Composition of two uniformly convergent sequences of functions is uniformly convergent?












3














I am trying to prove or provide a counter-example for the following:



Let $f_k$ and $g_k$ be sequences of continuous functions on $[0,1]to[0,1]$ converging uniformly to $f:[0,1]to mathbb{R}$ and $g:[0,1]to mathbb{R}$ respectively. Does $f_k circ g_k$ coverge uniformly to $fcirc g$?



What I've done so far:



I know I need to prove that $forall epsilon>0, exists Nin mathbb{N}$ such that $||f_k(g_k(x)) - f(g(x))|| < epsilon$ for all $x in [0,1]$ and $kgeq N$.



At first, I thought I can prove this easily since it follows trivially from the definition of $f_k$ uniformly converging to $f$. However, I noticed that is only true for all $x in [0,1]$ and $g$ maps onto all of $mathbb{R}$, not just $[0,1]$. So does that mean it's not necessarily true? Can anyone provide a counter-example?






real-analysis sequences-and-series uniform-convergence







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  • Without loss of generality, you may assume that the domain and range of both $f$ and $g$ are $[0,1]$. (Uniform limit of sequence of continuous functions with compact support, so you may rescale them without changing much of the problem.)
    – GNUSupporter 8964民主女神 地下教會
    Dec 2 '18 at 5:52
















3














I am trying to prove or provide a counter-example for the following:



Let $f_k$ and $g_k$ be sequences of continuous functions on $[0,1]to[0,1]$ converging uniformly to $f:[0,1]to mathbb{R}$ and $g:[0,1]to mathbb{R}$ respectively. Does $f_k circ g_k$ coverge uniformly to $fcirc g$?



What I've done so far:



I know I need to prove that $forall epsilon>0, exists Nin mathbb{N}$ such that $||f_k(g_k(x)) - f(g(x))|| < epsilon$ for all $x in [0,1]$ and $kgeq N$.



At first, I thought I can prove this easily since it follows trivially from the definition of $f_k$ uniformly converging to $f$. However, I noticed that is only true for all $x in [0,1]$ and $g$ maps onto all of $mathbb{R}$, not just $[0,1]$. So does that mean it's not necessarily true? Can anyone provide a counter-example?






real-analysis sequences-and-series uniform-convergence







share|cite|improve this question






















  • Without loss of generality, you may assume that the domain and range of both $f$ and $g$ are $[0,1]$. (Uniform limit of sequence of continuous functions with compact support, so you may rescale them without changing much of the problem.)
    – GNUSupporter 8964民主女神 地下教會
    Dec 2 '18 at 5:52














3












3








3







I am trying to prove or provide a counter-example for the following:



Let $f_k$ and $g_k$ be sequences of continuous functions on $[0,1]to[0,1]$ converging uniformly to $f:[0,1]to mathbb{R}$ and $g:[0,1]to mathbb{R}$ respectively. Does $f_k circ g_k$ coverge uniformly to $fcirc g$?



What I've done so far:



I know I need to prove that $forall epsilon>0, exists Nin mathbb{N}$ such that $||f_k(g_k(x)) - f(g(x))|| < epsilon$ for all $x in [0,1]$ and $kgeq N$.



At first, I thought I can prove this easily since it follows trivially from the definition of $f_k$ uniformly converging to $f$. However, I noticed that is only true for all $x in [0,1]$ and $g$ maps onto all of $mathbb{R}$, not just $[0,1]$. So does that mean it's not necessarily true? Can anyone provide a counter-example?






real-analysis sequences-and-series uniform-convergence







share|cite|improve this question













I am trying to prove or provide a counter-example for the following:



Let $f_k$ and $g_k$ be sequences of continuous functions on $[0,1]to[0,1]$ converging uniformly to $f:[0,1]to mathbb{R}$ and $g:[0,1]to mathbb{R}$ respectively. Does $f_k circ g_k$ coverge uniformly to $fcirc g$?



What I've done so far:



I know I need to prove that $forall epsilon>0, exists Nin mathbb{N}$ such that $||f_k(g_k(x)) - f(g(x))|| < epsilon$ for all $x in [0,1]$ and $kgeq N$.



