Is $f_{n}$ is analytic on $(a, b)$ and $f_{n} rightarrow f$ uniformly on $(a, b)$ then is $f$ analytic on...












1












$begingroup$



Is $f_{n}$ is analytic on $(a, b)$ and $f_{n} rightarrow f$ uniformly
on $(a, b)$ then is $f$ analytic on $(a, b)$?




Intuitively, I think that the answer is no. I know that the statement holds for integrability and continuity; however, I don't think it's necessary for analyticity. Am I correct?






real-analysis convergence uniform-convergence analytic-functions sequence-of-function







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  • 2




    $begingroup$
    Try $f_n(x)=frac1nsqrt{1+n^2x^2}$ on $(-1,1)$.
    $endgroup$
    – Did
    Dec 14 '18 at 15:30
















1












$begingroup$



Is $f_{n}$ is analytic on $(a, b)$ and $f_{n} rightarrow f$ uniformly
on $(a, b)$ then is $f$ analytic on $(a, b)$?




Intuitively, I think that the answer is no. I know that the statement holds for integrability and continuity; however, I don't think it's necessary for analyticity. Am I correct?






real-analysis convergence uniform-convergence analytic-functions sequence-of-function







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Try $f_n(x)=frac1nsqrt{1+n^2x^2}$ on $(-1,1)$.
    $endgroup$
    – Did
    Dec 14 '18 at 15:30














1












1








1





$begingroup$



Is $f_{n}$ is analytic on $(a, b)$ and $f_{n} rightarrow f$ uniformly
on $(a, b)$ then is $f$ analytic on $(a, b)$?




Intuitively, I think that the answer is no. I know that the statement holds for integrability and continuity; however, I don't think it's necessary for analyticity. Am I correct?






real-analysis convergence uniform-convergence analytic-functions sequence-of-function







share|cite|improve this question











$endgroup$





Is $f_{n}$ is analytic on $(a, b)$ and $f_{n} rightarrow f$ uniformly
on $(a, b)$ then is $f$ analytic on $(a, b)$?




Intuitively, I think that the answer is no. I know that the statement holds for integrability and continuity; however, I don't think it's necessary for analyticity. Am I correct?






real-analysis convergence uniform-convergence analytic-functions sequence-of-function




real-analysis convergence uniform-convergence analytic-functions sequence-of-function






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 15:32









José Carlos Santos

165k22132235




165k22132235










asked Dec 14 '18 at 15:24









josephjoseph

496111




496111








  • 2




    $begingroup$
    Try $f_n(x)=frac1nsqrt{1+n^2x^2}$ on $(-1,1)$.
    $endgroup$
    – Did
    Dec 14 '18 at 15:30














  • 2




    $begingroup$
    Try $f_n(x)=frac1nsqrt{1+n^2x^2}$ on $(-1,1)$.
    $endgroup$
    – Did
    Dec 14 '18 at 15:30








2




2




$begingroup$
Try $f_n(x)=frac1nsqrt{1+n^2x^2}$ on $(-1,1)$.
$endgroup$
– Did
Dec 14 '18 at 15:30




$begingroup$
Try $f_n(x)=frac1nsqrt{1+n^2x^2}$ on $(-1,1)$.
$endgroup$
– Did
Dec 14 '18 at 15:30










2 Answers
2






active

oldest

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3












$begingroup$

Yes, you are correct. Just consider$$begin{array}{rccc}f_ncolon&(-1,1)&longrightarrow&mathbb R\&x&mapsto&sqrt{x^2+frac1{n^2}}.end{array}$$The sequence $(f_n)_{ninmathbb N}$ is a sequence of analytic functions that converges uniformly to the absolute value functions, which isn't differentiable.






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  • $begingroup$
    Nice counterexample
    $endgroup$
    – joseph
    Dec 14 '18 at 15:33










  • $begingroup$
    @joseph If my answer was useful, perhaps that you could mark it as the accepted one.
    $endgroup$
    – José Carlos Santos
    Dec 14 '18 at 16:59



















0












$begingroup$

Stone-Weierstrass theorem indicates that any continuous function $f:[a,b]toBbb R$ is a uniform limit of a sequence of polynomials $p_n$. Polynomials are obviously analytic but $f$ need not be differentiable.






