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Proof of $limsup_ntoinfty x_kn leq limsup_ntoinfty x_n$
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Consider a bounded sequence $x_n$ of real numbers.
I want to proof that the $limsup$ of a subsequence $x_kn$ with $k in mathbbR$ is smaller or equal to the $limsup$ of the mainsequence $x_n$ or in symbols:
$limsup_ntoinfty x_kn leq limsup_ntoinfty x_n$.
I can see why this is correct, but don't know how to prove it. I think it is because $x_kn$ is a subsequence so if the mainsequence converges, so will the subsequence do, but I can't see how to fix it with the suprema.
Thanks in advance.
real-analysis limits supremum-and-infimum
asked Aug 6 at 12:54
Belgium_Physics
1367
add a comment |Â
up vote
2
down vote
favorite
Consider a bounded sequence $x_n$ of real numbers.
I want to proof that the $limsup$ of a subsequence $x_kn$ with $k in mathbbR$ is smaller or equal to the $limsup$ of the mainsequence $x_n$ or in symbols:
$limsup_ntoinfty x_kn leq limsup_ntoinfty x_n$.
I can see why this is correct, but don't know how to prove it. I think it is because $x_kn$ is a subsequence so if the mainsequence converges, so will the subsequence do, but I can't see how to fix it with the suprema.
Thanks in advance.
real-analysis limits supremum-and-infimum
asked Aug 6 at 12:54
Belgium_Physics
1367
This is not a question of convergence. Limsup exists independent of convergence of $x_n$. What is the definition of limsup? What would happen if you put the definitions there, I think looking it from that angle could help you.
– Kolja
Aug 6 at 12:58
Every term in the subsequence shows up as a term in the sequence. It doesn't matter what your definition of limsup is, you just need to unpack the definition and apply this observation.
– Robert Wolfe
Aug 6 at 13:08
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider a bounded sequence $x_n$ of real numbers.
I want to proof that the $limsup$ of a subsequence $x_kn$ with $k in mathbbR$ is smaller or equal to the $limsup$ of the mainsequence $x_n$ or in symbols:
$limsup_ntoinfty x_kn leq limsup_ntoinfty x_n$.
I can see why this is correct, but don't know how to prove it. I think it is because $x_kn$ is a subsequence so if the mainsequence converges, so will the subsequence do, but I can't see how to fix it with the suprema.
Thanks in advance.
real-analysis limits supremum-and-infimum
asked Aug 6 at 12:54
Belgium_Physics
1367
Consider a bounded sequence $x_n$ of real numbers.
I want to proof that the $limsup$ of a subsequence $x_kn$ with $k in mathbbR$ is smaller or equal to the $limsup$ of the mainsequence $x_n$ or in symbols:
$limsup_ntoinfty x_kn leq limsup_ntoinfty x_n$.
I can see why this is correct, but don't know how to prove it. I think it is because $x_kn$ is a subsequence so if the mainsequence converges, so will the subsequence do, but I can't see how to fix it with the suprema.
Thanks in advance.
real-analysis limits supremum-and-infimum
asked Aug 6 at 12:54
Belgium_Physics
1367
asked Aug 6 at 12:54
Belgium_Physics
1367
asked Aug 6 at 12:54
Belgium_Physics
1367
asked Aug 6 at 12:54
Belgium_Physics
1367
1367
This is not a question of convergence. Limsup exists independent of convergence of $x_n$. What is the definition of limsup? What would happen if you put the definitions there, I think looking it from that angle could help you.
– Kolja
Aug 6 at 12:58
Every term in the subsequence shows up as a term in the sequence. It doesn't matter what your definition of limsup is, you just need to unpack the definition and apply this observation.
– Robert Wolfe
Aug 6 at 13:08
add a comment |Â
This is not a question of convergence. Limsup exists independent of convergence of $x_n$. What is the definition of limsup? What would happen if you put the definitions there, I think looking it from that angle could help you.
– Kolja
Aug 6 at 12:58
Every term in the subsequence shows up as a term in the sequence. It doesn't matter what your definition of limsup is, you just need to unpack the definition and apply this observation.
