d _ { infty } ( f , g ) = sup | f ( x ) - g ( x ) |$, and let $S_1$ be the same set of continuous functions on $[0,1]$ but with the distance $d _ { 1 } ( f , g ) = int _ { 0 } ^ { 1 } | f ( x ) - g ( x ) | d x$ . Now let $F : S _ { infty } rightarrow S _ { 1 }$ be the identity map taking any function $f$ to itself. Are $F$ and $F^{-1}$ continuous in this given metric space?



    I believe that $F^{-1}$ will not be continuous due to a counter example of $f_n(x) = x^n$. Unless I am mistaken, this is because $f_n$ will converge to $0$ in $ (S_1,d_1),$ but won't converge in $(S_{infty},d_{infty})$. This would be enough proof that it does not converge, right? I am not sure how to prove that $F$ does or does not converge though. I would assume that it does converge, but I am not certain.



    By the way, this is from a class on Real Analysis so my knowledge of topology is fairly limited.






    real-analysis convergence metric-spaces







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      The full question is below:



      Let $S _ { infty }$ be the set of continuous functions $f : [ 0,1 ] rightarrow [0,1]$, with the distance $d _ { infty } ( f , g ) = sup | f ( x ) - g ( x ) |$, and let $S_1$ be the same set of continuous functions on $[0,1]$ but with the distance $d _ { 1 } ( f , g ) = int _ { 0 } ^ { 1 } | f ( x ) - g ( x ) | d x$ . Now let $F : S _ { infty } rightarrow S _ { 1 }$ be the identity map taking any function $f$ to itself. Are $F$ and $F^{-1}$ continuous in this given metric space?



      I believe that $F^{-1}$ will not be continuous due to a counter example of $f_n(x) = x^n$. Unless I am mistaken, this is because $f_n$ will converge to $0$ in $ (S_1,d_1),$ but won't converge in $(S_{infty},d_{infty})$. This would be enough proof that it does not converge, right? I am not sure how to prove that $F$ does or does not converge though. I would assume that it does converge, but I am not certain.



      By the way, this is from a class on Real Analysis so my knowledge of topology is fairly limited.






      real-analysis convergence metric-spaces







      share|cite|improve this question









      $endgroup$




      The full question is below:



      Let $S _ { infty }$ be the set of continuous functions $f : [ 0,1 ] rightarrow [0,1]$, with the distance $d _ { infty } ( f , g ) = sup | f ( x ) - g ( x ) |$, and let $S_1$ be the same set of continuous functions on $[0,1]$ but with the distance $d _ { 1 } ( f , g ) = int _ { 0 } ^ { 1 } | f ( x ) - g ( x ) | d x$ . Now let $F : S _ { infty } rightarrow S _ { 1 }$ be the identity map taking any function $f$ to itself. Are $F$ and $F^{-1}$ continuous in this given metric space?



      I believe that $F^{-1}$ will not be continuous due to a counter example of $f_n(x) = x^n$. Unless I am mistaken, this is because $f_n$ will converge to $0$ in $ (S_1,d_1),$ but won't converge in $(S_{infty},d_{infty})$. This would be enough proof that it does not converge, right? I am not sure how to prove that $F$ does or does not converge though. I would assume that it does converge, but I am not certain.



      By the way, this is from a class on Real Analysis so my knowledge of topology is fairly limited.






      real-analysis convergence metric-spaces




      real-analysis convergence metric-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 18 '18 at 22:09









      Mohammed ShahidMohammed Shahid

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          $begingroup$

          Your example is fine for $F^{-1}$; do show that $f_n to 0$ though under $d_1$.



          For $F$ it's easy to see that it's Lipschitz:



          $$d_1(f,g) le d_infty(f,g)$$



          (we bound $|f-g|$ in the integral above by $d_infty(f,g)$ and the length of the interval is $1$) and thus $F$ is (uniformly) continuous.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I have not learned about Lipschitz functions. Would there be any other way to do this problem? I am studying for an exam and would want to stay within the scope of the class.
            $endgroup$
            – Mohammed Shahid
            Dec 18 '18 at 22:45






          • 1




            $begingroup$
            @MohammedShahid The inequality is enough info. Take $delta=varepsilon$ in the (uniform) continuity proof.
            $endgroup$
            – Henno Brandsma
            Dec 18 '18 at 22:46










          • $begingroup$
            oh, I see. Thank you so much.
            $endgroup$
            – Mohammed Shahid
            Dec 18 '18 at 22:51






          • 1




            $begingroup$
            @MohammedShahid you're welcome. Good luck with the exam.
            $endgroup$
            – Henno Brandsma
            Dec 18 '18 at 22:53












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          1 Answer
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          1 Answer
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          active

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          1












          $begingroup$

          Your example is fine for $F^{-1}$; do show that $f_n to 0$ though under $d_1$.



          For $F$ it's easy to see that it's Lipschitz:



          $$d_1(f,g) le d_infty(f,g)$$



          (we bound $|f-g|$ in the integral above by $d_infty(f,g)$ and the length of the interval is $1$) and thus $F$ is (uniformly) continuous.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I have not learned about Lipschitz functions. Would there be any other way to do this problem? I am studying for an exam and would want to stay within the scope of the class.
            $endgroup$
            – Mohammed Shahid
            Dec 18 '18 at 22:45






          • 1




            $begingroup$
            @MohammedShahid The inequality is enough info. Take $delta=varepsilon$ in the (uniform) continuity proof.
            $endgroup$
            – Henno Brandsma
            Dec 18 '18 at 22:46










          • $begingroup$
            oh, I see. Thank you so much.
            $endgroup$
            – Mohammed Shahid
            Dec 18 '18 at 22:51






          • 1




            $begingroup$
            @MohammedShahid you're welcome. Good luck with the exam.
            $endgroup$
            – Henno Brandsma
            Dec 18 '18 at 22:53
















          1












          $begingroup$

          Your example is fine for $F^{-1}$; do show that $f_n to 0$ though under $d_1$.



