Does uniform convergence of $f$ imply convergance of derivatives?












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$begingroup$


Let $X$ denote the collection of all differentiable functions $f : [0, 1] rightarrow Bbb R$, such that $f(0)=0$ and $f'$ is continuous.



Let ${f_n}$ be a Cauchy sequence. By Cauchy criterion for uniform convergence, $f_n$ converges uniformly to some $f$.



Does that imply that $f'_n rightarrow f'$ uniformly?






real-analysis derivatives convergence uniform-convergence







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$endgroup$

















    1












    $begingroup$


    Let $X$ denote the collection of all differentiable functions $f : [0, 1] rightarrow Bbb R$, such that $f(0)=0$ and $f'$ is continuous.



    Let ${f_n}$ be a Cauchy sequence. By Cauchy criterion for uniform convergence, $f_n$ converges uniformly to some $f$.



    Does that imply that $f'_n rightarrow f'$ uniformly?






    real-analysis derivatives convergence uniform-convergence







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $X$ denote the collection of all differentiable functions $f : [0, 1] rightarrow Bbb R$, such that $f(0)=0$ and $f'$ is continuous.



      Let ${f_n}$ be a Cauchy sequence. By Cauchy criterion for uniform convergence, $f_n$ converges uniformly to some $f$.



      Does that imply that $f'_n rightarrow f'$ uniformly?






      real-analysis derivatives convergence uniform-convergence







      share|cite|improve this question











      $endgroup$




      Let $X$ denote the collection of all differentiable functions $f : [0, 1] rightarrow Bbb R$, such that $f(0)=0$ and $f'$ is continuous.



      Let ${f_n}$ be a Cauchy sequence. By Cauchy criterion for uniform convergence, $f_n$ converges uniformly to some $f$.



      Does that imply that $f'_n rightarrow f'$ uniformly?






      real-analysis derivatives convergence uniform-convergence




      real-analysis derivatives convergence uniform-convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 '18 at 14:37









      Cebiş Mellim

      13011




      13011










      asked Apr 29 '14 at 17:48









      KatestrophicalKatestrophical

      495




      495






















          2 Answers
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          $begingroup$

          No! Imagine the $f_n$ getting close to $f$ uniformly, but getting bumpier and bumpier. You should be able to use this idea to come up with a counterexample.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            As an idea, try looking at the sequence $f+frac{1}{n}phi(nx)$, where $phi$ is a bounded function with derivative $1$ at $0$.
            $endgroup$
            – Steven Gubkin
            Apr 29 '14 at 17:54










          • $begingroup$
            What if we flipped $f$ and $f'$ ? $$$$ Let ${f'_n}$ be a Cauchy sequence. By Cauchy criterion for uniform convergence, $f'_n$ converges uniformly to some $f'$. Does that imply that $f_n rightarrow f$ uniformly?
            $endgroup$
            – Katestrophical
            Apr 29 '14 at 18:12












          • $begingroup$
            Up to a constant, yes. This is just fundamental theorem of calculus + interchanging integration and uniform limits. Generally, integration behaves well numerically, differentiation does not.
            $endgroup$
            – Steven Gubkin
            Apr 29 '14 at 18:21










          • $begingroup$
            @Katestrophical see math.stackexchange.com/q/214218/99325
            $endgroup$
            – derivative
            Apr 29 '14 at 18:23



















          0












          $begingroup$

          This is a counterexample. Take e.g. $f_n(x) = frac{1}{n} (sqrt{1+(n x)^2} - 1)$. We have that $f_n(0) = 0$, and $f'_n(x) = dfrac{nx}{sqrt{1+(nx)^2}}$ is continuous. We also have that $lim_{ntoinfty} f_n(x) = |x|$, and
          $$
          |x| - f_n(x) = frac{1}{n} left(1 - frac{1}{n|x| + sqrt{1 + (nx)^2}}right)
          $$
          hence $0 leq |x| - f_n(x) < frac{1}{n}$ for every $x$, which implies that $f_n(x) to |x|$ uniformly. On the other hand
          $$
          lim_{ntoinfty} f'_n(x) =
          begin{cases}
          1 & text{ if $x>0$} \
          0 & text{ if $x=0$} \
          -1 & text{ if $x<0$}
          end{cases}
          $$
          So we have that $f'_n$ is a sequence of continuous functions that converge to a discontinuous function, so its convergence cannot be uniform.






