Proof verification - If $f$ is continuous, then $f_{n}$ is continuous in at least one $n in mathbb{N}$












1














Given $f_{n} : [a, b] mapsto mathbb{R}$ a sequence of functions that converge uniformaly to $f : [a, b] mapsto mathbb{R}$



Prove or disprove: If $f$ is continuous, then $f_{n}$ is continuous in at least one $n in mathbb{N}$



My attempt at a proof:



Let $epsilon > 0$ and $x_{0} in [a,b]$ arbitrary



Since $f$ is continuous on $[a,b]$ there exists a $delta_{epsilon} > 0$ such that for all $x in [a,b]$
$$ |x - x_{0}| < delta_{epsilon} Longrightarrow |f(x) - f(x_{0})| < frac{epsilon}{3}$$



Because $f_{n}$ converges uniformaly to $f$ we can find an $N_{epsilon} in mathbb{N}$ such that for all $n geq N_{epsilon}$ and all $x in [a,b]$
$$|f_{n}(x) - f(x)| < frac{epsilon}{3} $$



which means that for $n geq N_{epsilon}$ : if $|x - x_{0}| < delta_{epsilon}$ then
$$ |f_{n}(x) - f_{n}(x_{0})| leq |f_{n}(x) - f(x)| + |f(x) - f(x_{0})| + |f(x_{0}) -f_{n}(x_{0})| < epsilon $$



Therefore $f_{n}$ is continuous for at least one $n$



Is this proof correct? Thanks.






real-analysis proof-verification continuity uniform-convergence







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    1














    Given $f_{n} : [a, b] mapsto mathbb{R}$ a sequence of functions that converge uniformaly to $f : [a, b] mapsto mathbb{R}$



    Prove or disprove: If $f$ is continuous, then $f_{n}$ is continuous in at least one $n in mathbb{N}$



    My attempt at a proof:



    Let $epsilon > 0$ and $x_{0} in [a,b]$ arbitrary



    Since $f$ is continuous on $[a,b]$ there exists a $delta_{epsilon} > 0$ such that for all $x in [a,b]$
    $$ |x - x_{0}| < delta_{epsilon} Longrightarrow |f(x) - f(x_{0})| < frac{epsilon}{3}$$



    Because $f_{n}$ converges uniformaly to $f$ we can find an $N_{epsilon} in mathbb{N}$ such that for all $n geq N_{epsilon}$ and all $x in [a,b]$
    $$|f_{n}(x) - f(x)| < frac{epsilon}{3} $$



    which means that for $n geq N_{epsilon}$ : if $|x - x_{0}| < delta_{epsilon}$ then
    $$ |f_{n}(x) - f_{n}(x_{0})| leq |f_{n}(x) - f(x)| + |f(x) - f(x_{0})| + |f(x_{0}) -f_{n}(x_{0})| < epsilon $$



    Therefore $f_{n}$ is continuous for at least one $n$



    Is this proof correct? Thanks.






    real-analysis proof-verification continuity uniform-convergence







    share|cite|improve this question



























      1












      1








      1







      Given $f_{n} : [a, b] mapsto mathbb{R}$ a sequence of functions that converge uniformaly to $f : [a, b] mapsto mathbb{R}$



      Prove or disprove: If $f$ is continuous, then $f_{n}$ is continuous in at least one $n in mathbb{N}$



      My attempt at a proof:



      Let $epsilon > 0$ and $x_{0} in [a,b]$ arbitrary



      Since $f$ is continuous on $[a,b]$ there exists a $delta_{epsilon} > 0$ such that for all $x in [a,b]$
      $$ |x - x_{0}| < delta_{epsilon} Longrightarrow |f(x) - f(x_{0})| < frac{epsilon}{3}$$