          For $F$ it's easy to see that it's Lipschitz:



          $$d_1(f,g) le d_infty(f,g)$$



          (we bound $|f-g|$ in the integral above by $d_infty(f,g)$ and the length of the interval is $1$) and thus $F$ is (uniformly) continuous.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I have not learned about Lipschitz functions. Would there be any other way to do this problem? I am studying for an exam and would want to stay within the scope of the class.
            $endgroup$
            – Mohammed Shahid
            Dec 18 '18 at 22:45






          • 1




            $begingroup$
            @MohammedShahid The inequality is enough info. Take $delta=varepsilon$ in the (uniform) continuity proof.
            $endgroup$
            – Henno Brandsma
            Dec 18 '18 at 22:46










          • $begingroup$
            oh, I see. Thank you so much.
            $endgroup$
            – Mohammed Shahid
            Dec 18 '18 at 22:51






          • 1




            $begingroup$
            @MohammedShahid you're welcome. Good luck with the exam.
            $endgroup$
            – Henno Brandsma
            Dec 18 '18 at 22:53














          1












          1








          1





          $begingroup$

          Your example is fine for $F^{-1}$; do show that $f_n to 0$ though under $d_1$.



          For $F$ it's easy to see that it's Lipschitz:



          $$d_1(f,g) le d_infty(f,g)$$



          (we bound $|f-g|$ in the integral above by $d_infty(f,g)$ and the length of the interval is $1$) and thus $F$ is (uniformly) continuous.






          share|cite|improve this answer









          $endgroup$



          Your example is fine for $F^{-1}$; do show that $f_n to 0$ though under $d_1$.



          For $F$ it's easy to see that it's Lipschitz:



          $$d_1(f,g) le d_infty(f,g)$$



          (we bound $|f-g|$ in the integral above by $d_infty(f,g)$ and the length of the interval is $1$) and thus $F$ is (uniformly) continuous.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 22:33









          Henno BrandsmaHenno Brandsma

          115k348124




          115k348124












          • $begingroup$
            I have not learned about Lipschitz functions. Would there be any other way to do this problem? I am studying for an exam and would want to stay within the scope of the class.
            $endgroup$
            – Mohammed Shahid
            Dec 18 '18 at 22:45






          • 1




            $begingroup$
            @MohammedShahid The inequality is enough info. Take $delta=varepsilon$ in the (uniform) continuity proof.
            $endgroup$
            – Henno Brandsma
            Dec 18 '18 at 22:46










          • $begingroup$
            oh, I see. Thank you so much.
            $endgroup$
            – Mohammed Shahid
            Dec 18 '18 at 22:51






          • 1




            $begingroup$
            @MohammedShahid you're welcome. Good luck with the exam.
            $endgroup$
            – Henno Brandsma
            Dec 18 '18 at 22:53


















          • $begingroup$
            I have not learned about Lipschitz functions. Would there be any other way to do this problem? I am studying for an exam and would want to stay within the scope of the class.
            $endgroup$
            – Mohammed Shahid
            Dec 18 '18 at 22:45






          • 1




            $begingroup$
            @MohammedShahid The inequality is enough info. Take $delta=varepsilon$ in the (uniform) continuity proof.
            $endgroup$
            – Henno Brandsma
            Dec 18 '18 at 22:46










          • $begingroup$
            oh, I see. Thank you so much.
            $endgroup$
            – Mohammed Shahid
            Dec 18 '18 at 22:51






          • 1




            $begingroup$
            @MohammedShahid you're welcome. Good luck with the exam.
            $endgroup$
            – Henno Brandsma
            Dec 18 '18 at 22:53
















          $begingroup$
          I have not learned about Lipschitz functions. Would there be any other way to do this problem? I am studying for an exam and would want to stay within the scope of the class.
          $endgroup$
          – Mohammed Shahid
          Dec 18 '18 at 22:45




          $begingroup$
          I have not learned about Lipschitz functions. Would there be any other way to do this problem? I am studying for an exam and would want to stay within the scope of the class.
          $endgroup$
          – Mohammed Shahid
          Dec 18 '18 at 22:45




          1




          1




          $begingroup$
          @MohammedShahid The inequality is enough info. Take $delta=varepsilon$ in the (uniform) continuity proof.
          $endgroup$
          – Henno Brandsma
          Dec 18 '18 at 22:46




          $begingroup$
          @MohammedShahid The inequality is enough info. Take $delta=varepsilon$ in the (uniform) continuity proof.
          $endgroup$
          – Henno Brandsma
          Dec 18 '18 at 22:46












          $begingroup$
          oh, I see. Thank you so much.
          $endgroup$
          – Mohammed Shahid
          Dec 18 '18 at 22:51




          $begingroup$
          oh, I see. Thank you so much.
          $endgroup$
          – Mohammed Shahid
          Dec 18 '18 at 22:51




          1




          1




          $begingroup$
          @MohammedShahid you're welcome. Good luck with the exam.
          $endgroup$
          – Henno Brandsma
          Dec 18 '18 at 22:53




          $begingroup$
          @MohammedShahid you're welcome. Good luck with the exam.
          $endgroup$
          – Henno Brandsma
          Dec 18 '18 at 22:53


















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