          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









            2












            $begingroup$

            No! Imagine the $f_n$ getting close to $f$ uniformly, but getting bumpier and bumpier. You should be able to use this idea to come up with a counterexample.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              As an idea, try looking at the sequence $f+frac{1}{n}phi(nx)$, where $phi$ is a bounded function with derivative $1$ at $0$.
              $endgroup$
              – Steven Gubkin
              Apr 29 '14 at 17:54










            • $begingroup$
              What if we flipped $f$ and $f'$ ? $$$$ Let ${f'_n}$ be a Cauchy sequence. By Cauchy criterion for uniform convergence, $f'_n$ converges uniformly to some $f'$. Does that imply that $f_n rightarrow f$ uniformly?
              $endgroup$
              – Katestrophical
              Apr 29 '14 at 18:12












            • $begingroup$
              Up to a constant, yes. This is just fundamental theorem of calculus + interchanging integration and uniform limits. Generally, integration behaves well numerically, differentiation does not.
              $endgroup$
              – Steven Gubkin
              Apr 29 '14 at 18:21










            • $begingroup$
              @Katestrophical see math.stackexchange.com/q/214218/99325
              $endgroup$
              – derivative
              Apr 29 '14 at 18:23
















            2












            $begingroup$

            No! Imagine the $f_n$ getting close to $f$ uniformly, but getting bumpier and bumpier. You should be able to use this idea to come up with a counterexample.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              As an idea, try looking at the sequence $f+frac{1}{n}phi(nx)$, where $phi$ is a bounded function with derivative $1$ at $0$.
              $endgroup$
              – Steven Gubkin
              Apr 29 '14 at 17:54










            • $begingroup$
              What if we flipped $f$ and $f'$ ? $$$$ Let ${f'_n}$ be a Cauchy sequence. By Cauchy criterion for uniform convergence, $f'_n$ converges uniformly to some $f'$. Does that imply that $f_n rightarrow f$ uniformly?
              $endgroup$
              – Katestrophical
              Apr 29 '14 at 18:12












            • $begingroup$
              Up to a constant, yes. This is just fundamental theorem of calculus + interchanging integration and uniform limits. Generally, integration behaves well numerically, differentiation does not.
              $endgroup$
              – Steven Gubkin
              Apr 29 '14 at 18:21










            • $begingroup$
              @Katestrophical see math.stackexchange.com/q/214218/99325
              $endgroup$
              – derivative
              Apr 29 '14 at 18:23














            2












            2








            2





            $begingroup$

            No! Imagine the $f_n$ getting close to $f$ uniformly, but getting bumpier and bumpier. You should be able to use this idea to come up with a counterexample.






            share|cite|improve this answer









            $endgroup$



            No! Imagine the $f_n$ getting close to $f$ uniformly, but getting bumpier and bumpier. You should be able to use this idea to come up with a counterexample.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 29 '14 at 17:52









            Steven GubkinSteven Gubkin

            5,6241531




            5,6241531












            • $begingroup$
              As an idea, try looking at the sequence $f+frac{1}{n}phi(nx)$, where $phi$ is a bounded function with derivative $1$ at $0$.
              $endgroup$
              – Steven Gubkin
              Apr 29 '14 at 17:54










            • $begingroup$
              What if we flipped $f$ and $f'$ ? $$$$ Let ${f'_n}$ be a Cauchy sequence. By Cauchy criterion for uniform convergence, $f'_n$ converges uniformly to some $f'$. Does that imply that $f_n rightarrow f$ uniformly?
              $endgroup$
              – Katestrophical
              Apr 29 '14 at 18:12












