Sequence of smooth equicontinuous functions with unbounded derivatives
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I found a very elegant construction in this link: https://math.stackexchange.com/a/311289/444015, I have no questions about this example. However, at least for me, it doesnt look like a natural construction.
I tried to find more direct examples, but I didnt succeed. Every smooth equicontinuous sequence that I found is bounded.
I was wondering if there is any more obvious example than this, or if indeed to find any example, it is necessary to construct the sequence of functions in a similar way.
real-analysis equicontinuity
asked Jan 11 at 14:39
Lucas CorrêaLucas Corrêa
1,6181321
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add a comment |
$begingroup$
I found a very elegant construction in this link: https://math.stackexchange.com/a/311289/444015, I have no questions about this example. However, at least for me, it doesnt look like a natural construction.
I tried to find more direct examples, but I didnt succeed. Every smooth equicontinuous sequence that I found is bounded.
I was wondering if there is any more obvious example than this, or if indeed to find any example, it is necessary to construct the sequence of functions in a similar way.
real-analysis equicontinuity
asked Jan 11 at 14:39
Lucas CorrêaLucas Corrêa
1,6181321
$endgroup$
add a comment |
$begingroup$
I found a very elegant construction in this link: https://math.stackexchange.com/a/311289/444015, I have no questions about this example. However, at least for me, it doesnt look like a natural construction.
I tried to find more direct examples, but I didnt succeed. Every smooth equicontinuous sequence that I found is bounded.
I was wondering if there is any more obvious example than this, or if indeed to find any example, it is necessary to construct the sequence of functions in a similar way.
real-analysis equicontinuity
asked Jan 11 at 14:39
Lucas CorrêaLucas Corrêa
1,6181321
$endgroup$
I found a very elegant construction in this link: https://math.stackexchange.com/a/311289/444015, I have no questions about this example. However, at least for me, it doesnt look like a natural construction.
I tried to find more direct examples, but I didnt succeed. Every smooth equicontinuous sequence that I found is bounded.
I was wondering if there is any more obvious example than this, or if indeed to find any example, it is necessary to construct the sequence of functions in a similar way.
real-analysis equicontinuity
real-analysis equicontinuity
asked Jan 11 at 14:39
Lucas CorrêaLucas Corrêa
1,6181321
asked Jan 11 at 14:39
Lucas CorrêaLucas Corrêa
1,6181321
asked Jan 11 at 14:39
Lucas CorrêaLucas Corrêa
1,6181321
asked Jan 11 at 14:39
Lucas CorrêaLucas Corrêa
1,6181321
asked Jan 11 at 14:39
Lucas CorrêaLucas Corrêa
1,6181321
1,6181321
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $f_n(x)=frac 1nsin(x^2)$. Then $f_nto0$ uniformly, so $(f_n)$ is equicontinuous.
(That gives $sup_x|f_n'(x)|=infty$. "With unbounded derivatives" is somewhat ambiguous; if you want $sup_n|f_n'(0)|=infty$ use $frac 1nsin(n^2x)$.)
answered Jan 11 at 14:44
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
$begingroup$
I got it! I believe that the question asks for $sup_{x}$. Thank you!
$endgroup$
– Lucas Corrêa
Jan 11 at 14:55
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
Let $f_n(x)=frac 1nsin(x^2)$. Then $f_nto0$ uniformly, so $(f_n)$ is equicontinuous.
(That gives $sup_x|f_n'(x)|=infty$. "With unbounded derivatives" is somewhat ambiguous; if you want $sup_n|f_n'(0)|=infty$ use $frac 1nsin(n^2x)$.)
answered Jan 11 at 14:44
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
$begingroup$
I got it! I believe that the question asks for $sup_{x}$. Thank you!
$endgroup$
– Lucas Corrêa
Jan 11 at 14:55
add a comment |
$begingroup$
Let $f_n(x)=frac 1nsin(x^2)$. Then $f_nto0$ uniformly, so $(f_n)$ is equicontinuous.
(That gives $sup_x|f_n'(x)|=infty$. "With unbounded derivatives" is somewhat ambiguous; if you want $sup_n|f_n'(0)|=infty$ use $frac 1nsin(n^2x)$.)
answered Jan 11 at 14:44
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
$begingroup$
I got it! I believe that the question asks for $sup_{x}$. Thank you!
$endgroup$
– Lucas Corrêa
Jan 11 at 14:55
add a comment |
$begingroup$
Let $f_n(x)=frac 1nsin(x^2)$. Then $f_nto0$ uniformly, so $(f_n)$ is equicontinuous.
(That gives $sup_x|f_n'(x)|=infty$. "With unbounded derivatives" is somewhat ambiguous; if you want $sup_n|f_n'(0)|=infty$ use $frac 1nsin(n^2x)$.)
answered Jan 11 at 14:44
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
Let $f_n(x)=frac 1nsin(x^2)$. Then $f_nto0$ uniformly, so $(f_n)$ is equicontinuous.
(That gives $sup_x|f_n'(x)|=infty$. "With unbounded derivatives" is somewhat ambiguous; if you want $sup_n|f_n'(0)|=infty$ use $frac 1nsin(n^2x)$.)
answered Jan 11 at 14:44
David C. UllrichDavid C. Ullrich
60k43994
answered Jan 11 at 14:44
David C. UllrichDavid C. Ullrich
60k43994
answered Jan 11 at 14:44
David C. UllrichDavid C. Ullrich
60k43994
answered Jan 11 at 14:44
David C. UllrichDavid C. Ullrich
60k43994
60k43994
$begingroup$
I got it! I believe that the question asks for $sup_{x}$. Thank you!
$endgroup$
– Lucas Corrêa
Jan 11 at 14:55
add a comment |
$begingroup$
I got it! I believe that the question asks for $sup_{x}$. Thank you!
$endgroup$
– Lucas Corrêa
Jan 11 at 14:55
$begingroup$
I got it! I believe that the question asks for $sup_{x}$. Thank you!
$endgroup$
– Lucas Corrêa
Jan 11 at 14:55
$begingroup$
I got it! I believe that the question asks for $sup_{x}$. Thank you!
$endgroup$
– Lucas Corrêa
Jan 11 at 14:55
add a comment |
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