      Because $f_{n}$ converges uniformaly to $f$ we can find an $N_{epsilon} in mathbb{N}$ such that for all $n geq N_{epsilon}$ and all $x in [a,b]$
      $$|f_{n}(x) - f(x)| < frac{epsilon}{3} $$



      which means that for $n geq N_{epsilon}$ : if $|x - x_{0}| < delta_{epsilon}$ then
      $$ |f_{n}(x) - f_{n}(x_{0})| leq |f_{n}(x) - f(x)| + |f(x) - f(x_{0})| + |f(x_{0}) -f_{n}(x_{0})| < epsilon $$



      Therefore $f_{n}$ is continuous for at least one $n$



      Is this proof correct? Thanks.






      real-analysis proof-verification continuity uniform-convergence







      share|cite|improve this question















      Given $f_{n} : [a, b] mapsto mathbb{R}$ a sequence of functions that converge uniformaly to $f : [a, b] mapsto mathbb{R}$



      Prove or disprove: If $f$ is continuous, then $f_{n}$ is continuous in at least one $n in mathbb{N}$



      My attempt at a proof:



      Let $epsilon > 0$ and $x_{0} in [a,b]$ arbitrary



      Since $f$ is continuous on $[a,b]$ there exists a $delta_{epsilon} > 0$ such that for all $x in [a,b]$
      $$ |x - x_{0}| < delta_{epsilon} Longrightarrow |f(x) - f(x_{0})| < frac{epsilon}{3}$$



      Because $f_{n}$ converges uniformaly to $f$ we can find an $N_{epsilon} in mathbb{N}$ such that for all $n geq N_{epsilon}$ and all $x in [a,b]$
      $$|f_{n}(x) - f(x)| < frac{epsilon}{3} $$



      which means that for $n geq N_{epsilon}$ : if $|x - x_{0}| < delta_{epsilon}$ then
      $$ |f_{n}(x) - f_{n}(x_{0})| leq |f_{n}(x) - f(x)| + |f(x) - f(x_{0})| + |f(x_{0}) -f_{n}(x_{0})| < epsilon $$



      Therefore $f_{n}$ is continuous for at least one $n$



      Is this proof correct? Thanks.






      real-analysis proof-verification continuity uniform-convergence




      real-analysis proof-verification continuity uniform-convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 '18 at 21:39









      José Carlos Santos

      150k22121221




      150k22121221










      asked Nov 27 '18 at 21:15









      9Sp

      244




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          2 Answers
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          7














          No, it is not correct and it could not possibly be correct, since the statement is false. For each $ninmathbb N$, define $f_ncolon[-1,1]longrightarrowmathbb R$ by$$f_n(x)=begin{cases}frac1n&text{ if }x>0\0&text{ otherwise.}end{cases}$$Then $(f_n)_{ninmathbb N}$ converges uniformly to the null function, but no $f_n$ is continuous.



          Concerning your proof, your goal was to prove thatbegin{multline}(exists ninmathbb{N})(forall x_0in[a,b])(forallvarepsilon>0)(existsdelta>0):xin[x_0-delta,x_0+delta]cap[a,b]implies\impliesbigllvert f_n(x)-f_n(x_0bigrrvert<varepsilon,end{multline}but what you actually proved was thatbegin{multline}(forall x_0in[a,b])(forallvarepsilon>0)(exists ninmathbb{N})(existsdelta>0):xin[x_0-delta,x_0+delta]cap[a,b]implies\impliesbigllvert f_n(x)-f_n(x_0bigrrvert<varepsilon.end{multline}






          share|cite|improve this answer























          • Thanks. I have seen other counterexamples too now. Could you please point out where I have gone wrong in my "proof"?
            – 9Sp
            Nov 27 '18 at 21:26






          • 3




            It is because the positive integer $N_{epsilon}$ that you found depends on $epsilon>0$. So while the equality you end up with is true for the given value of $epsilon$, it won't necessarily hold for other $epsilon>0$.
            – Matt A Pelto
            Nov 27 '18 at 21:35










          • rather inequality*, not equality
            – Matt A Pelto
            Nov 27 '18 at 21:46



















