Compactness of $A:=${$f in C[0,1], |f|_infty le K, |f'|_infty le M$} The 2019 Stack Overflow...

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Compactness of $A:=${$f in C[0,1], |f|_infty le K, |f'|_infty le M$}



The 2019 Stack Overflow Developer Survey Results Are InUniform convergence of subsequences implying uniform convergenceCompactness in $C([0,1])$Arzela-Ascoli Theorem in $L^p[0,1]$Compactness of the $K subset C[0,1]$About the proof of a corollary of Arzela-Ascoli Theorem.Folland Problem 4.64 relating to Hölder continuity and compactness.Arzela Ascoli counterexamplescharacterizing compactness in function space.For which n differential operator $C^{(n)}[0, 1] rightarrow C[0,1]$ is compactArzela-Ascoli Theorem for a subset of $C([0,1])$












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Here we use infinity norm as metrics for $C[0,1]$. The professor claims that this set is compact. I can show this set is relative compact by Arzela Ascoli, i.e., for each subsequence there is a further sub-subsequence that uniformly converges to a continuous function on $[0,1]$. But it is not intuitive to me why this set is compact, i.e. the limiting function also satisfies $|f'|_infty le M$. Is it true that $A$ is close and therefore compact?






real-analysis functional-analysis analysis







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Here we use infinity norm as metrics for $C[0,1]$. The professor claims that this set is compact. I can show this set is relative compact by Arzela Ascoli, i.e., for each subsequence there is a further sub-subsequence that uniformly converges to a continuous function on $[0,1]$. But it is not intuitive to me why this set is compact, i.e. the limiting function also satisfies $|f'|_infty le M$. Is it true that $A$ is close and therefore compact?






    real-analysis functional-analysis analysis







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Here we use infinity norm as metrics for $C[0,1]$. The professor claims that this set is compact. I can show this set is relative compact by Arzela Ascoli, i.e., for each subsequence there is a further sub-subsequence that uniformly converges to a continuous function on $[0,1]$. But it is not intuitive to me why this set is compact, i.e. the limiting function also satisfies $|f'|_infty le M$. Is it true that $A$ is close and therefore compact?






      real-analysis functional-analysis analysis







      share|cite|improve this question









      $endgroup$




      Here we use infinity norm as metrics for $C[0,1]$. The professor claims that this set is compact. I can show this set is relative compact by Arzela Ascoli, i.e., for each subsequence there is a further sub-subsequence that uniformly converges to a continuous function on $[0,1]$. But it is not intuitive to me why this set is compact, i.e. the limiting function also satisfies $|f'|_infty le M$. Is it true that $A$ is close and therefore compact?






      real-analysis functional-analysis analysis




      real-analysis functional-analysis analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 21 at 15:39









      Daniel LiDaniel Li

      787414




      787414






















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          $begingroup$

          No it is not compact. For example, $A$ contains



          $$ f_n (x) = sqrt{(x-1/2)^2 + 1/n}.$$



          since



          $$ |f_n'(x)| = frac{2|x-1/2|}{sqrt{(x-1/2)^2 + 1/n}}le 2. $$



          But $f_n$ converges to $|x-1/2|$ which is not differentiable.



          Remark: If instead one consider
          $$A' = { fin C[0,1] : |f|_inftyle K, operatorname{Lip}f le M},$$



          here
          $$operatorname{Lip}f := sup_{x, yin [0,1], xneq y} frac{|f(x) - f(y)|}{|x-y|},$$



          then $Asubset A'$ and $A'$ is compact.






          share|cite|improve this answer









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            $begingroup$

            No it is not compact. For example, $A$ contains



            $$ f_n (x) = sqrt{(x-1/2)^2 + 1/n}.$$



            since



            $$ |f_n'(x)| = frac{2|x-1/2|}{sqrt{(x-1/2)^2 + 1/n}}le 2. $$



            But $f_n$ converges to $|x-1/2|$ which is not differentiable.



            Remark: If instead one consider
            $$A' = { fin C[0,1] : |f|_inftyle K, operatorname{Lip}f le M},$$



            here
            $$operatorname{Lip}f := sup_{x, yin [0,1], xneq y} frac{|f(x) - f(y)|}{|x-y|},$$



            then $Asubset A'$ and $A'$ is compact.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              No it is not compact. For example, $A$ contains



              $$ f_n (x) = sqrt{(x-1/2)^2 + 1/n}.$$



              since



              $$ |f_n'(x)| = frac{2|x-1/2|}{sqrt{(x-1/2)^2 + 1/n}}le 2. $$



              But $f_n$ converges to $|x-1/2|$ which is not differentiable.



              Remark: If instead one consider
              $$A' = { fin C[0,1] : |f|_inftyle K, operatorname{Lip}f le M},$$



              here
              $$operatorname{Lip}f := sup_{x, yin [0,1], xneq y} frac{|f(x) - f(y)|}{|x-y|},$$



              then $Asubset A'$ and $A'$ is compact.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                No it is not compact. For example, $A$ contains



                $$ f_n (x) = sqrt{(x-1/2)^2 + 1/n}.$$



                since



                $$ |f_n'(x)| = frac{2|x-1/2|}{sqrt{(x-1/2)^2 + 1/n}}le 2. $$



                But $f_n$ converges to $|x-1/2|$ which is not differentiable.



                Remark: If instead one consider
                $$A' = { fin C[0,1] : |f|_inftyle K, operatorname{Lip}f le M},$$



                here
                $$operatorname{Lip}f := sup_{x, yin [0,1], xneq y} frac{|f(x) - f(y)|}{|x-y|},$$



                then $Asubset A'$ and $A'$ is compact.






                share|cite|improve this answer









                $endgroup$



                No it is not compact. For example, $A$ contains



                $$ f_n (x) = sqrt{(x-1/2)^2 + 1/n}.$$



                since



                $$ |f_n'(x)| = frac{2|x-1/2|}{sqrt{(x-1/2)^2 + 1/n}}le 2. $$



                But $f_n$ converges to $|x-1/2|$ which is not differentiable.



                Remark: If instead one consider
                $$A' = { fin C[0,1] : |f|_inftyle K, operatorname{Lip}f le M},$$



                here
                $$operatorname{Lip}f := sup_{x, yin [0,1], xneq y} frac{|f(x) - f(y)|}{|x-y|},$$



                then $Asubset A'$ and $A'$ is compact.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 21 at 17:19









                Arctic CharArctic Char

                471115




                471115






























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