Proof that pointwise convergence can disrupt convergence
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I'm trying to get a grasp on point-wise convergence and am hoping to prove something to give a concrete example of why it's weak. The lemma goes as follows .
Suppose $f _ { n } : [ a , b ] rightarrow mathbb { R }$ is a sequence of continuous functions
that converge point wise , but NOT UNIFORMLY to a continuous function $ f : [ a , b ] rightarrow mathbb { R } $ .
Then there exists a convergence sequence $ x _ { n } rightarrow x operatorname { in } [ a , b ] $ such that $f _ { n } left( x _ { n } right)$ does not converge to $ f(x) $ .
real-analysis analysis pointwise-convergence
asked Dec 16 '18 at 2:17
Andrew HardyAndrew Hardy
125
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add a comment |
$begingroup$
I'm trying to get a grasp on point-wise convergence and am hoping to prove something to give a concrete example of why it's weak. The lemma goes as follows .
Suppose $f _ { n } : [ a , b ] rightarrow mathbb { R }$ is a sequence of continuous functions
that converge point wise , but NOT UNIFORMLY to a continuous function $ f : [ a , b ] rightarrow mathbb { R } $ .
Then there exists a convergence sequence $ x _ { n } rightarrow x operatorname { in } [ a , b ] $ such that $f _ { n } left( x _ { n } right)$ does not converge to $ f(x) $ .
real-analysis analysis pointwise-convergence
asked Dec 16 '18 at 2:17
Andrew HardyAndrew Hardy
125
$endgroup$
add a comment |
$begingroup$
I'm trying to get a grasp on point-wise convergence and am hoping to prove something to give a concrete example of why it's weak. The lemma goes as follows .
Suppose $f _ { n } : [ a , b ] rightarrow mathbb { R }$ is a sequence of continuous functions
that converge point wise , but NOT UNIFORMLY to a continuous function $ f : [ a , b ] rightarrow mathbb { R } $ .
Then there exists a convergence sequence $ x _ { n } rightarrow x operatorname { in } [ a , b ] $ such that $f _ { n } left( x _ { n } right)$ does not converge to $ f(x) $ .
real-analysis analysis pointwise-convergence
asked Dec 16 '18 at 2:17
Andrew HardyAndrew Hardy
125
$endgroup$
I'm trying to get a grasp on point-wise convergence and am hoping to prove something to give a concrete example of why it's weak. The lemma goes as follows .
Suppose $f _ { n } : [ a , b ] rightarrow mathbb { R }$ is a sequence of continuous functions
that converge point wise , but NOT UNIFORMLY to a continuous function $ f : [ a , b ] rightarrow mathbb { R } $ .
Then there exists a convergence sequence $ x _ { n } rightarrow x operatorname { in } [ a , b ] $ such that $f _ { n } left( x _ { n } right)$ does not converge to $ f(x) $ .
real-analysis analysis pointwise-convergence
real-analysis analysis pointwise-convergence
asked Dec 16 '18 at 2:17
Andrew HardyAndrew Hardy
125
asked Dec 16 '18 at 2:17
Andrew HardyAndrew Hardy
125
edited Dec 16 '18 at 2:22
Andrew Hardy
asked Dec 16 '18 at 2:17
Andrew HardyAndrew Hardy
125
asked Dec 16 '18 at 2:17
Andrew HardyAndrew Hardy
125
asked Dec 16 '18 at 2:17
Andrew HardyAndrew Hardy
125
125
add a comment |
add a comment |
1 Answer
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Since the convergence is not uniform there exists $epsilon>0$ and a subsequence $f_{n_i}(x_{n_i})$ (where we can assume the $x_{n_i}$ are convergent to some $x$ because $[a,b]$ is compact) such that $|f_{n_i}(x_{n_i})-f(x_{n_i})|ge epsilon$. For $i$ large enough, $|f(x)-f(x_{n_i})|<epsilon/2$. By the Reverse Triangle Inequality:
$$ |f_{n_i}(x_{n_i})-f(x)|ge||f_{n_i}(x_{n_i})-f(x_{n_i})| -|f(x_{n_i})-f(x)||ge epsilon/2$$
So $f_{n_i}(x_{n_i})$ cannot converge to $f(x)$.
