Mixing
$((x/c)^n)$ has no convergent subsequence in C[a,b]
$begingroup$
Consider the metric space C[a,b] equipped with the metric $d(f,g)=max|f(x)-g(x)|$. I would like to show that the sequence $((x/c)^n)$ has no convergent subsequence, where c is a constant such that $x/c in [0,1] $.
My attempt:
Assume it does, then there is a continuous function $f$ such that a subsequence $((x/c)^{n_k})$ converges to it. Then $((x/c)^{n_k})$ cauchy. Thus for all $epsilon gt 0$ there exists K such that for any $k_n, k_m gt K$, $d(f_{k_n},f_{k_m}) lt epsilon$. Then $max_{x in[a,b]}|x^{k_n}/c^{k_n}-x^{k_m}/c^{k_m}|lt epsilon$. I would like to derive a contradiction from this, perhaps using the fact that $(x/c)^n$ is not uniformly continuous in the reals. However, I am uncertain of how to proceed.
Any help is appreciated
sequences-and-series continuity metric-spaces uniform-continuity
asked Feb 3 at 5:05
mmmmommmmo
1347
$endgroup$
add a comment |
$begingroup$
Consider the metric space C[a,b] equipped with the metric $d(f,g)=max|f(x)-g(x)|$. I would like to show that the sequence $((x/c)^n)$ has no convergent subsequence, where c is a constant such that $x/c in [0,1] $.
My attempt:
Assume it does, then there is a continuous function $f$ such that a subsequence $((x/c)^{n_k})$ converges to it. Then $((x/c)^{n_k})$ cauchy. Thus for all $epsilon gt 0$ there exists K such that for any $k_n, k_m gt K$, $d(f_{k_n},f_{k_m}) lt epsilon$. Then $max_{x in[a,b]}|x^{k_n}/c^{k_n}-x^{k_m}/c^{k_m}|lt epsilon$. I would like to derive a contradiction from this, perhaps using the fact that $(x/c)^n$ is not uniformly continuous in the reals. However, I am uncertain of how to proceed.
Any help is appreciated
sequences-and-series continuity metric-spaces uniform-continuity
asked Feb 3 at 5:05
mmmmommmmo
1347
$endgroup$
$begingroup$
If yuo are considering continous functions on $[a,b]$ why are talking about uniform continuity on the whole line. Perhaps, you have not stated the question correctly.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 12:36
add a comment |
$begingroup$
Consider the metric space C[a,b] equipped with the metric $d(f,g)=max|f(x)-g(x)|$. I would like to show that the sequence $((x/c)^n)$ has no convergent subsequence, where c is a constant such that $x/c in [0,1] $.
My attempt:
Assume it does, then there is a continuous function $f$ such that a subsequence $((x/c)^{n_k})$ converges to it. Then $((x/c)^{n_k})$ cauchy. Thus for all $epsilon gt 0$ there exists K such that for any $k_n, k_m gt K$, $d(f_{k_n},f_{k_m}) lt epsilon$. Then $max_{x in[a,b]}|x^{k_n}/c^{k_n}-x^{k_m}/c^{k_m}|lt epsilon$. I would like to derive a contradiction from this, perhaps using the fact that $(x/c)^n$ is not uniformly continuous in the reals. However, I am uncertain of how to proceed.
Any help is appreciated
sequences-and-series continuity metric-spaces uniform-continuity
asked Feb 3 at 5:05
mmmmommmmo
1347
$endgroup$
Consider the metric space C[a,b] equipped with the metric $d(f,g)=max|f(x)-g(x)|$. I would like to show that the sequence $((x/c)^n)$ has no convergent subsequence, where c is a constant such that $x/c in [0,1] $.
My attempt:
Assume it does, then there is a continuous function $f$ such that a subsequence $((x/c)^{n_k})$ converges to it. Then $((x/c)^{n_k})$ cauchy. Thus for all $epsilon gt 0$ there exists K such that for any $k_n, k_m gt K$, $d(f_{k_n},f_{k_m}) lt epsilon$. Then $max_{x in[a,b]}|x^{k_n}/c^{k_n}-x^{k_m}/c^{k_m}|lt epsilon$. I would like to derive a contradiction from this, perhaps using the fact that $(x/c)^n$ is not uniformly continuous in the reals. However, I am uncertain of how to proceed.
