Pointwise convergence of average of continuous functions
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
edited Dec 7 '18 at 20:44
Alex Ravsky
40.4k32282
40.4k32282
asked Oct 22 '18 at 12:45
mo15mo15
1768
1768
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
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$begingroup$
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
|
show 2 more comments
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1 Answer
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1 Answer
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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.
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
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.
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
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.
$endgroup$
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.
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
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