          0














          Another counterexample:



          For each $n in mathbb N$ define $f_n: [a,b] to mathbb R$ by
          $$f_n(x)=begin{cases}frac1n&text{ if }x in mathbb Q \0&text{ otherwise.}end{cases}$$



          Given $n in mathbb N$ and $x in [a,b]$ we may show that $f_n$ does not satisfy the definition of a continuous function by selecting $ varepsilon=frac{1}{n+1}$. So if $delta>0$, then there is $hat{x}$ such that $|hat x - x|<delta$ but $|,f_n(hat x) - f_n(x)|geq varepsilon$.



          Call the limit function $f(x)=lim_{n to infty} f_n(x)$ a degenerate case of a Dirichlet function (kind of a bad joke).






          share|cite|improve this answer























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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            7














            No, it is not correct and it could not possibly be correct, since the statement is false. For each $ninmathbb N$, define $f_ncolon[-1,1]longrightarrowmathbb R$ by$$f_n(x)=begin{cases}frac1n&text{ if }x>0\0&text{ otherwise.}end{cases}$$Then $(f_n)_{ninmathbb N}$ converges uniformly to the null function, but no $f_n$ is continuous.



            Concerning your proof, your goal was to prove thatbegin{multline}(exists ninmathbb{N})(forall x_0in[a,b])(forallvarepsilon>0)(existsdelta>0):xin[x_0-delta,x_0+delta]cap[a,b]implies\impliesbigllvert f_n(x)-f_n(x_0bigrrvert<varepsilon,end{multline}but what you actually proved was thatbegin{multline}(forall x_0in[a,b])(forallvarepsilon>0)(exists ninmathbb{N})(existsdelta>0):xin[x_0-delta,x_0+delta]cap[a,b]implies\impliesbigllvert f_n(x)-f_n(x_0bigrrvert<varepsilon.end{multline}






            share|cite|improve this answer























            • Thanks. I have seen other counterexamples too now. Could you please point out where I have gone wrong in my "proof"?
              – 9Sp
              Nov 27 '18 at 21:26






            • 3




              It is because the positive integer $N_{epsilon}$ that you found depends on $epsilon>0$. So while the equality you end up with is true for the given value of $epsilon$, it won't necessarily hold for other $epsilon>0$.
              – Matt A Pelto
              Nov 27 '18 at 21:35










            • rather inequality*, not equality
              – Matt A Pelto
              Nov 27 '18 at 21:46
















            7














            No, it is not correct and it could not possibly be correct, since the statement is false. For each $ninmathbb N$, define $f_ncolon[-1,1]longrightarrowmathbb R$ by$$f_n(x)=begin{cases}frac1n&text{ if }x>0\0&text{ otherwise.}end{cases}$$Then $(f_n)_{ninmathbb N}$ converges uniformly to the null function, but no $f_n$ is continuous.



            Concerning your proof, your goal was to prove thatbegin{multline}(exists ninmathbb{N})(forall x_0in[a,b])(forallvarepsilon>0)(existsdelta>0):xin[x_0-delta,x_0+delta]cap[a,b]implies\impliesbigllvert f_n(x)-f_n(x_0bigrrvert<varepsilon,end{multline}but what you actually proved was thatbegin{multline}(forall x_0in[a,b])(forallvarepsilon>0)(exists ninmathbb{N})(existsdelta>0):xin[x_0-delta,x_0+delta]cap[a,b]implies\impliesbigllvert f_n(x)-f_n(x_0bigrrvert<varepsilon.end{multline}






            share|cite|improve this answer























            • Thanks. I have seen other counterexamples too now. Could you please point out where I have gone wrong in my "proof"?
              – 9Sp
              Nov 27 '18 at 21:26






            • 3




              It is because the positive integer $N_{epsilon}$ that you found depends on $epsilon>0$. So while the equality you end up with is true for the given value of $epsilon$, it won't necessarily hold for other $epsilon>0$.
              – Matt A Pelto
              Nov 27 '18 at 21:35










            • rather inequality*, not equality
              – Matt A Pelto
              Nov 27 '18 at 21:46














            7












            7








            7






            No, it is not correct and it could not possibly be correct, since the statement is false. For each $ninmathbb N$, define $f_ncolon[-1,1]longrightarrowmathbb R$ by$$f_n(x)=begin{cases}frac1n&text{ if }x>0\0&text{ otherwise.}end{cases}$$Then $(f_n)_{ninmathbb N}$ converges uniformly to the null function, but no $f_n$ is continuous.



