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Existence of an antiderivative for a continuous function on an arbitrary subset of $mathbbR$ [duplicate]
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Existence of antiderivative on a part $I$ of $mathbbR$ [closed]
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Let $f$ be a continuous function on a set $Isubset mathbbR$. Does there always exist a function $F$ differentiable on an open set $J$ containing $I$ such that $F'=f$ on $I$ ?
The case where J is an interval has been studied in this thread Existence of antiderivative on a part $I$ of $mathbbR$
real-analysis integration
asked Jul 16 at 16:31
Tina
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marked as duplicate by user21820, Delta-u, Taroccoesbrocco, amWhy integration
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Jul 25 at 11:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
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This question already has an answer here:
Existence of antiderivative on a part $I$ of $mathbbR$ [closed]
1 answer
Let $f$ be a continuous function on a set $Isubset mathbbR$. Does there always exist a function $F$ differentiable on an open set $J$ containing $I$ such that $F'=f$ on $I$ ?
The case where J is an interval has been studied in this thread Existence of antiderivative on a part $I$ of $mathbbR$
real-analysis integration
asked Jul 16 at 16:31
Tina
355112
marked as duplicate by user21820, Delta-u, Taroccoesbrocco, amWhy integration
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Jul 25 at 11:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Almost a direct repost of this question last hour.
– T. Bongers
Jul 16 at 16:33
but now J is an open set
– Tina
Jul 16 at 16:34
add a comment |Â
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This question already has an answer here:
Existence of antiderivative on a part $I$ of $mathbbR$ [closed]
1 answer
Let $f$ be a continuous function on a set $Isubset mathbbR$. Does there always exist a function $F$ differentiable on an open set $J$ containing $I$ such that $F'=f$ on $I$ ?
The case where J is an interval has been studied in this thread Existence of antiderivative on a part $I$ of $mathbbR$
real-analysis integration
asked Jul 16 at 16:31
Tina
355112
This question already has an answer here:
Existence of antiderivative on a part $I$ of $mathbbR$ [closed]
1 answer
Let $f$ be a continuous function on a set $Isubset mathbbR$. Does there always exist a function $F$ differentiable on an open set $J$ containing $I$ such that $F'=f$ on $I$ ?
The case where J is an interval has been studied in this thread Existence of antiderivative on a part $I$ of $mathbbR$
This question already has an answer here:
Existence of antiderivative on a part $I$ of $mathbbR$ [closed]
1 answer
real-analysis integration
asked Jul 16 at 16:31
Tina
355112
edited Jul 26 at 10:31
asked Jul 16 at 16:31
Tina
355112
asked Jul 16 at 16:31
Tina
355112
asked Jul 16 at 16:31
Tina
355112
355112
marked as duplicate by user21820, Delta-u, Taroccoesbrocco, amWhy integration
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Jul 25 at 11:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by user21820, Delta-u, Taroccoesbrocco, amWhy integration
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Jul 25 at 11:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Almost a direct repost of this question last hour.
– T. Bongers
Jul 16 at 16:33
but now J is an open set
– Tina
Jul 16 at 16:34
add a comment |Â
Almost a direct repost of this question last hour.
– T. Bongers
Jul 16 at 16:33
but now J is an open set
– Tina
Jul 16 at 16:34
Almost a direct repost of this question last hour.
– T. Bongers
Jul 16 at 16:33
Almost a direct repost of this question last hour.
– T. Bongers
Jul 16 at 16:33
but now J is an open set
– Tina
Jul 16 at 16:34
but now J is an open set
– Tina
Jul 16 at 16:34
add a comment |Â
1 Answer
1
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up vote
3
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accepted
Hint: There exists a strictly increasing function $f:mathbb R to mathbb R$ that is continuous at each irrational, and discontinuous at each rational. Let $I$ be the irrationals. So we have $f$ continuous on $I.$ Show that if there were such an $F$ on an open $J$ containing $I,$ then $F'$ would violate the Darboux property on each connected component of $J.$
Added later: Here is such a function: Set $g(x) = 0$ for $x<0,$ $g(x) = 1$ for $xge 0.$ Let $r_1,r_2,dots$ be the rationals. For $xin mathbb R ,$ define
$$f(x) = sum_n=1^inftyfracg(x-r_n)2^n.$$
answered Jul 16 at 17:37
zhw.
65.8k42870
the function that you think is the Thomae's function? fr.wikipedia.org/wiki/Fonction_de_Thomae
– Tina
Jul 16 at 20:11
No, not all. Thomae function is $0$ on the irrationals.
– zhw.
Jul 16 at 20:13
@Tina I added to my answer.
– zhw.
Jul 16 at 20:19
thank you so much
– Tina
Jul 16 at 21:52
@zhw. Ugh, you are right. Differentiable, not continuously differentiable. Sorry.
– Fimpellizieri
Jul 16 at 21:59
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Hint: There exists a strictly increasing function $f:mathbb R to mathbb R$ that is continuous at each irrational, and discontinuous at each rational. Let $I$ be the irrationals. So we have $f$ continuous on $I.$ Show that if there were such an $F$ on an open $J$ containing $I,$ then $F'$ would violate the Darboux property on each connected component of $J.$
Added later: Here is such a function: Set $g(x) = 0$ for $x<0,$ $g(x) = 1$ for $xge 0.$ Let $r_1,r_2,dots$ be the rationals. For $xin mathbb R ,$ define
$$f(x) = sum_n=1^inftyfracg(x-r_n)2^n.$$
answered Jul 16 at 17:37
zhw.
65.8k42870
the function that you think is the Thomae's function? fr.wikipedia.org/wiki/Fonction_de_Thomae
– Tina
Jul 16 at 20:11
No, not all. Thomae function is $0$ on the irrationals.
