Nested uniform convergence

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












Suppose $f^m,n$ converges to $f^n$ uniformly as $mtoinfty$ and $f^n$ converges to $f$ uniformly as $ntoinfty$. Is it true that $f^m_n,n$ converges to $f$ uniformly as $ntoinfty$ where $m_n$ is strictly increasing in $n$?




analysis uniform-convergence




share|cite|improve this question















  • 1




    Is $m_n$ a given sequence or is something we can choose?
    – Kavi Rama Murthy
    Jul 30 at 10:23










  • Maybe try to estimate $|f^m_n,n - f| leqslant |f^m_n, n - f^n| + |f^n - f|$.
    – xbh
    Jul 30 at 10:24














up vote
0
down vote

favorite












Suppose $f^m,n$ converges to $f^n$ uniformly as $mtoinfty$ and $f^n$ converges to $f$ uniformly as $ntoinfty$. Is it true that $f^m_n,n$ converges to $f$ uniformly as $ntoinfty$ where $m_n$ is strictly increasing in $n$?




analysis uniform-convergence




share|cite|improve this question















  • 1




    Is $m_n$ a given sequence or is something we can choose?
    – Kavi Rama Murthy
    Jul 30 at 10:23










  • Maybe try to estimate $|f^m_n,n - f| leqslant |f^m_n, n - f^n| + |f^n - f|$.
    – xbh
    Jul 30 at 10:24












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Suppose $f^m,n$ converges to $f^n$ uniformly as $mtoinfty$ and $f^n$ converges to $f$ uniformly as $ntoinfty$. Is it true that $f^m_n,n$ converges to $f$ uniformly as $ntoinfty$ where $m_n$ is strictly increasing in $n$?




analysis uniform-convergence




share|cite|improve this question











Suppose $f^m,n$ converges to $f^n$ uniformly as $mtoinfty$ and $f^n$ converges to $f$ uniformly as $ntoinfty$. Is it true that $f^m_n,n$ converges to $f$ uniformly as $ntoinfty$ where $m_n$ is strictly increasing in $n$?





analysis uniform-convergence





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 30 at 10:19









jerom

647




647







  • 1




    Is $m_n$ a given sequence or is something we can choose?
    – Kavi Rama Murthy
    Jul 30 at 10:23










  • Maybe try to estimate $|f^m_n,n - f| leqslant |f^m_n, n - f^n| + |f^n - f|$.
    – xbh
    Jul 30 at 10:24












  • 1




    Is $m_n$ a given sequence or is something we can choose?
    – Kavi Rama Murthy
    Jul 30 at 10:23










  • Maybe try to estimate $|f^m_n,n - f| leqslant |f^m_n, n - f^n| + |f^n - f|$.
    – xbh
    Jul 30 at 10:24







1




1




Is $m_n$ a given sequence or is something we can choose?
– Kavi Rama Murthy
Jul 30 at 10:23




Is $m_n$ a given sequence or is something we can choose?
– Kavi Rama Murthy
Jul 30 at 10:23












Maybe try to estimate $|f^m_n,n - f| leqslant |f^m_n, n - f^n| + |f^n - f|$.
– xbh
Jul 30 at 10:24




Maybe try to estimate $|f^m_n,n - f| leqslant |f^m_n, n - f^n| + |f^n - f|$.
– xbh
Jul 30 at 10:24










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










Yes, for some sequence $m_n%$ strictly increasing to $infty$; false if $m_n$ is given in advance. In either case the arguments boil down to the numbers $||f^m_n,n-f||$ etc so the question does not have much to do with functions and uniform limits. Some details: for each $n$ choose $m_n$ such that $||f^m,n-f^n||<frac 1 n$ for $m geq m_n$. Then $||f^m_n,n-f||to 0$. [ Norm is the sup norm]. For a counterexampe to show that this does not hold for any $m_n$ take $f^m,nequiv frac m m+n$,$m_n=n$.






share|cite|improve this answer






























    up vote
    1
    down vote













    No. Take$$beginarrayrcccf^m,ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begincasest^n&text if tleqslant 1-frac1m\left(1-frac1mright)^n&text otherwise.endcasesendarray$$Then $(f^m,n)_minmathbb N$ converges uniformly $f_n(t)=t^n$. However. $(f^n,n)_ninmathbb N$ doesn't converge uniformly.






    share|cite|improve this answer





















      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',
      convertImagesToLinks: true,
      noModals: false,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      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%2f2866865%2fnested-uniform-convergence%23new-answer', 'question_page');

      );

