Pointwise convergence of average of continuous functions












3












$begingroup$


Suppose that $(f^n)$ is a sequence of continuous functions from a compact metric space $A$ into a convex compact subset of the reals $B$. Is there a subsequence, $(f^{n_k})$, such that the sequence



$f^{n_1},(1/2)f^{n_1}+(1/2)f^{n_2},(1/3)f^{n_1}+(1/3)f^{n_2}+(1/3)f^{n_3},...$



converges pointwise to some map $f:Ato B$?






functional-analysis metric-spaces compactness uniform-convergence pointwise-convergence







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Suppose that $(f^n)$ is a sequence of continuous functions from a compact metric space $A$ into a convex compact subset of the reals $B$. Is there a subsequence, $(f^{n_k})$, such that the sequence



    $f^{n_1},(1/2)f^{n_1}+(1/2)f^{n_2},(1/3)f^{n_1}+(1/3)f^{n_2}+(1/3)f^{n_3},...$



    converges pointwise to some map $f:Ato B$?






    functional-analysis metric-spaces compactness uniform-convergence pointwise-convergence







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Suppose that $(f^n)$ is a sequence of continuous functions from a compact metric space $A$ into a convex compact subset of the reals $B$. Is there a subsequence, $(f^{n_k})$, such that the sequence



      $f^{n_1},(1/2)f^{n_1}+(1/2)f^{n_2},(1/3)f^{n_1}+(1/3)f^{n_2}+(1/3)f^{n_3},...$



      converges pointwise to some map $f:Ato B$?






      functional-analysis metric-spaces compactness uniform-convergence pointwise-convergence







      share|cite|improve this question











      $endgroup$




      Suppose that $(f^n)$ is a sequence of continuous functions from a compact metric space $A$ into a convex compact subset of the reals $B$. Is there a subsequence, $(f^{n_k})$, such that the sequence



      $f^{n_1},(1/2)f^{n_1}+(1/2)f^{n_2},(1/3)f^{n_1}+(1/3)f^{n_2}+(1/3)f^{n_3},...$



      converges pointwise to some map $f:Ato B$?






      functional-analysis metric-spaces compactness uniform-convergence pointwise-convergence




      functional-analysis metric-spaces compactness uniform-convergence pointwise-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 20:44









      Alex Ravsky

      40.4k32282




      40.4k32282










      asked Oct 22 '18 at 12:45









      mo15mo15

      1768




      1768






















          1 Answer
          1






          active

          oldest

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          1












          $begingroup$

          The answer depends on the sequence $(f^n)$.



          Since the sequence $(f^n)$ is uniformly bounded, if it is equicontinuous then by Arzelà–Ascoli theorem for compact Hausdorff spaces, it is relatively compact in the space $C(A)$ of real-valued continuous functions on $A$. So it contains a subsequence $(f^{n_i})$, uniformly convergent to a function $fin C(A)$. Then a sequence $left(frac 1ksum_{i=1}^k f^{n_i}right)$ uniformly converges to $f$ too.



          But for general case the claim may fail. Consider the following example. Let $A$ be the Cantor cube, that is a countable power of a discrete space ${0,1}$ endowed with the Tychonoff product topology. For each $n$ let $f^n:Ato [0,1]$ be the projection at the $n$-th coordinate. Let $(n_i)$ be an arbitrary increasing sequence of natural numbers. It is easy to construct a set $DsubsetBbb N$ with undefined density, that is such that the limit $d(D)=lim_{ktoinfty} frac 1k|{ ain D: ale k}|$ does not exist. Pick any point $x=(x_n)in A$ such that for each natural $k$ we have $x_{n_k}=1$, if $kin D$ and $x_{n_k}=0$, otherwise. Since for each $k$ we have $|{ ain D: ale k}|=sum_{i=1}^k f^{n_i}(x)$, the limit $frac 1ksum_{i=1}^k f^{n_i}(x)$ does not exist.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer. I am little confused with the construction. Do you mean a set $Dsubseteq{0,1}^N$ such that $lim_{ktoinfty}frac{1}{k}(d_1+dots+d_k)$ does not exist? And then pick $xin A$ such that $x_{n_k}=d_k$ for each $k$?
            $endgroup$
            – mo15
            Dec 8 '18 at 22:02










          • $begingroup$
            @mo15 Oops, I sometimes denoted the set $D$ as $A$. Sorry. Corrected.
            $endgroup$
            – Alex Ravsky
            Dec 8 '18 at 23:03










          • $begingroup$
            But $D$ must be a subset of $A$, not $N$, correct? An the limit $d(D)$ must be defined differently, I think.
            $endgroup$
            – mo15
            Dec 8 '18 at 23:15












          • $begingroup$
            In view of this example, a natural question would be whether the claim is true when $A$ is a compact subset of Euclidean space $mathbb R^n$.
            $endgroup$
            – mo15
            Dec 9 '18 at 0:42












          • $begingroup$
            No, $D$ is a subset of $Bbb N$.
            $endgroup$
            – Alex Ravsky
            Dec 9 '18 at 3:48











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          $begingroup$

          The answer depends on the sequence $(f^n)$.