Any help is appreciated
sequences-and-series continuity metric-spaces uniform-continuity
sequences-and-series continuity metric-spaces uniform-continuity
asked Feb 3 at 5:05
mmmmommmmo
1347
asked Feb 3 at 5:05
mmmmommmmo
1347
asked Feb 3 at 5:05
mmmmommmmo
1347
asked Feb 3 at 5:05
mmmmommmmo
1347
asked Feb 3 at 5:05
mmmmommmmo
1347
1347
$begingroup$
If yuo are considering continous functions on $[a,b]$ why are talking about uniform continuity on the whole line. Perhaps, you have not stated the question correctly.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 12:36
add a comment |
$begingroup$
If yuo are considering continous functions on $[a,b]$ why are talking about uniform continuity on the whole line. Perhaps, you have not stated the question correctly.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 12:36
$begingroup$
If yuo are considering continous functions on $[a,b]$ why are talking about uniform continuity on the whole line. Perhaps, you have not stated the question correctly.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 12:36
$begingroup$
If yuo are considering continous functions on $[a,b]$ why are talking about uniform continuity on the whole line. Perhaps, you have not stated the question correctly.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 12:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is false. Take $a=0,b=1$. If $c$ is large enough then $sup_x |frac x c| <frac 1 2$ so the entire sequence $(frac x c)^{n}$ converges uniformly to $0$.
If your question is to find a $c$ such that no subsequence of $(frac x c)^{n}$ converges uniformly you can do the following: suppose $0<a<b$. take $c=b$. Then $(frac x c)^{n}$ converges to $0$ at each point $x in [a,b)$ and to $1$ at $x=b$. Since the limitimg function is not continuous it follows that no subsequence of $(frac x c)^{n}$ converges uniformly. I will let you handle the other cases.
answered Feb 3 at 5:19
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is false. Take $a=0,b=1$. If $c$ is large enough then $sup_x |frac x c| <frac 1 2$ so the entire sequence $(frac x c)^{n}$ converges uniformly to $0$.
If your question is to find a $c$ such that no subsequence of $(frac x c)^{n}$ converges uniformly you can do the following: suppose $0<a<b$. take $c=b$. Then $(frac x c)^{n}$ converges to $0$ at each point $x in [a,b)$ and to $1$ at $x=b$. Since the limitimg function is not continuous it follows that no subsequence of $(frac x c)^{n}$ converges uniformly. I will let you handle the other cases.
answered Feb 3 at 5:19
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
add a comment |
$begingroup$
This is false. Take $a=0,b=1$. If $c$ is large enough then $sup_x |frac x c| <frac 1 2$ so the entire sequence $(frac x c)^{n}$ converges uniformly to $0$.
If your question is to find a $c$ such that no subsequence of $(frac x c)^{n}$ converges uniformly you can do the following: suppose $0<a<b$. take $c=b$. Then $(frac x c)^{n}$ converges to $0$ at each point $x in [a,b)$ and to $1$ at $x=b$. Since the limitimg function is not continuous it follows that no subsequence of $(frac x c)^{n}$ converges uniformly. I will let you handle the other cases.
answered Feb 3 at 5:19
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
add a comment |
$begingroup$
This is false. Take $a=0,b=1$. If $c$ is large enough then $sup_x |frac x c| <frac 1 2$ so the entire sequence $(frac x c)^{n}$ converges uniformly to $0$.
If your question is to find a $c$ such that no subsequence of $(frac x c)^{n}$ converges uniformly you can do the following: suppose $0<a<b$. take $c=b$. Then $(frac x c)^{n}$ converges to $0$ at each point $x in [a,b)$ and to $1$ at $x=b$. Since the limitimg function is not continuous it follows that no subsequence of $(frac x c)^{n}$ converges uniformly. I will let you handle the other cases.
answered Feb 3 at 5:19
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
This is false. Take $a=0,b=1$. If $c$ is large enough then $sup_x |frac x c| <frac 1 2$ so the entire sequence $(frac x c)^{n}$ converges uniformly to $0$.
If your question is to find a $c$ such that no subsequence of $(frac x c)^{n}$ converges uniformly you can do the following: suppose $0<a<b$. take $c=b$. Then $(frac x c)^{n}$ converges to $0$ at each point $x in [a,b)$ and to $1$ at $x=b$. Since the limitimg function is not continuous it follows that no subsequence of $(frac x c)^{n}$ converges uniformly. I will let you handle the other cases.
answered Feb 3 at 5:19
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
edited Feb 3 at 12:41
answered Feb 3 at 5:19
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Feb 3 at 5:19
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Feb 3 at 5:19
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
74.9k53270
add a comment |
add a comment |
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$begingroup$
If yuo are considering continous functions on $[a,b]$ why are talking about uniform continuity on the whole line. Perhaps, you have not stated the question correctly.
$endgroup$
– Kavi Rama Murthy
Feb 3 at 12:36