answered Dec 16 '18 at 3:19
Guacho PerezGuacho Perez
3,92911132
$endgroup$
$begingroup$
Thanks, I think this is 99% of it! Do you mind clarifying the initial claim. From lacking uniform convergence, we're finding a small enough $ epsilon $ so that the norm is not always less than it for all the elements in the sequence? That doesn't seem trivial to me. I'm also confused why you seemed to take two norms in your final inequality. Finally a proposed an edit about redefining $ 2 epsilon $ so your final inequality is more obviously sufficient?
$endgroup$
– Andrew Hardy
Dec 16 '18 at 3:50
$begingroup$
$f_nto f$ uniformly iff $lim max_{xin [a,b]} |f_n(x)-f(x)|=0$. If we do not have uniform convergence, this limit does not exist so there must be some $epsilon>0$ s.t. for any $N$, there is some $nge N$ with $max_{xin [a,b]} |f_n(x)-f(x)|ge epsilon$, so you can find an $x_n$ with $|f_n(x_n)-f(x_n)|geepsilon$. You can then repeat this process to obtain the subsequence in the initial claim.
$endgroup$
– Guacho Perez
Dec 16 '18 at 4:19
add a comment |
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$begingroup$
Since the convergence is not uniform there exists $epsilon>0$ and a subsequence $f_{n_i}(x_{n_i})$ (where we can assume the $x_{n_i}$ are convergent to some $x$ because $[a,b]$ is compact) such that $|f_{n_i}(x_{n_i})-f(x_{n_i})|ge epsilon$. For $i$ large enough, $|f(x)-f(x_{n_i})|<epsilon/2$. By the Reverse Triangle Inequality:
$$ |f_{n_i}(x_{n_i})-f(x)|ge||f_{n_i}(x_{n_i})-f(x_{n_i})| -|f(x_{n_i})-f(x)||ge epsilon/2$$
So $f_{n_i}(x_{n_i})$ cannot converge to $f(x)$.
answered Dec 16 '18 at 3:19
Guacho PerezGuacho Perez
3,92911132
$endgroup$
$begingroup$
Thanks, I think this is 99% of it! Do you mind clarifying the initial claim. From lacking uniform convergence, we're finding a small enough $ epsilon $ so that the norm is not always less than it for all the elements in the sequence? That doesn't seem trivial to me. I'm also confused why you seemed to take two norms in your final inequality. Finally a proposed an edit about redefining $ 2 epsilon $ so your final inequality is more obviously sufficient?
$endgroup$
– Andrew Hardy
Dec 16 '18 at 3:50
$begingroup$
$f_nto f$ uniformly iff $lim max_{xin [a,b]} |f_n(x)-f(x)|=0$. If we do not have uniform convergence, this limit does not exist so there must be some $epsilon>0$ s.t. for any $N$, there is some $nge N$ with $max_{xin [a,b]} |f_n(x)-f(x)|ge epsilon$, so you can find an $x_n$ with $|f_n(x_n)-f(x_n)|geepsilon$. You can then repeat this process to obtain the subsequence in the initial claim.