          Since the sequence $(f^n)$ is uniformly bounded, if it is equicontinuous then by Arzelà–Ascoli theorem for compact Hausdorff spaces, it is relatively compact in the space $C(A)$ of real-valued continuous functions on $A$. So it contains a subsequence $(f^{n_i})$, uniformly convergent to a function $fin C(A)$. Then a sequence $left(frac 1ksum_{i=1}^k f^{n_i}right)$ uniformly converges to $f$ too.



          But for general case the claim may fail. Consider the following example. Let $A$ be the Cantor cube, that is a countable power of a discrete space ${0,1}$ endowed with the Tychonoff product topology. For each $n$ let $f^n:Ato [0,1]$ be the projection at the $n$-th coordinate. Let $(n_i)$ be an arbitrary increasing sequence of natural numbers. It is easy to construct a set $DsubsetBbb N$ with undefined density, that is such that the limit $d(D)=lim_{ktoinfty} frac 1k|{ ain D: ale k}|$ does not exist. Pick any point $x=(x_n)in A$ such that for each natural $k$ we have $x_{n_k}=1$, if $kin D$ and $x_{n_k}=0$, otherwise. Since for each $k$ we have $|{ ain D: ale k}|=sum_{i=1}^k f^{n_i}(x)$, the limit $frac 1ksum_{i=1}^k f^{n_i}(x)$ does not exist.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer. I am little confused with the construction. Do you mean a set $Dsubseteq{0,1}^N$ such that $lim_{ktoinfty}frac{1}{k}(d_1+dots+d_k)$ does not exist? And then pick $xin A$ such that $x_{n_k}=d_k$ for each $k$?
            $endgroup$
            – mo15
            Dec 8 '18 at 22:02










          • $begingroup$
            @mo15 Oops, I sometimes denoted the set $D$ as $A$. Sorry. Corrected.
            $endgroup$
            – Alex Ravsky
            Dec 8 '18 at 23:03










          • $begingroup$
            But $D$ must be a subset of $A$, not $N$, correct? An the limit $d(D)$ must be defined differently, I think.
            $endgroup$
            – mo15
            Dec 8 '18 at 23:15












          • $begingroup$
            In view of this example, a natural question would be whether the claim is true when $A$ is a compact subset of Euclidean space $mathbb R^n$.
            $endgroup$
            – mo15
            Dec 9 '18 at 0:42












          • $begingroup$
            No, $D$ is a subset of $Bbb N$.
            $endgroup$
            – Alex Ravsky
            Dec 9 '18 at 3:48
















          1












          $begingroup$

          The answer depends on the sequence $(f^n)$.



          Since the sequence $(f^n)$ is uniformly bounded, if it is equicontinuous then by Arzelà–Ascoli theorem for compact Hausdorff spaces, it is relatively compact in the space $C(A)$ of real-valued continuous functions on $A$. So it contains a subsequence $(f^{n_i})$, uniformly convergent to a function $fin C(A)$. Then a sequence $left(frac 1ksum_{i=1}^k f^{n_i}right)$ uniformly converges to $f$ too.



          But for general case the claim may fail. Consider the following example. Let $A$ be the Cantor cube, that is a countable power of a discrete space ${0,1}$ endowed with the Tychonoff product topology. For each $n$ let $f^n:Ato [0,1]$ be the projection at the $n$-th coordinate. Let $(n_i)$ be an arbitrary increasing sequence of natural numbers. It is easy to construct a set $DsubsetBbb N$ with undefined density, that is such that the limit $d(D)=lim_{ktoinfty} frac 1k|{ ain D: ale k}|$ does not exist. Pick any point $x=(x_n)in A$ such that for each natural $k$ we have $x_{n_k}=1$, if $kin D$ and $x_{n_k}=0$, otherwise. Since for each $k$ we have $|{ ain D: ale k}|=sum_{i=1}^k f^{n_i}(x)$, the limit $frac 1ksum_{i=1}^k f^{n_i}(x)$ does not exist.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer. I am little confused with the construction. Do you mean a set $Dsubseteq{0,1}^N$ such that $lim_{ktoinfty}frac{1}{k}(d_1+dots+d_k)$ does not exist? And then pick $xin A$ such that $x_{n_k}=d_k$ for each $k$?
            $endgroup$
            – mo15
            Dec 8 '18 at 22:02