            Concerning your proof, your goal was to prove thatbegin{multline}(exists ninmathbb{N})(forall x_0in[a,b])(forallvarepsilon>0)(existsdelta>0):xin[x_0-delta,x_0+delta]cap[a,b]implies\impliesbigllvert f_n(x)-f_n(x_0bigrrvert<varepsilon,end{multline}but what you actually proved was thatbegin{multline}(forall x_0in[a,b])(forallvarepsilon>0)(exists ninmathbb{N})(existsdelta>0):xin[x_0-delta,x_0+delta]cap[a,b]implies\impliesbigllvert f_n(x)-f_n(x_0bigrrvert<varepsilon.end{multline}






            share|cite|improve this answer














            No, it is not correct and it could not possibly be correct, since the statement is false. For each $ninmathbb N$, define $f_ncolon[-1,1]longrightarrowmathbb R$ by$$f_n(x)=begin{cases}frac1n&text{ if }x>0\0&text{ otherwise.}end{cases}$$Then $(f_n)_{ninmathbb N}$ converges uniformly to the null function, but no $f_n$ is continuous.



            Concerning your proof, your goal was to prove thatbegin{multline}(exists ninmathbb{N})(forall x_0in[a,b])(forallvarepsilon>0)(existsdelta>0):xin[x_0-delta,x_0+delta]cap[a,b]implies\impliesbigllvert f_n(x)-f_n(x_0bigrrvert<varepsilon,end{multline}but what you actually proved was thatbegin{multline}(forall x_0in[a,b])(forallvarepsilon>0)(exists ninmathbb{N})(existsdelta>0):xin[x_0-delta,x_0+delta]cap[a,b]implies\impliesbigllvert f_n(x)-f_n(x_0bigrrvert<varepsilon.end{multline}







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 27 '18 at 22:18

























            answered Nov 27 '18 at 21:19









            José Carlos Santos

            150k22121221




            150k22121221












            • Thanks. I have seen other counterexamples too now. Could you please point out where I have gone wrong in my "proof"?
              – 9Sp
              Nov 27 '18 at 21:26






            • 3




              It is because the positive integer $N_{epsilon}$ that you found depends on $epsilon>0$. So while the equality you end up with is true for the given value of $epsilon$, it won't necessarily hold for other $epsilon>0$.
              – Matt A Pelto
              Nov 27 '18 at 21:35










            • rather inequality*, not equality
              – Matt A Pelto
              Nov 27 '18 at 21:46


















            • Thanks. I have seen other counterexamples too now. Could you please point out where I have gone wrong in my "proof"?
              – 9Sp
              Nov 27 '18 at 21:26






            • 3




              It is because the positive integer $N_{epsilon}$ that you found depends on $epsilon>0$. So while the equality you end up with is true for the given value of $epsilon$, it won't necessarily hold for other $epsilon>0$.
              – Matt A Pelto
              Nov 27 '18 at 21:35










            • rather inequality*, not equality
              – Matt A Pelto
              Nov 27 '18 at 21:46
















            Thanks. I have seen other counterexamples too now. Could you please point out where I have gone wrong in my "proof"?
            – 9Sp
            Nov 27 '18 at 21:26




            Thanks. I have seen other counterexamples too now. Could you please point out where I have gone wrong in my "proof"?
            – 9Sp
            Nov 27 '18 at 21:26