At first, I thought I can prove this easily since it follows trivially from the definition of $f_k$ uniformly converging to $f$. However, I noticed that is only true for all $x in [0,1]$ and $g$ maps onto all of $mathbb{R}$, not just $[0,1]$. So does that mean it's not necessarily true? Can anyone provide a counter-example?






real-analysis sequences-and-series uniform-convergence




real-analysis sequences-and-series uniform-convergence






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asked Dec 2 '18 at 5:41









darcy

311




311












  • Without loss of generality, you may assume that the domain and range of both $f$ and $g$ are $[0,1]$. (Uniform limit of sequence of continuous functions with compact support, so you may rescale them without changing much of the problem.)
    – GNUSupporter 8964民主女神 地下教會
    Dec 2 '18 at 5:52


















  • Without loss of generality, you may assume that the domain and range of both $f$ and $g$ are $[0,1]$. (Uniform limit of sequence of continuous functions with compact support, so you may rescale them without changing much of the problem.)
    – GNUSupporter 8964民主女神 地下教會
    Dec 2 '18 at 5:52
















Without loss of generality, you may assume that the domain and range of both $f$ and $g$ are $[0,1]$. (Uniform limit of sequence of continuous functions with compact support, so you may rescale them without changing much of the problem.)
– GNUSupporter 8964民主女神 地下教會
Dec 2 '18 at 5:52




Without loss of generality, you may assume that the domain and range of both $f$ and $g$ are $[0,1]$. (Uniform limit of sequence of continuous functions with compact support, so you may rescale them without changing much of the problem.)
– GNUSupporter 8964民主女神 地下教會
Dec 2 '18 at 5:52










2 Answers
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1














Let $|h|_infty = sup_{x in [0,1]} |h(x)|$.



Note that $f$ is uniformly continuous since $[0,1]$ is compact. Hence since $|g-g_k|_infty to 0$, we see that $|f circ g-f circ g_k|_infty to 0$.



Then
begin{eqnarray}
|f circ g(x)-f_k circ g_k (x)| &le& |f circ g(x)-f circ g_k (x)| + |f circ g_k(x)-f_k circ g_k (x)| \
&le & |f circ g-f circ g_k|_infty + |f-f_k|_infty
end{eqnarray}
Hence $|f circ g-f_k circ g_k|_infty to 0$.






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    0














    For each $x in [0,1]$, $f_n(x) to f(x)$. Since $f_n(x) in [0,1]$, $f(x) in [0,1]$. Idem for $g$.



    Let $epsilon > 0$.



    begin{align}
    & exists N_1 in Bbb{N}: forall n ge N_1, forall y in [0,1], |f_n(y)-f(y)| < epsilon tag1 label1 \
    & exists delta > 0 : forall |y - y'| le delta, |f(y) - f(y')| < epsilon tag2 label2 \
    & exists N_2 in Bbb{N}: forall k ge N_2, forall x in [0,1], |g_k(x)-g(x)| < delta tag3 label3
    end{align}



    eqref{1} and eqref{3} are the definition of uniform continuity; eqref{2} is due to the facts that the uniform limit $f$ of a sequence of continuous functions $(f_n)_n$ is continuous, and that $f$ is uniformly continuous on closed and bounded interval $[0,1]$.



    Put eqref{1}-eqref{3} together. Take $N = max{N_1,N_2}$. For all $n ge N$,



    begin{align}
    & quad |f_n(g_n(x)) - f(g(x))| \
    &le |f_n(g_n(x)) - f(g_n(x))| + |f(g_n(x)) - f(g(x))| \
    &le epsilon + epsilon = 2epsilon.
    end{align}



    In the last inequality, we applied eqref{1} with $y = g_n(x)$ in the first term, and eqref{3} with $k = n$ composed with eqref{2} with $y = g_n(x)$ and $y' = g(x)$ in the second term.



    Hence $f_n circ g_n$ converges uniformly to $f circ g$.






    share|cite|improve this answer





















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      2 Answers
      2






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      2 Answers
      2






      active

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      active

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      1














      Let $|h|_infty = sup_{x in [0,1]} |h(x)|$.



      Note that $f$ is uniformly continuous since $[0,1]$ is compact. Hence since $|g-g_k|_infty to 0$, we see that $|f circ g-f circ g_k|_infty to 0$.



      Then
      begin{eqnarray}
      |f circ g(x)-f_k circ g_k (x)| &le& |f circ g(x)-f circ g_k (x)| + |f circ g_k(x)-f_k circ g_k (x)| \
      &le & |f circ g-f circ g_k|_infty + |f-f_k|_infty
      end{eqnarray}
      Hence $|f circ g-f_k circ g_k|_infty to 0$.






      share|cite|improve this answer


























        1














        Let $|h|_infty = sup_{x in [0,1]} |h(x)|$.