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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

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    active

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    3












    $begingroup$

    Yes, you are correct. Just consider$$begin{array}{rccc}f_ncolon&(-1,1)&longrightarrow&mathbb R\&x&mapsto&sqrt{x^2+frac1{n^2}}.end{array}$$The sequence $(f_n)_{ninmathbb N}$ is a sequence of analytic functions that converges uniformly to the absolute value functions, which isn't differentiable.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Nice counterexample
      $endgroup$
      – joseph
      Dec 14 '18 at 15:33










    • $begingroup$
      @joseph If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      Dec 14 '18 at 16:59
















    3












    $begingroup$

    Yes, you are correct. Just consider$$begin{array}{rccc}f_ncolon&(-1,1)&longrightarrow&mathbb R\&x&mapsto&sqrt{x^2+frac1{n^2}}.end{array}$$The sequence $(f_n)_{ninmathbb N}$ is a sequence of analytic functions that converges uniformly to the absolute value functions, which isn't differentiable.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Nice counterexample
      $endgroup$
      – joseph
      Dec 14 '18 at 15:33










    • $begingroup$
      @joseph If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      Dec 14 '18 at 16:59














    3












    3








    3





    $begingroup$

    Yes, you are correct. Just consider$$begin{array}{rccc}f_ncolon&(-1,1)&longrightarrow&mathbb R\&x&mapsto&sqrt{x^2+frac1{n^2}}.end{array}$$The sequence $(f_n)_{ninmathbb N}$ is a sequence of analytic functions that converges uniformly to the absolute value functions, which isn't differentiable.






    share|cite|improve this answer









    $endgroup$



    Yes, you are correct. Just consider$$begin{array}{rccc}f_ncolon&(-1,1)&longrightarrow&mathbb R\&x&mapsto&sqrt{x^2+frac1{n^2}}.end{array}$$The sequence $(f_n)_{ninmathbb N}$ is a sequence of analytic functions that converges uniformly to the absolute value functions, which isn't differentiable.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 14 '18 at 15:32









    José Carlos SantosJosé Carlos Santos

    165k22132235




    165k22132235












    • $begingroup$
      Nice counterexample
      $endgroup$
      – joseph
      Dec 14 '18 at 15:33










    • $begingroup$
      @joseph If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      Dec 14 '18 at 16:59


















    • $begingroup$
      Nice counterexample
      $endgroup$
      – joseph
      Dec 14 '18 at 15:33










    • $begingroup$
      @joseph If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      Dec 14 '18 at 16:59
















    $begingroup$
    Nice counterexample
    $endgroup$
    – joseph
    Dec 14 '18 at 15:33




    $begingroup$
    Nice counterexample
    $endgroup$
    – joseph
    Dec 14 '18 at 15:33












    $begingroup$
    @joseph If my answer was useful, perhaps that you could mark it as the accepted one.
    $endgroup$
    – José Carlos Santos
    Dec 14 '18 at 16:59




    $begingroup$
    @joseph If my answer was useful, perhaps that you could mark it as the accepted one.
    $endgroup$
    – José Carlos Santos
    Dec 14 '18 at 16:59











    0












    $begingroup$

    Stone-Weierstrass theorem indicates that any continuous function $f:[a,b]toBbb R$ is a uniform limit of a sequence of polynomials $p_n$. Polynomials are obviously analytic but $f$ need not be differentiable.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Stone-Weierstrass theorem indicates that any continuous function $f:[a,b]toBbb R$ is a uniform limit of a sequence of polynomials $p_n$. Polynomials are obviously analytic but $f$ need not be differentiable.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Stone-Weierstrass theorem indicates that any continuous function $f:[a,b]toBbb R$ is a uniform limit of a sequence of polynomials $p_n$. Polynomials are obviously analytic but $f$ need not be differentiable.






        share|cite|improve this answer









        $endgroup$



        Stone-Weierstrass theorem indicates that any continuous function $f:[a,b]toBbb R$ is a uniform limit of a sequence of polynomials $p_n$. Polynomials are obviously analytic but $f$ need not be differentiable.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 16:46









        BigbearZzzBigbearZzz

        8,88821652




        8,88821652






























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