– zhw.
Jul 16 at 20:13
@Tina I added to my answer.
– zhw.
Jul 16 at 20:19
thank you so much
– Tina
Jul 16 at 21:52
@zhw. Ugh, you are right. Differentiable, not continuously differentiable. Sorry.
– Fimpellizieri
Jul 16 at 21:59
 |Â
show 4 more comments
up vote
3
down vote
accepted
Hint: There exists a strictly increasing function $f:mathbb R to mathbb R$ that is continuous at each irrational, and discontinuous at each rational. Let $I$ be the irrationals. So we have $f$ continuous on $I.$ Show that if there were such an $F$ on an open $J$ containing $I,$ then $F'$ would violate the Darboux property on each connected component of $J.$
Added later: Here is such a function: Set $g(x) = 0$ for $x<0,$ $g(x) = 1$ for $xge 0.$ Let $r_1,r_2,dots$ be the rationals. For $xin mathbb R ,$ define
$$f(x) = sum_n=1^inftyfracg(x-r_n)2^n.$$
answered Jul 16 at 17:37
zhw.
65.8k42870
the function that you think is the Thomae's function? fr.wikipedia.org/wiki/Fonction_de_Thomae
– Tina
Jul 16 at 20:11
No, not all. Thomae function is $0$ on the irrationals.
– zhw.
Jul 16 at 20:13
@Tina I added to my answer.
– zhw.
Jul 16 at 20:19
thank you so much
– Tina
Jul 16 at 21:52
@zhw. Ugh, you are right. Differentiable, not continuously differentiable. Sorry.
– Fimpellizieri
Jul 16 at 21:59
 |Â
show 4 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Hint: There exists a strictly increasing function $f:mathbb R to mathbb R$ that is continuous at each irrational, and discontinuous at each rational. Let $I$ be the irrationals. So we have $f$ continuous on $I.$ Show that if there were such an $F$ on an open $J$ containing $I,$ then $F'$ would violate the Darboux property on each connected component of $J.$
Added later: Here is such a function: Set $g(x) = 0$ for $x<0,$ $g(x) = 1$ for $xge 0.$ Let $r_1,r_2,dots$ be the rationals. For $xin mathbb R ,$ define
$$f(x) = sum_n=1^inftyfracg(x-r_n)2^n.$$
answered Jul 16 at 17:37
zhw.
65.8k42870
Hint: There exists a strictly increasing function $f:mathbb R to mathbb R$ that is continuous at each irrational, and discontinuous at each rational. Let $I$ be the irrationals. So we have $f$ continuous on $I.$ Show that if there were such an $F$ on an open $J$ containing $I,$ then $F'$ would violate the Darboux property on each connected component of $J.$
Added later: Here is such a function: Set $g(x) = 0$ for $x<0,$ $g(x) = 1$ for $xge 0.$ Let $r_1,r_2,dots$ be the rationals. For $xin mathbb R ,$ define
$$f(x) = sum_n=1^inftyfracg(x-r_n)2^n.$$
answered Jul 16 at 17:37
zhw.
65.8k42870
edited Jul 16 at 20:19
answered Jul 16 at 17:37
zhw.
65.8k42870
answered Jul 16 at 17:37
zhw.
65.8k42870
answered Jul 16 at 17:37
zhw.
65.8k42870
65.8k42870
the function that you think is the Thomae's function? fr.wikipedia.org/wiki/Fonction_de_Thomae
– Tina
Jul 16 at 20:11
No, not all. Thomae function is $0$ on the irrationals.
– zhw.
Jul 16 at 20:13
@Tina I added to my answer.
– zhw.
Jul 16 at 20:19
thank you so much
– Tina
Jul 16 at 21:52
@zhw. Ugh, you are right. Differentiable, not continuously differentiable. Sorry.
– Fimpellizieri
Jul 16 at 21:59
 |Â
show 4 more comments
the function that you think is the Thomae's function? fr.wikipedia.org/wiki/Fonction_de_Thomae
– Tina
Jul 16 at 20:11
No, not all. Thomae function is $0$ on the irrationals.
– zhw.
Jul 16 at 20:13
@Tina I added to my answer.
– zhw.
Jul 16 at 20:19
thank you so much
– Tina
Jul 16 at 21:52
@zhw. Ugh, you are right. Differentiable, not continuously differentiable. Sorry.
– Fimpellizieri
Jul 16 at 21:59
the function that you think is the Thomae's function? fr.wikipedia.org/wiki/Fonction_de_Thomae
– Tina
Jul 16 at 20:11
the function that you think is the Thomae's function? fr.wikipedia.org/wiki/Fonction_de_Thomae
– Tina
Jul 16 at 20:11
No, not all. Thomae function is $0$ on the irrationals.
– zhw.
Jul 16 at 20:13
No, not all. Thomae function is $0$ on the irrationals.
– zhw.
Jul 16 at 20:13
@Tina I added to my answer.
– zhw.
Jul 16 at 20:19
@Tina I added to my answer.
– zhw.
Jul 16 at 20:19
thank you so much
– Tina
Jul 16 at 21:52
thank you so much
– Tina
Jul 16 at 21:52
@zhw. Ugh, you are right. Differentiable, not continuously differentiable. Sorry.
– Fimpellizieri
Jul 16 at 21:59
@zhw. Ugh, you are right. Differentiable, not continuously differentiable. Sorry.
– Fimpellizieri
Jul 16 at 21:59
 |Â
show 4 more comments
Almost a direct repost of this question last hour.
– T. Bongers
Jul 16 at 16:33
but now J is an open set
– Tina
Jul 16 at 16:34