$endgroup$
– Guacho Perez
Dec 16 '18 at 4:19
add a comment |
$begingroup$
Since the convergence is not uniform there exists $epsilon>0$ and a subsequence $f_{n_i}(x_{n_i})$ (where we can assume the $x_{n_i}$ are convergent to some $x$ because $[a,b]$ is compact) such that $|f_{n_i}(x_{n_i})-f(x_{n_i})|ge epsilon$. For $i$ large enough, $|f(x)-f(x_{n_i})|<epsilon/2$. By the Reverse Triangle Inequality:
$$ |f_{n_i}(x_{n_i})-f(x)|ge||f_{n_i}(x_{n_i})-f(x_{n_i})| -|f(x_{n_i})-f(x)||ge epsilon/2$$
So $f_{n_i}(x_{n_i})$ cannot converge to $f(x)$.
answered Dec 16 '18 at 3:19
Guacho PerezGuacho Perez
3,92911132
$endgroup$
$begingroup$
Thanks, I think this is 99% of it! Do you mind clarifying the initial claim. From lacking uniform convergence, we're finding a small enough $ epsilon $ so that the norm is not always less than it for all the elements in the sequence? That doesn't seem trivial to me. I'm also confused why you seemed to take two norms in your final inequality. Finally a proposed an edit about redefining $ 2 epsilon $ so your final inequality is more obviously sufficient?
$endgroup$
– Andrew Hardy
Dec 16 '18 at 3:50
$begingroup$
$f_nto f$ uniformly iff $lim max_{xin [a,b]} |f_n(x)-f(x)|=0$. If we do not have uniform convergence, this limit does not exist so there must be some $epsilon>0$ s.t. for any $N$, there is some $nge N$ with $max_{xin [a,b]} |f_n(x)-f(x)|ge epsilon$, so you can find an $x_n$ with $|f_n(x_n)-f(x_n)|geepsilon$. You can then repeat this process to obtain the subsequence in the initial claim.
$endgroup$
– Guacho Perez
Dec 16 '18 at 4:19
add a comment |
$begingroup$
Since the convergence is not uniform there exists $epsilon>0$ and a subsequence $f_{n_i}(x_{n_i})$ (where we can assume the $x_{n_i}$ are convergent to some $x$ because $[a,b]$ is compact) such that $|f_{n_i}(x_{n_i})-f(x_{n_i})|ge epsilon$. For $i$ large enough, $|f(x)-f(x_{n_i})|<epsilon/2$. By the Reverse Triangle Inequality:
$$ |f_{n_i}(x_{n_i})-f(x)|ge||f_{n_i}(x_{n_i})-f(x_{n_i})| -|f(x_{n_i})-f(x)||ge epsilon/2$$
So $f_{n_i}(x_{n_i})$ cannot converge to $f(x)$.
answered Dec 16 '18 at 3:19
Guacho PerezGuacho Perez
3,92911132
$endgroup$
Since the convergence is not uniform there exists $epsilon>0$ and a subsequence $f_{n_i}(x_{n_i})$ (where we can assume the $x_{n_i}$ are convergent to some $x$ because $[a,b]$ is compact) such that $|f_{n_i}(x_{n_i})-f(x_{n_i})|ge epsilon$. For $i$ large enough, $|f(x)-f(x_{n_i})|<epsilon/2$. By the Reverse Triangle Inequality:
$$ |f_{n_i}(x_{n_i})-f(x)|ge||f_{n_i}(x_{n_i})-f(x_{n_i})| -|f(x_{n_i})-f(x)||ge epsilon/2$$
So $f_{n_i}(x_{n_i})$ cannot converge to $f(x)$.
answered Dec 16 '18 at 3:19
Guacho PerezGuacho Perez
3,92911132
answered Dec 16 '18 at 3:19
Guacho PerezGuacho Perez
3,92911132
answered Dec 16 '18 at 3:19
Guacho PerezGuacho Perez
3,92911132
answered Dec 16 '18 at 3:19
Guacho PerezGuacho Perez
3,92911132
3,92911132
$begingroup$
Thanks, I think this is 99% of it! Do you mind clarifying the initial claim. From lacking uniform convergence, we're finding a small enough $ epsilon $ so that the norm is not always less than it for all the elements in the sequence? That doesn't seem trivial to me. I'm also confused why you seemed to take two norms in your final inequality. Finally a proposed an edit about redefining $ 2 epsilon $ so your final inequality is more obviously sufficient?
