Uniform limit points of a sequence of oscillating functions
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The sequence $$f_n(x):=sin(x+n), qquad xin [0, 2pi],$$ is relatively compact in $C([0, 2pi])$ by the theorem of Ascoli-Arzelà. This means that there exist sequences $n_kinmathbb N$ and functions $gin C([0, 2pi])$ such that $|f_{n_k}-g|_inftyto 0$ as $kto infty$.
Can you give some concrete examples of such functions $g$? Is $g=0$ a limit point of $f_n$?
This example could shed some light on the mechanism of proof of the theorem of Ascoli-Arzelà. (Notice that showing that $0$ is a limit point of the numerical sequence $sin n$ is already nontrivial).
functional-analysis compactness arzela-ascoli
asked Dec 25 '18 at 15:18
Giuseppe NegroGiuseppe Negro
17.1k330124
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$begingroup$
The sequence $$f_n(x):=sin(x+n), qquad xin [0, 2pi],$$ is relatively compact in $C([0, 2pi])$ by the theorem of Ascoli-Arzelà. This means that there exist sequences $n_kinmathbb N$ and functions $gin C([0, 2pi])$ such that $|f_{n_k}-g|_inftyto 0$ as $kto infty$.
Can you give some concrete examples of such functions $g$? Is $g=0$ a limit point of $f_n$?
This example could shed some light on the mechanism of proof of the theorem of Ascoli-Arzelà. (Notice that showing that $0$ is a limit point of the numerical sequence $sin n$ is already nontrivial).
functional-analysis compactness arzela-ascoli
asked Dec 25 '18 at 15:18
Giuseppe NegroGiuseppe Negro
17.1k330124
$endgroup$
add a comment |
$begingroup$
The sequence $$f_n(x):=sin(x+n), qquad xin [0, 2pi],$$ is relatively compact in $C([0, 2pi])$ by the theorem of Ascoli-Arzelà. This means that there exist sequences $n_kinmathbb N$ and functions $gin C([0, 2pi])$ such that $|f_{n_k}-g|_inftyto 0$ as $kto infty$.
Can you give some concrete examples of such functions $g$? Is $g=0$ a limit point of $f_n$?
This example could shed some light on the mechanism of proof of the theorem of Ascoli-Arzelà. (Notice that showing that $0$ is a limit point of the numerical sequence $sin n$ is already nontrivial).
functional-analysis compactness arzela-ascoli
asked Dec 25 '18 at 15:18
Giuseppe NegroGiuseppe Negro
17.1k330124
$endgroup$
The sequence $$f_n(x):=sin(x+n), qquad xin [0, 2pi],$$ is relatively compact in $C([0, 2pi])$ by the theorem of Ascoli-Arzelà. This means that there exist sequences $n_kinmathbb N$ and functions $gin C([0, 2pi])$ such that $|f_{n_k}-g|_inftyto 0$ as $kto infty$.
Can you give some concrete examples of such functions $g$? Is $g=0$ a limit point of $f_n$?
This example could shed some light on the mechanism of proof of the theorem of Ascoli-Arzelà. (Notice that showing that $0$ is a limit point of the numerical sequence $sin n$ is already nontrivial).
functional-analysis compactness arzela-ascoli
functional-analysis compactness arzela-ascoli
asked Dec 25 '18 at 15:18
Giuseppe NegroGiuseppe Negro
17.1k330124
asked Dec 25 '18 at 15:18
Giuseppe NegroGiuseppe Negro
17.1k330124
asked Dec 25 '18 at 15:18
Giuseppe NegroGiuseppe Negro
17.1k330124
asked Dec 25 '18 at 15:18
Giuseppe NegroGiuseppe Negro
17.1k330124
asked Dec 25 '18 at 15:18
Giuseppe NegroGiuseppe Negro
17.1k330124
17.1k330124
add a comment |
add a comment |
1 Answer
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We certainly know that it cannot be the case that $gequiv0$; the quantity $||f_{n_k}-g||_{infty}=1$ in that case. I suspect that $g(x)=sin(x)$ is a concrete example of the functions you are looking for, mostly because we know that $2kpi$ is equidistributed modulo $1$; there exist $k$ such that $2kpi$ is arbitrarily close to an integer $n_k$, and so $f_{n_k}$ will be arbitrarily close to $g$.