– Robert Wolfe
Aug 6 at 13:08
This is not a question of convergence. Limsup exists independent of convergence of $x_n$. What is the definition of limsup? What would happen if you put the definitions there, I think looking it from that angle could help you.
– Kolja
Aug 6 at 12:58
This is not a question of convergence. Limsup exists independent of convergence of $x_n$. What is the definition of limsup? What would happen if you put the definitions there, I think looking it from that angle could help you.
– Kolja
Aug 6 at 12:58
Every term in the subsequence shows up as a term in the sequence. It doesn't matter what your definition of limsup is, you just need to unpack the definition and apply this observation.
– Robert Wolfe
Aug 6 at 13:08
Every term in the subsequence shows up as a term in the sequence. It doesn't matter what your definition of limsup is, you just need to unpack the definition and apply this observation.
– Robert Wolfe
Aug 6 at 13:08
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
If this weren't the case, i.e.
beginalign*
limsup_ktoinfty x_n_k > limsup_ntoinfty x_n doteq c,
endalign*
then for any $N$, we can find an $m(N) > N$ such that $x_n_m(N) > c+varepsilon$. Since each $x_n_m(N)$ is a member of the original sequence, we have constructed a further subsequence $x_n_m(N)_N=1^infty$ whose limit is strictly greater than $c$ (which by definition is the supremum of all subsequential limits).
answered Aug 6 at 13:01
Daniel Xiang
1,823414
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If this weren't the case, i.e.
beginalign*
limsup_ktoinfty x_n_k > limsup_ntoinfty x_n doteq c,
endalign*
then for any $N$, we can find an $m(N) > N$ such that $x_n_m(N) > c+varepsilon$. Since each $x_n_m(N)$ is a member of the original sequence, we have constructed a further subsequence $x_n_m(N)_N=1^infty$ whose limit is strictly greater than $c$ (which by definition is the supremum of all subsequential limits).
answered Aug 6 at 13:01
Daniel Xiang
1,823414
add a comment |Â
up vote
2
down vote
If this weren't the case, i.e.
beginalign*
limsup_ktoinfty x_n_k > limsup_ntoinfty x_n doteq c,
endalign*
then for any $N$, we can find an $m(N) > N$ such that $x_n_m(N) > c+varepsilon$. Since each $x_n_m(N)$ is a member of the original sequence, we have constructed a further subsequence $x_n_m(N)_N=1^infty$ whose limit is strictly greater than $c$ (which by definition is the supremum of all subsequential limits).
answered Aug 6 at 13:01
Daniel Xiang
1,823414
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If this weren't the case, i.e.
beginalign*
limsup_ktoinfty x_n_k > limsup_ntoinfty x_n doteq c,
endalign*
then for any $N$, we can find an $m(N) > N$ such that $x_n_m(N) > c+varepsilon$. Since each $x_n_m(N)$ is a member of the original sequence, we have constructed a further subsequence $x_n_m(N)_N=1^infty$ whose limit is strictly greater than $c$ (which by definition is the supremum of all subsequential limits).
answered Aug 6 at 13:01
Daniel Xiang
1,823414
If this weren't the case, i.e.
beginalign*
limsup_ktoinfty x_n_k > limsup_ntoinfty x_n doteq c,
endalign*
then for any $N$, we can find an $m(N) > N$ such that $x_n_m(N) > c+varepsilon$. Since each $x_n_m(N)$ is a member of the original sequence, we have constructed a further subsequence $x_n_m(N)_N=1^infty$ whose limit is strictly greater than $c$ (which by definition is the supremum of all subsequential limits).
answered Aug 6 at 13:01
Daniel Xiang
1,823414
answered Aug 6 at 13:01
Daniel Xiang
1,823414
answered Aug 6 at 13:01
Daniel Xiang
1,823414
answered Aug 6 at 13:01
Daniel Xiang
1,823414
1,823414
add a comment |Â
add a comment |Â
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This is not a question of convergence. Limsup exists independent of convergence of $x_n$. What is the definition of limsup? What would happen if you put the definitions there, I think looking it from that angle could help you.
– Kolja
Aug 6 at 12:58
Every term in the subsequence shows up as a term in the sequence. It doesn't matter what your definition of limsup is, you just need to unpack the definition and apply this observation.
– Robert Wolfe
Aug 6 at 13:08