          • $begingroup$
            @mo15 Oops, I sometimes denoted the set $D$ as $A$. Sorry. Corrected.
            $endgroup$
            – Alex Ravsky
            Dec 8 '18 at 23:03










          • $begingroup$
            But $D$ must be a subset of $A$, not $N$, correct? An the limit $d(D)$ must be defined differently, I think.
            $endgroup$
            – mo15
            Dec 8 '18 at 23:15












          • $begingroup$
            In view of this example, a natural question would be whether the claim is true when $A$ is a compact subset of Euclidean space $mathbb R^n$.
            $endgroup$
            – mo15
            Dec 9 '18 at 0:42












          • $begingroup$
            No, $D$ is a subset of $Bbb N$.
            $endgroup$
            – Alex Ravsky
            Dec 9 '18 at 3:48














          1












          1








          1





          $begingroup$

          The answer depends on the sequence $(f^n)$.



          Since the sequence $(f^n)$ is uniformly bounded, if it is equicontinuous then by Arzelà–Ascoli theorem for compact Hausdorff spaces, it is relatively compact in the space $C(A)$ of real-valued continuous functions on $A$. So it contains a subsequence $(f^{n_i})$, uniformly convergent to a function $fin C(A)$. Then a sequence $left(frac 1ksum_{i=1}^k f^{n_i}right)$ uniformly converges to $f$ too.



          But for general case the claim may fail. Consider the following example. Let $A$ be the Cantor cube, that is a countable power of a discrete space ${0,1}$ endowed with the Tychonoff product topology. For each $n$ let $f^n:Ato [0,1]$ be the projection at the $n$-th coordinate. Let $(n_i)$ be an arbitrary increasing sequence of natural numbers. It is easy to construct a set $DsubsetBbb N$ with undefined density, that is such that the limit $d(D)=lim_{ktoinfty} frac 1k|{ ain D: ale k}|$ does not exist. Pick any point $x=(x_n)in A$ such that for each natural $k$ we have $x_{n_k}=1$, if $kin D$ and $x_{n_k}=0$, otherwise. Since for each $k$ we have $|{ ain D: ale k}|=sum_{i=1}^k f^{n_i}(x)$, the limit $frac 1ksum_{i=1}^k f^{n_i}(x)$ does not exist.






          share|cite|improve this answer











          $endgroup$



          The answer depends on the sequence $(f^n)$.



          Since the sequence $(f^n)$ is uniformly bounded, if it is equicontinuous then by Arzelà–Ascoli theorem for compact Hausdorff spaces, it is relatively compact in the space $C(A)$ of real-valued continuous functions on $A$. So it contains a subsequence $(f^{n_i})$, uniformly convergent to a function $fin C(A)$. Then a sequence $left(frac 1ksum_{i=1}^k f^{n_i}right)$ uniformly converges to $f$ too.



          But for general case the claim may fail. Consider the following example. Let $A$ be the Cantor cube, that is a countable power of a discrete space ${0,1}$ endowed with the Tychonoff product topology. For each $n$ let $f^n:Ato [0,1]$ be the projection at the $n$-th coordinate. Let $(n_i)$ be an arbitrary increasing sequence of natural numbers. It is easy to construct a set $DsubsetBbb N$ with undefined density, that is such that the limit $d(D)=lim_{ktoinfty} frac 1k|{ ain D: ale k}|$ does not exist. Pick any point $x=(x_n)in A$ such that for each natural $k$ we have $x_{n_k}=1$, if $kin D$ and $x_{n_k}=0$, otherwise. Since for each $k$ we have $|{ ain D: ale k}|=sum_{i=1}^k f^{n_i}(x)$, the limit $frac 1ksum_{i=1}^k f^{n_i}(x)$ does not exist.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 3:47

























          answered Dec 7 '18 at 20:38









          Alex RavskyAlex Ravsky

          40.4k32282




          40.4k32282












          • $begingroup$
            Thanks for your answer. I am little confused with the construction. Do you mean a set $Dsubseteq{0,1}^N$ such that $lim_{ktoinfty}frac{1}{k}(d_1+dots+d_k)$ does not exist? And then pick $xin A$ such that $x_{n_k}=d_k$ for each $k$?
            $endgroup$
            – mo15
            Dec 8 '18 at 22:02