In fact, using the same kind of argument, you can leverage the fact that the sequence $2kpi+alpha$ is also equidistributed modulo $1$ to conclude that $g_alpha(x)=sin(x+alpha)$ is an example for any real $alpha$.
answered Dec 25 '18 at 15:25
ImNotTheGuyImNotTheGuy
38516
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1 Answer
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$begingroup$
We certainly know that it cannot be the case that $gequiv0$; the quantity $||f_{n_k}-g||_{infty}=1$ in that case. I suspect that $g(x)=sin(x)$ is a concrete example of the functions you are looking for, mostly because we know that $2kpi$ is equidistributed modulo $1$; there exist $k$ such that $2kpi$ is arbitrarily close to an integer $n_k$, and so $f_{n_k}$ will be arbitrarily close to $g$.
In fact, using the same kind of argument, you can leverage the fact that the sequence $2kpi+alpha$ is also equidistributed modulo $1$ to conclude that $g_alpha(x)=sin(x+alpha)$ is an example for any real $alpha$.
answered Dec 25 '18 at 15:25
ImNotTheGuyImNotTheGuy
38516
$endgroup$
add a comment |
$begingroup$
We certainly know that it cannot be the case that $gequiv0$; the quantity $||f_{n_k}-g||_{infty}=1$ in that case. I suspect that $g(x)=sin(x)$ is a concrete example of the functions you are looking for, mostly because we know that $2kpi$ is equidistributed modulo $1$; there exist $k$ such that $2kpi$ is arbitrarily close to an integer $n_k$, and so $f_{n_k}$ will be arbitrarily close to $g$.
In fact, using the same kind of argument, you can leverage the fact that the sequence $2kpi+alpha$ is also equidistributed modulo $1$ to conclude that $g_alpha(x)=sin(x+alpha)$ is an example for any real $alpha$.
answered Dec 25 '18 at 15:25
ImNotTheGuyImNotTheGuy
38516
$endgroup$
add a comment |
$begingroup$
We certainly know that it cannot be the case that $gequiv0$; the quantity $||f_{n_k}-g||_{infty}=1$ in that case. I suspect that $g(x)=sin(x)$ is a concrete example of the functions you are looking for, mostly because we know that $2kpi$ is equidistributed modulo $1$; there exist $k$ such that $2kpi$ is arbitrarily close to an integer $n_k$, and so $f_{n_k}$ will be arbitrarily close to $g$.
In fact, using the same kind of argument, you can leverage the fact that the sequence $2kpi+alpha$ is also equidistributed modulo $1$ to conclude that $g_alpha(x)=sin(x+alpha)$ is an example for any real $alpha$.
answered Dec 25 '18 at 15:25
ImNotTheGuyImNotTheGuy
38516
$endgroup$
We certainly know that it cannot be the case that $gequiv0$; the quantity $||f_{n_k}-g||_{infty}=1$ in that case. I suspect that $g(x)=sin(x)$ is a concrete example of the functions you are looking for, mostly because we know that $2kpi$ is equidistributed modulo $1$; there exist $k$ such that $2kpi$ is arbitrarily close to an integer $n_k$, and so $f_{n_k}$ will be arbitrarily close to $g$.
In fact, using the same kind of argument, you can leverage the fact that the sequence $2kpi+alpha$ is also equidistributed modulo $1$ to conclude that $g_alpha(x)=sin(x+alpha)$ is an example for any real $alpha$.
answered Dec 25 '18 at 15:25
ImNotTheGuyImNotTheGuy
38516
edited Dec 25 '18 at 15:33
answered Dec 25 '18 at 15:25
ImNotTheGuyImNotTheGuy
38516
answered Dec 25 '18 at 15:25
ImNotTheGuyImNotTheGuy
38516
answered Dec 25 '18 at 15:25
ImNotTheGuyImNotTheGuy
38516
38516
add a comment |
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