A sequence of bounded $C^1$ functions whose derivatives are unbounded.
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What is an example of a sequence of functions, $(f_n)_{n=1}^inftysubset C^1([a,b])$, which are bounded in $C^1([a,b])$ under $|cdot|_{infty}$ but are such that their first derivatives $|f'_n|_{infty}toinfty$ as $ntoinfty$?
There are many examples on this site for a single function, but I am looking for an intance when we have a sequence of functions. Are there any particularly simple instances which can be provided to demonstrate the above point?
calculus functional-analysis functions banach-spaces normed-spaces
asked Jan 21 at 12:55
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
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add a comment |
$begingroup$
What is an example of a sequence of functions, $(f_n)_{n=1}^inftysubset C^1([a,b])$, which are bounded in $C^1([a,b])$ under $|cdot|_{infty}$ but are such that their first derivatives $|f'_n|_{infty}toinfty$ as $ntoinfty$?
There are many examples on this site for a single function, but I am looking for an intance when we have a sequence of functions. Are there any particularly simple instances which can be provided to demonstrate the above point?
calculus functional-analysis functions banach-spaces normed-spaces
asked Jan 21 at 12:55
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
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5
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$f_n(x)=sin(nx)$?
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– Mindlack
Jan 21 at 12:56
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Indeed particularly simple! Cheers!
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 12:59
add a comment |
$begingroup$
What is an example of a sequence of functions, $(f_n)_{n=1}^inftysubset C^1([a,b])$, which are bounded in $C^1([a,b])$ under $|cdot|_{infty}$ but are such that their first derivatives $|f'_n|_{infty}toinfty$ as $ntoinfty$?
There are many examples on this site for a single function, but I am looking for an intance when we have a sequence of functions. Are there any particularly simple instances which can be provided to demonstrate the above point?
calculus functional-analysis functions banach-spaces normed-spaces
asked Jan 21 at 12:55
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
$endgroup$
What is an example of a sequence of functions, $(f_n)_{n=1}^inftysubset C^1([a,b])$, which are bounded in $C^1([a,b])$ under $|cdot|_{infty}$ but are such that their first derivatives $|f'_n|_{infty}toinfty$ as $ntoinfty$?
There are many examples on this site for a single function, but I am looking for an intance when we have a sequence of functions. Are there any particularly simple instances which can be provided to demonstrate the above point?
calculus functional-analysis functions banach-spaces normed-spaces
calculus functional-analysis functions banach-spaces normed-spaces
asked Jan 21 at 12:55
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
asked Jan 21 at 12:55
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
asked Jan 21 at 12:55
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
asked Jan 21 at 12:55
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
asked Jan 21 at 12:55
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
992715
5
$begingroup$
$f_n(x)=sin(nx)$?
$endgroup$
– Mindlack
Jan 21 at 12:56
$begingroup$
Indeed particularly simple! Cheers!
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 12:59
add a comment |
5
$begingroup$
$f_n(x)=sin(nx)$?
$endgroup$
– Mindlack
Jan 21 at 12:56
$begingroup$
Indeed particularly simple! Cheers!
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 12:59
5
5
$begingroup$
$f_n(x)=sin(nx)$?
$endgroup$
– Mindlack
Jan 21 at 12:56
$begingroup$
$f_n(x)=sin(nx)$?
$endgroup$
– Mindlack
Jan 21 at 12:56
$begingroup$
Indeed particularly simple! Cheers!
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 12:59
$begingroup$
Indeed particularly simple! Cheers!
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 12:59
add a comment |
1 Answer
1
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oldest
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Take $f_n(x)=sin(nx)$ as per the comment of Mindlack.
The sequence $(sin(nx))_{n=1}^inftysubset C^1([a,b])$ is bounded since for all $ninmathbb N$ one has that $|sin(nx)|_{infty}=1$.
Additionally, for each $ninmathbb N$ we have that $f_n'(x)=ncos(nx)$. Then,
$$|f_n'|_infty=|ncos(nx)|_infty=|n||cos(nx)|_infty= n,$$
which implies that $lim_{ntoinfty}|f_n'|_infty=infty$. Thus, the derivatives of $(sin(nx))_{n=1}^inftysubset C^1([a,b])$ are unbounded.
answered Jan 22 at 10:53
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
$endgroup$
$begingroup$
Could you at least type up a little bit more? Otherwise you leave people no choice but to vote-to-delete this "answer".
$endgroup$
– Lee David Chung Lin
Jan 22 at 12:31
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Take $f_n(x)=sin(nx)$ as per the comment of Mindlack.