            3




            3




            It is because the positive integer $N_{epsilon}$ that you found depends on $epsilon>0$. So while the equality you end up with is true for the given value of $epsilon$, it won't necessarily hold for other $epsilon>0$.
            – Matt A Pelto
            Nov 27 '18 at 21:35




            It is because the positive integer $N_{epsilon}$ that you found depends on $epsilon>0$. So while the equality you end up with is true for the given value of $epsilon$, it won't necessarily hold for other $epsilon>0$.
            – Matt A Pelto
            Nov 27 '18 at 21:35












            rather inequality*, not equality
            – Matt A Pelto
            Nov 27 '18 at 21:46




            rather inequality*, not equality
            – Matt A Pelto
            Nov 27 '18 at 21:46











            0














            Another counterexample:



            For each $n in mathbb N$ define $f_n: [a,b] to mathbb R$ by
            $$f_n(x)=begin{cases}frac1n&text{ if }x in mathbb Q \0&text{ otherwise.}end{cases}$$



            Given $n in mathbb N$ and $x in [a,b]$ we may show that $f_n$ does not satisfy the definition of a continuous function by selecting $ varepsilon=frac{1}{n+1}$. So if $delta>0$, then there is $hat{x}$ such that $|hat x - x|<delta$ but $|,f_n(hat x) - f_n(x)|geq varepsilon$.



            Call the limit function $f(x)=lim_{n to infty} f_n(x)$ a degenerate case of a Dirichlet function (kind of a bad joke).






            share|cite|improve this answer




























              0














              Another counterexample:



              For each $n in mathbb N$ define $f_n: [a,b] to mathbb R$ by
              $$f_n(x)=begin{cases}frac1n&text{ if }x in mathbb Q \0&text{ otherwise.}end{cases}$$



              Given $n in mathbb N$ and $x in [a,b]$ we may show that $f_n$ does not satisfy the definition of a continuous function by selecting $ varepsilon=frac{1}{n+1}$. So if $delta>0$, then there is $hat{x}$ such that $|hat x - x|<delta$ but $|,f_n(hat x) - f_n(x)|geq varepsilon$.



              Call the limit function $f(x)=lim_{n to infty} f_n(x)$ a degenerate case of a Dirichlet function (kind of a bad joke).






              share|cite|improve this answer


























                0












                0








                0






                Another counterexample:



                For each $n in mathbb N$ define $f_n: [a,b] to mathbb R$ by
                $$f_n(x)=begin{cases}frac1n&text{ if }x in mathbb Q \0&text{ otherwise.}end{cases}$$



                Given $n in mathbb N$ and $x in [a,b]$ we may show that $f_n$ does not satisfy the definition of a continuous function by selecting $ varepsilon=frac{1}{n+1}$. So if $delta>0$, then there is $hat{x}$ such that $|hat x - x|<delta$ but $|,f_n(hat x) - f_n(x)|geq varepsilon$.



                Call the limit function $f(x)=lim_{n to infty} f_n(x)$ a degenerate case of a Dirichlet function (kind of a bad joke).






                share|cite|improve this answer














                Another counterexample:



                For each $n in mathbb N$ define $f_n: [a,b] to mathbb R$ by
                $$f_n(x)=begin{cases}frac1n&text{ if }x in mathbb Q \0&text{ otherwise.}end{cases}$$



                Given $n in mathbb N$ and $x in [a,b]$ we may show that $f_n$ does not satisfy the definition of a continuous function by selecting $ varepsilon=frac{1}{n+1}$. So if $delta>0$, then there is $hat{x}$ such that $|hat x - x|<delta$ but $|,f_n(hat x) - f_n(x)|geq varepsilon$.



                Call the limit function $f(x)=lim_{n to infty} f_n(x)$ a degenerate case of a Dirichlet function (kind of a bad joke).







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 29 '18 at 20:46

























                answered Nov 27 '18 at 21:29









                Matt A Pelto

                2,367620




                2,367620






























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