            • $begingroup$
              Up to a constant, yes. This is just fundamental theorem of calculus + interchanging integration and uniform limits. Generally, integration behaves well numerically, differentiation does not.
              $endgroup$
              – Steven Gubkin
              Apr 29 '14 at 18:21










            • $begingroup$
              @Katestrophical see math.stackexchange.com/q/214218/99325
              $endgroup$
              – derivative
              Apr 29 '14 at 18:23


















            • $begingroup$
              As an idea, try looking at the sequence $f+frac{1}{n}phi(nx)$, where $phi$ is a bounded function with derivative $1$ at $0$.
              $endgroup$
              – Steven Gubkin
              Apr 29 '14 at 17:54










            • $begingroup$
              What if we flipped $f$ and $f'$ ? $$$$ Let ${f'_n}$ be a Cauchy sequence. By Cauchy criterion for uniform convergence, $f'_n$ converges uniformly to some $f'$. Does that imply that $f_n rightarrow f$ uniformly?
              $endgroup$
              – Katestrophical
              Apr 29 '14 at 18:12












            • $begingroup$
              Up to a constant, yes. This is just fundamental theorem of calculus + interchanging integration and uniform limits. Generally, integration behaves well numerically, differentiation does not.
              $endgroup$
              – Steven Gubkin
              Apr 29 '14 at 18:21










            • $begingroup$
              @Katestrophical see math.stackexchange.com/q/214218/99325
              $endgroup$
              – derivative
              Apr 29 '14 at 18:23
















            $begingroup$
            As an idea, try looking at the sequence $f+frac{1}{n}phi(nx)$, where $phi$ is a bounded function with derivative $1$ at $0$.
            $endgroup$
            – Steven Gubkin
            Apr 29 '14 at 17:54




            $begingroup$
            As an idea, try looking at the sequence $f+frac{1}{n}phi(nx)$, where $phi$ is a bounded function with derivative $1$ at $0$.
            $endgroup$
            – Steven Gubkin
            Apr 29 '14 at 17:54












            $begingroup$
            What if we flipped $f$ and $f'$ ? $$$$ Let ${f'_n}$ be a Cauchy sequence. By Cauchy criterion for uniform convergence, $f'_n$ converges uniformly to some $f'$. Does that imply that $f_n rightarrow f$ uniformly?
            $endgroup$
            – Katestrophical
            Apr 29 '14 at 18:12






            $begingroup$
            What if we flipped $f$ and $f'$ ? $$$$ Let ${f'_n}$ be a Cauchy sequence. By Cauchy criterion for uniform convergence, $f'_n$ converges uniformly to some $f'$. Does that imply that $f_n rightarrow f$ uniformly?
            $endgroup$
            – Katestrophical
            Apr 29 '14 at 18:12














            $begingroup$
            Up to a constant, yes. This is just fundamental theorem of calculus + interchanging integration and uniform limits. Generally, integration behaves well numerically, differentiation does not.
            $endgroup$
            – Steven Gubkin
            Apr 29 '14 at 18:21




            $begingroup$
            Up to a constant, yes. This is just fundamental theorem of calculus + interchanging integration and uniform limits. Generally, integration behaves well numerically, differentiation does not.
            $endgroup$
            – Steven Gubkin
            Apr 29 '14 at 18:21












            $begingroup$
            @Katestrophical see math.stackexchange.com/q/214218/99325
            $endgroup$
            – derivative
            Apr 29 '14 at 18:23




            $begingroup$
            @Katestrophical see math.stackexchange.com/q/214218/99325
            $endgroup$
            – derivative
            Apr 29 '14 at 18:23