        Note that $f$ is uniformly continuous since $[0,1]$ is compact. Hence since $|g-g_k|_infty to 0$, we see that $|f circ g-f circ g_k|_infty to 0$.



        Then
        begin{eqnarray}
        |f circ g(x)-f_k circ g_k (x)| &le& |f circ g(x)-f circ g_k (x)| + |f circ g_k(x)-f_k circ g_k (x)| \
        &le & |f circ g-f circ g_k|_infty + |f-f_k|_infty
        end{eqnarray}
        Hence $|f circ g-f_k circ g_k|_infty to 0$.






        share|cite|improve this answer
























          1












          1








          1






          Let $|h|_infty = sup_{x in [0,1]} |h(x)|$.



          Note that $f$ is uniformly continuous since $[0,1]$ is compact. Hence since $|g-g_k|_infty to 0$, we see that $|f circ g-f circ g_k|_infty to 0$.



          Then
          begin{eqnarray}
          |f circ g(x)-f_k circ g_k (x)| &le& |f circ g(x)-f circ g_k (x)| + |f circ g_k(x)-f_k circ g_k (x)| \
          &le & |f circ g-f circ g_k|_infty + |f-f_k|_infty
          end{eqnarray}
          Hence $|f circ g-f_k circ g_k|_infty to 0$.






          share|cite|improve this answer












          Let $|h|_infty = sup_{x in [0,1]} |h(x)|$.



          Note that $f$ is uniformly continuous since $[0,1]$ is compact. Hence since $|g-g_k|_infty to 0$, we see that $|f circ g-f circ g_k|_infty to 0$.



          Then
          begin{eqnarray}
          |f circ g(x)-f_k circ g_k (x)| &le& |f circ g(x)-f circ g_k (x)| + |f circ g_k(x)-f_k circ g_k (x)| \
          &le & |f circ g-f circ g_k|_infty + |f-f_k|_infty
          end{eqnarray}
          Hence $|f circ g-f_k circ g_k|_infty to 0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 7:14









          copper.hat

          126k559159




          126k559159























              0














              For each $x in [0,1]$, $f_n(x) to f(x)$. Since $f_n(x) in [0,1]$, $f(x) in [0,1]$. Idem for $g$.



              Let $epsilon > 0$.



              begin{align}
              & exists N_1 in Bbb{N}: forall n ge N_1, forall y in [0,1], |f_n(y)-f(y)| < epsilon tag1 label1 \
              & exists delta > 0 : forall |y - y'| le delta, |f(y) - f(y')| < epsilon tag2 label2 \
              & exists N_2 in Bbb{N}: forall k ge N_2, forall x in [0,1], |g_k(x)-g(x)| < delta tag3 label3
              end{align}



              eqref{1} and eqref{3} are the definition of uniform continuity; eqref{2} is due to the facts that the uniform limit $f$ of a sequence of continuous functions $(f_n)_n$ is continuous, and that $f$ is uniformly continuous on closed and bounded interval $[0,1]$.



              Put eqref{1}-eqref{3} together. Take $N = max{N_1,N_2}$. For all $n ge N$,



              begin{align}
              & quad |f_n(g_n(x)) - f(g(x))| \
              &le |f_n(g_n(x)) - f(g_n(x))| + |f(g_n(x)) - f(g(x))| \
              &le epsilon + epsilon = 2epsilon.
              end{align}



              In the last inequality, we applied eqref{1} with $y = g_n(x)$ in the first term, and eqref{3} with $k = n$ composed with eqref{2} with $y = g_n(x)$ and $y' = g(x)$ in the second term.



              Hence $f_n circ g_n$ converges uniformly to $f circ g$.






              share|cite|improve this answer


























                0














                For each $x in [0,1]$, $f_n(x) to f(x)$. Since $f_n(x) in [0,1]$, $f(x) in [0,1]$. Idem for $g$.



                Let $epsilon > 0$.



                begin{align}
                & exists N_1 in Bbb{N}: forall n ge N_1, forall y in [0,1], |f_n(y)-f(y)| < epsilon tag1 label1 \
                & exists delta > 0 : forall |y - y'| le delta, |f(y) - f(y')| < epsilon tag2 label2 \
                & exists N_2 in Bbb{N}: forall k ge N_2, forall x in [0,1], |g_k(x)-g(x)| < delta tag3 label3
                end{align}



                eqref{1} and eqref{3} are the definition of uniform continuity; eqref{2} is due to the facts that the uniform limit $f$ of a sequence of continuous functions $(f_n)_n$ is continuous, and that $f$ is uniformly continuous on closed and bounded interval $[0,1]$.