      Post as a guest

















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      2
      down vote



      accepted










      Yes, for some sequence $m_n%$ strictly increasing to $infty$; false if $m_n$ is given in advance. In either case the arguments boil down to the numbers $||f^m_n,n-f||$ etc so the question does not have much to do with functions and uniform limits. Some details: for each $n$ choose $m_n$ such that $||f^m,n-f^n||<frac 1 n$ for $m geq m_n$. Then $||f^m_n,n-f||to 0$. [ Norm is the sup norm]. For a counterexampe to show that this does not hold for any $m_n$ take $f^m,nequiv frac m m+n$,$m_n=n$.






      share|cite|improve this answer



























        up vote
        2
        down vote



        accepted










        Yes, for some sequence $m_n%$ strictly increasing to $infty$; false if $m_n$ is given in advance. In either case the arguments boil down to the numbers $||f^m_n,n-f||$ etc so the question does not have much to do with functions and uniform limits. Some details: for each $n$ choose $m_n$ such that $||f^m,n-f^n||<frac 1 n$ for $m geq m_n$. Then $||f^m_n,n-f||to 0$. [ Norm is the sup norm]. For a counterexampe to show that this does not hold for any $m_n$ take $f^m,nequiv frac m m+n$,$m_n=n$.






        share|cite|improve this answer

























          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          Yes, for some sequence $m_n%$ strictly increasing to $infty$; false if $m_n$ is given in advance. In either case the arguments boil down to the numbers $||f^m_n,n-f||$ etc so the question does not have much to do with functions and uniform limits. Some details: for each $n$ choose $m_n$ such that $||f^m,n-f^n||<frac 1 n$ for $m geq m_n$. Then $||f^m_n,n-f||to 0$. [ Norm is the sup norm]. For a counterexampe to show that this does not hold for any $m_n$ take $f^m,nequiv frac m m+n$,$m_n=n$.






          share|cite|improve this answer















          Yes, for some sequence $m_n%$ strictly increasing to $infty$; false if $m_n$ is given in advance. In either case the arguments boil down to the numbers $||f^m_n,n-f||$ etc so the question does not have much to do with functions and uniform limits. Some details: for each $n$ choose $m_n$ such that $||f^m,n-f^n||<frac 1 n$ for $m geq m_n$. Then $||f^m_n,n-f||to 0$. [ Norm is the sup norm]. For a counterexampe to show that this does not hold for any $m_n$ take $f^m,nequiv frac m m+n$,$m_n=n$.







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 30 at 10:34


























          answered Jul 30 at 10:26









          Kavi Rama Murthy

          19.7k2829




          19.7k2829




















              up vote
              1
              down vote













              No. Take$$beginarrayrcccf^m,ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begincasest^n&text if tleqslant 1-frac1m\left(1-frac1mright)^n&text otherwise.endcasesendarray$$Then $(f^m,n)_minmathbb N$ converges uniformly $f_n(t)=t^n$. However. $(f^n,n)_ninmathbb N$ doesn't converge uniformly.






              share|cite|improve this answer

























                up vote
                1
                down vote













                No. Take$$beginarrayrcccf^m,ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begincasest^n&text if tleqslant 1-frac1m\left(1-frac1mright)^n&text otherwise.endcasesendarray$$Then $(f^m,n)_minmathbb N$ converges uniformly $f_n(t)=t^n$. However. $(f^n,n)_ninmathbb N$ doesn't converge uniformly.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  No. Take$$beginarrayrcccf^m,ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begincasest^n&text if tleqslant 1-frac1m\left(1-frac1mright)^n&text otherwise.endcasesendarray$$Then $(f^m,n)_minmathbb N$ converges uniformly $f_n(t)=t^n$. However. $(f^n,n)_ninmathbb N$ doesn't converge uniformly.






                  share|cite|improve this answer













                  No. Take$$beginarrayrcccf^m,ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begincasest^n&text if tleqslant 1-frac1m\left(1-frac1mright)^n&text otherwise.endcasesendarray$$Then $(f^m,n)_minmathbb N$ converges uniformly $f_n(t)=t^n$. However. $(f^n,n)_ninmathbb N$ doesn't converge uniformly.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Jul 30 at 10:29









                  José Carlos Santos

                  112k1696172




                  112k1696172






















                       

                      draft saved


                      draft discarded


























                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866865%2fnested-uniform-convergence%23new-answer', 'question_page');

                      );

                      Post as a guest
































































                      Comments

                      Post a Comment

                      Popular posts from this blog

                      What is the equation of a 3D cone with generalised tilt?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
                      Read more

                      Color the edges and diagonals of a regular polygon

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
                      Read more

                      Relationship between determinant of matrix and determinant of adjoint?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
                      Read more