            0












            $begingroup$

            This is a counterexample. Take e.g. $f_n(x) = frac{1}{n} (sqrt{1+(n x)^2} - 1)$. We have that $f_n(0) = 0$, and $f'_n(x) = dfrac{nx}{sqrt{1+(nx)^2}}$ is continuous. We also have that $lim_{ntoinfty} f_n(x) = |x|$, and
            $$
            |x| - f_n(x) = frac{1}{n} left(1 - frac{1}{n|x| + sqrt{1 + (nx)^2}}right)
            $$
            hence $0 leq |x| - f_n(x) < frac{1}{n}$ for every $x$, which implies that $f_n(x) to |x|$ uniformly. On the other hand
            $$
            lim_{ntoinfty} f'_n(x) =
            begin{cases}
            1 & text{ if $x>0$} \
            0 & text{ if $x=0$} \
            -1 & text{ if $x<0$}
            end{cases}
            $$
            So we have that $f'_n$ is a sequence of continuous functions that converge to a discontinuous function, so its convergence cannot be uniform.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              This is a counterexample. Take e.g. $f_n(x) = frac{1}{n} (sqrt{1+(n x)^2} - 1)$. We have that $f_n(0) = 0$, and $f'_n(x) = dfrac{nx}{sqrt{1+(nx)^2}}$ is continuous. We also have that $lim_{ntoinfty} f_n(x) = |x|$, and
              $$
              |x| - f_n(x) = frac{1}{n} left(1 - frac{1}{n|x| + sqrt{1 + (nx)^2}}right)
              $$
              hence $0 leq |x| - f_n(x) < frac{1}{n}$ for every $x$, which implies that $f_n(x) to |x|$ uniformly. On the other hand
              $$
              lim_{ntoinfty} f'_n(x) =
              begin{cases}
              1 & text{ if $x>0$} \
              0 & text{ if $x=0$} \
              -1 & text{ if $x<0$}
              end{cases}
              $$
              So we have that $f'_n$ is a sequence of continuous functions that converge to a discontinuous function, so its convergence cannot be uniform.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                This is a counterexample. Take e.g. $f_n(x) = frac{1}{n} (sqrt{1+(n x)^2} - 1)$. We have that $f_n(0) = 0$, and $f'_n(x) = dfrac{nx}{sqrt{1+(nx)^2}}$ is continuous. We also have that $lim_{ntoinfty} f_n(x) = |x|$, and
                $$
                |x| - f_n(x) = frac{1}{n} left(1 - frac{1}{n|x| + sqrt{1 + (nx)^2}}right)
                $$
                hence $0 leq |x| - f_n(x) < frac{1}{n}$ for every $x$, which implies that $f_n(x) to |x|$ uniformly. On the other hand
                $$
                lim_{ntoinfty} f'_n(x) =
                begin{cases}
                1 & text{ if $x>0$} \
                0 & text{ if $x=0$} \
                -1 & text{ if $x<0$}
                end{cases}
                $$
                So we have that $f'_n$ is a sequence of continuous functions that converge to a discontinuous function, so its convergence cannot be uniform.






                share|cite|improve this answer









                $endgroup$



                This is a counterexample. Take e.g. $f_n(x) = frac{1}{n} (sqrt{1+(n x)^2} - 1)$. We have that $f_n(0) = 0$, and $f'_n(x) = dfrac{nx}{sqrt{1+(nx)^2}}$ is continuous. We also have that $lim_{ntoinfty} f_n(x) = |x|$, and
                $$
                |x| - f_n(x) = frac{1}{n} left(1 - frac{1}{n|x| + sqrt{1 + (nx)^2}}right)
                $$
                hence $0 leq |x| - f_n(x) < frac{1}{n}$ for every $x$, which implies that $f_n(x) to |x|$ uniformly. On the other hand
                $$
                lim_{ntoinfty} f'_n(x) =
                begin{cases}
                1 & text{ if $x>0$} \
                0 & text{ if $x=0$} \
                -1 & text{ if $x<0$}
                end{cases}
                $$
                So we have that $f'_n$ is a sequence of continuous functions that converge to a discontinuous function, so its convergence cannot be uniform.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 7 '18 at 16:46









                mlerma54mlerma54

                1,177148




                1,177148






























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