          • $begingroup$
            @mo15 Oops, I sometimes denoted the set $D$ as $A$. Sorry. Corrected.
            $endgroup$
            – Alex Ravsky
            Dec 8 '18 at 23:03










          • $begingroup$
            But $D$ must be a subset of $A$, not $N$, correct? An the limit $d(D)$ must be defined differently, I think.
            $endgroup$
            – mo15
            Dec 8 '18 at 23:15












          • $begingroup$
            In view of this example, a natural question would be whether the claim is true when $A$ is a compact subset of Euclidean space $mathbb R^n$.
            $endgroup$
            – mo15
            Dec 9 '18 at 0:42












          • $begingroup$
            No, $D$ is a subset of $Bbb N$.
            $endgroup$
            – Alex Ravsky
            Dec 9 '18 at 3:48


















          • $begingroup$
            Thanks for your answer. I am little confused with the construction. Do you mean a set $Dsubseteq{0,1}^N$ such that $lim_{ktoinfty}frac{1}{k}(d_1+dots+d_k)$ does not exist? And then pick $xin A$ such that $x_{n_k}=d_k$ for each $k$?
            $endgroup$
            – mo15
            Dec 8 '18 at 22:02










          • $begingroup$
            @mo15 Oops, I sometimes denoted the set $D$ as $A$. Sorry. Corrected.
            $endgroup$
            – Alex Ravsky
            Dec 8 '18 at 23:03










          • $begingroup$
            But $D$ must be a subset of $A$, not $N$, correct? An the limit $d(D)$ must be defined differently, I think.
            $endgroup$
            – mo15
            Dec 8 '18 at 23:15












          • $begingroup$
            In view of this example, a natural question would be whether the claim is true when $A$ is a compact subset of Euclidean space $mathbb R^n$.
            $endgroup$
            – mo15
            Dec 9 '18 at 0:42












          • $begingroup$
            No, $D$ is a subset of $Bbb N$.
            $endgroup$
            – Alex Ravsky
            Dec 9 '18 at 3:48
















          $begingroup$
          Thanks for your answer. I am little confused with the construction. Do you mean a set $Dsubseteq{0,1}^N$ such that $lim_{ktoinfty}frac{1}{k}(d_1+dots+d_k)$ does not exist? And then pick $xin A$ such that $x_{n_k}=d_k$ for each $k$?
          $endgroup$
          – mo15
          Dec 8 '18 at 22:02




          $begingroup$
          Thanks for your answer. I am little confused with the construction. Do you mean a set $Dsubseteq{0,1}^N$ such that $lim_{ktoinfty}frac{1}{k}(d_1+dots+d_k)$ does not exist? And then pick $xin A$ such that $x_{n_k}=d_k$ for each $k$?
          $endgroup$
          – mo15
          Dec 8 '18 at 22:02












          $begingroup$
          @mo15 Oops, I sometimes denoted the set $D$ as $A$. Sorry. Corrected.
          $endgroup$
          – Alex Ravsky
          Dec 8 '18 at 23:03




          $begingroup$
          @mo15 Oops, I sometimes denoted the set $D$ as $A$. Sorry. Corrected.
          $endgroup$
          – Alex Ravsky
          Dec 8 '18 at 23:03












          $begingroup$
          But $D$ must be a subset of $A$, not $N$, correct? An the limit $d(D)$ must be defined differently, I think.
          $endgroup$
          – mo15
          Dec 8 '18 at 23:15






          $begingroup$
          But $D$ must be a subset of $A$, not $N$, correct? An the limit $d(D)$ must be defined differently, I think.
          $endgroup$
          – mo15
          Dec 8 '18 at 23:15














          $begingroup$
          In view of this example, a natural question would be whether the claim is true when $A$ is a compact subset of Euclidean space $mathbb R^n$.
          $endgroup$
          – mo15
          Dec 9 '18 at 0:42






          $begingroup$
          In view of this example, a natural question would be whether the claim is true when $A$ is a compact subset of Euclidean space $mathbb R^n$.
          $endgroup$
          – mo15
          Dec 9 '18 at 0:42














          $begingroup$
          No, $D$ is a subset of $Bbb N$.
          $endgroup$
          – Alex Ravsky
          Dec 9 '18 at 3:48




          $begingroup$
          No, $D$ is a subset of $Bbb N$.
          $endgroup$
          – Alex Ravsky
          Dec 9 '18 at 3:48


















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