The sequence $(sin(nx))_{n=1}^inftysubset C^1([a,b])$ is bounded since for all $ninmathbb N$ one has that $|sin(nx)|_{infty}=1$.
Additionally, for each $ninmathbb N$ we have that $f_n'(x)=ncos(nx)$. Then,
$$|f_n'|_infty=|ncos(nx)|_infty=|n||cos(nx)|_infty= n,$$
which implies that $lim_{ntoinfty}|f_n'|_infty=infty$. Thus, the derivatives of $(sin(nx))_{n=1}^inftysubset C^1([a,b])$ are unbounded.
answered Jan 22 at 10:53
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
$endgroup$
$begingroup$
Could you at least type up a little bit more? Otherwise you leave people no choice but to vote-to-delete this "answer".
$endgroup$
– Lee David Chung Lin
Jan 22 at 12:31
add a comment |
$begingroup$
Take $f_n(x)=sin(nx)$ as per the comment of Mindlack.
The sequence $(sin(nx))_{n=1}^inftysubset C^1([a,b])$ is bounded since for all $ninmathbb N$ one has that $|sin(nx)|_{infty}=1$.
Additionally, for each $ninmathbb N$ we have that $f_n'(x)=ncos(nx)$. Then,
$$|f_n'|_infty=|ncos(nx)|_infty=|n||cos(nx)|_infty= n,$$
which implies that $lim_{ntoinfty}|f_n'|_infty=infty$. Thus, the derivatives of $(sin(nx))_{n=1}^inftysubset C^1([a,b])$ are unbounded.
answered Jan 22 at 10:53
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
$endgroup$
$begingroup$
Could you at least type up a little bit more? Otherwise you leave people no choice but to vote-to-delete this "answer".
$endgroup$
– Lee David Chung Lin
Jan 22 at 12:31
add a comment |
$begingroup$
Take $f_n(x)=sin(nx)$ as per the comment of Mindlack.
The sequence $(sin(nx))_{n=1}^inftysubset C^1([a,b])$ is bounded since for all $ninmathbb N$ one has that $|sin(nx)|_{infty}=1$.
Additionally, for each $ninmathbb N$ we have that $f_n'(x)=ncos(nx)$. Then,
$$|f_n'|_infty=|ncos(nx)|_infty=|n||cos(nx)|_infty= n,$$
which implies that $lim_{ntoinfty}|f_n'|_infty=infty$. Thus, the derivatives of $(sin(nx))_{n=1}^inftysubset C^1([a,b])$ are unbounded.
answered Jan 22 at 10:53
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
$endgroup$
Take $f_n(x)=sin(nx)$ as per the comment of Mindlack.
The sequence $(sin(nx))_{n=1}^inftysubset C^1([a,b])$ is bounded since for all $ninmathbb N$ one has that $|sin(nx)|_{infty}=1$.
Additionally, for each $ninmathbb N$ we have that $f_n'(x)=ncos(nx)$. Then,
$$|f_n'|_infty=|ncos(nx)|_infty=|n||cos(nx)|_infty= n,$$
which implies that $lim_{ntoinfty}|f_n'|_infty=infty$. Thus, the derivatives of $(sin(nx))_{n=1}^inftysubset C^1([a,b])$ are unbounded.
answered Jan 22 at 10:53
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
edited Jan 22 at 13:12
answered Jan 22 at 10:53
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
answered Jan 22 at 10:53
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
answered Jan 22 at 10:53
Jeremy Jeffrey JamesJeremy Jeffrey James
992715
992715
$begingroup$
Could you at least type up a little bit more? Otherwise you leave people no choice but to vote-to-delete this "answer".
$endgroup$
– Lee David Chung Lin
Jan 22 at 12:31
add a comment |
$begingroup$
Could you at least type up a little bit more? Otherwise you leave people no choice but to vote-to-delete this "answer".
$endgroup$
– Lee David Chung Lin
Jan 22 at 12:31
$begingroup$
Could you at least type up a little bit more? Otherwise you leave people no choice but to vote-to-delete this "answer".
$endgroup$
– Lee David Chung Lin
Jan 22 at 12:31
$begingroup$
Could you at least type up a little bit more? Otherwise you leave people no choice but to vote-to-delete this "answer".
$endgroup$
– Lee David Chung Lin
Jan 22 at 12:31
add a comment |
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5
$begingroup$
$f_n(x)=sin(nx)$?
$endgroup$
– Mindlack
Jan 21 at 12:56
$begingroup$
Indeed particularly simple! Cheers!
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 12:59