                Put eqref{1}-eqref{3} together. Take $N = max{N_1,N_2}$. For all $n ge N$,



                begin{align}
                & quad |f_n(g_n(x)) - f(g(x))| \
                &le |f_n(g_n(x)) - f(g_n(x))| + |f(g_n(x)) - f(g(x))| \
                &le epsilon + epsilon = 2epsilon.
                end{align}



                In the last inequality, we applied eqref{1} with $y = g_n(x)$ in the first term, and eqref{3} with $k = n$ composed with eqref{2} with $y = g_n(x)$ and $y' = g(x)$ in the second term.



                Hence $f_n circ g_n$ converges uniformly to $f circ g$.






                share|cite|improve this answer
























                  0












                  0








                  0






                  For each $x in [0,1]$, $f_n(x) to f(x)$. Since $f_n(x) in [0,1]$, $f(x) in [0,1]$. Idem for $g$.



                  Let $epsilon > 0$.



                  begin{align}
                  & exists N_1 in Bbb{N}: forall n ge N_1, forall y in [0,1], |f_n(y)-f(y)| < epsilon tag1 label1 \
                  & exists delta > 0 : forall |y - y'| le delta, |f(y) - f(y')| < epsilon tag2 label2 \
                  & exists N_2 in Bbb{N}: forall k ge N_2, forall x in [0,1], |g_k(x)-g(x)| < delta tag3 label3
                  end{align}



                  eqref{1} and eqref{3} are the definition of uniform continuity; eqref{2} is due to the facts that the uniform limit $f$ of a sequence of continuous functions $(f_n)_n$ is continuous, and that $f$ is uniformly continuous on closed and bounded interval $[0,1]$.



                  Put eqref{1}-eqref{3} together. Take $N = max{N_1,N_2}$. For all $n ge N$,



                  begin{align}
                  & quad |f_n(g_n(x)) - f(g(x))| \
                  &le |f_n(g_n(x)) - f(g_n(x))| + |f(g_n(x)) - f(g(x))| \
                  &le epsilon + epsilon = 2epsilon.
                  end{align}



                  In the last inequality, we applied eqref{1} with $y = g_n(x)$ in the first term, and eqref{3} with $k = n$ composed with eqref{2} with $y = g_n(x)$ and $y' = g(x)$ in the second term.



                  Hence $f_n circ g_n$ converges uniformly to $f circ g$.






                  share|cite|improve this answer












                  For each $x in [0,1]$, $f_n(x) to f(x)$. Since $f_n(x) in [0,1]$, $f(x) in [0,1]$. Idem for $g$.



                  Let $epsilon > 0$.



                  begin{align}
                  & exists N_1 in Bbb{N}: forall n ge N_1, forall y in [0,1], |f_n(y)-f(y)| < epsilon tag1 label1 \
                  & exists delta > 0 : forall |y - y'| le delta, |f(y) - f(y')| < epsilon tag2 label2 \
                  & exists N_2 in Bbb{N}: forall k ge N_2, forall x in [0,1], |g_k(x)-g(x)| < delta tag3 label3
                  end{align}



                  eqref{1} and eqref{3} are the definition of uniform continuity; eqref{2} is due to the facts that the uniform limit $f$ of a sequence of continuous functions $(f_n)_n$ is continuous, and that $f$ is uniformly continuous on closed and bounded interval $[0,1]$.



                  Put eqref{1}-eqref{3} together. Take $N = max{N_1,N_2}$. For all $n ge N$,



                  begin{align}
                  & quad |f_n(g_n(x)) - f(g(x))| \
                  &le |f_n(g_n(x)) - f(g_n(x))| + |f(g_n(x)) - f(g(x))| \
                  &le epsilon + epsilon = 2epsilon.
                  end{align}



                  In the last inequality, we applied eqref{1} with $y = g_n(x)$ in the first term, and eqref{3} with $k = n$ composed with eqref{2} with $y = g_n(x)$ and $y' = g(x)$ in the second term.



                  Hence $f_n circ g_n$ converges uniformly to $f circ g$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 2 '18 at 6:48









                  GNUSupporter 8964民主女神 地下教會

                  12.8k72445




                  12.8k72445






























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