$endgroup$
– Andrew Hardy
Dec 16 '18 at 3:50
$begingroup$
$f_nto f$ uniformly iff $lim max_{xin [a,b]} |f_n(x)-f(x)|=0$. If we do not have uniform convergence, this limit does not exist so there must be some $epsilon>0$ s.t. for any $N$, there is some $nge N$ with $max_{xin [a,b]} |f_n(x)-f(x)|ge epsilon$, so you can find an $x_n$ with $|f_n(x_n)-f(x_n)|geepsilon$. You can then repeat this process to obtain the subsequence in the initial claim.
$endgroup$
– Guacho Perez
Dec 16 '18 at 4:19
add a comment |
$begingroup$
Thanks, I think this is 99% of it! Do you mind clarifying the initial claim. From lacking uniform convergence, we're finding a small enough $ epsilon $ so that the norm is not always less than it for all the elements in the sequence? That doesn't seem trivial to me. I'm also confused why you seemed to take two norms in your final inequality. Finally a proposed an edit about redefining $ 2 epsilon $ so your final inequality is more obviously sufficient?
$endgroup$
– Andrew Hardy
Dec 16 '18 at 3:50
$begingroup$
$f_nto f$ uniformly iff $lim max_{xin [a,b]} |f_n(x)-f(x)|=0$. If we do not have uniform convergence, this limit does not exist so there must be some $epsilon>0$ s.t. for any $N$, there is some $nge N$ with $max_{xin [a,b]} |f_n(x)-f(x)|ge epsilon$, so you can find an $x_n$ with $|f_n(x_n)-f(x_n)|geepsilon$. You can then repeat this process to obtain the subsequence in the initial claim.
$endgroup$
– Guacho Perez
Dec 16 '18 at 4:19
$begingroup$
Thanks, I think this is 99% of it! Do you mind clarifying the initial claim. From lacking uniform convergence, we're finding a small enough $ epsilon $ so that the norm is not always less than it for all the elements in the sequence? That doesn't seem trivial to me. I'm also confused why you seemed to take two norms in your final inequality. Finally a proposed an edit about redefining $ 2 epsilon $ so your final inequality is more obviously sufficient?
$endgroup$
– Andrew Hardy
Dec 16 '18 at 3:50
$begingroup$
Thanks, I think this is 99% of it! Do you mind clarifying the initial claim. From lacking uniform convergence, we're finding a small enough $ epsilon $ so that the norm is not always less than it for all the elements in the sequence? That doesn't seem trivial to me. I'm also confused why you seemed to take two norms in your final inequality. Finally a proposed an edit about redefining $ 2 epsilon $ so your final inequality is more obviously sufficient?
$endgroup$
– Andrew Hardy
Dec 16 '18 at 3:50
$begingroup$
$f_nto f$ uniformly iff $lim max_{xin [a,b]} |f_n(x)-f(x)|=0$. If we do not have uniform convergence, this limit does not exist so there must be some $epsilon>0$ s.t. for any $N$, there is some $nge N$ with $max_{xin [a,b]} |f_n(x)-f(x)|ge epsilon$, so you can find an $x_n$ with $|f_n(x_n)-f(x_n)|geepsilon$. You can then repeat this process to obtain the subsequence in the initial claim.
$endgroup$
– Guacho Perez
Dec 16 '18 at 4:19
$begingroup$
$f_nto f$ uniformly iff $lim max_{xin [a,b]} |f_n(x)-f(x)|=0$. If we do not have uniform convergence, this limit does not exist so there must be some $epsilon>0$ s.t. for any $N$, there is some $nge N$ with $max_{xin [a,b]} |f_n(x)-f(x)|ge epsilon$, so you can find an $x_n$ with $|f_n(x_n)-f(x_n)|geepsilon$. You can then repeat this process to obtain the subsequence in the initial claim.
$endgroup$
– Guacho Perez
Dec 16 '18 at 4:19
add a comment |
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