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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 part $I$ of $mathbbR$. Is they exist always a function $F$ differentiable on an interval $J$ containing I such that $F'=f$ on $I$?
If $I$ is an interval , it's ok
if $I$ is an open set , it's ok
But if $I$ is only a part of $mathbbR$ ?
real-analysis integration
edited Jul 19 at 14:26
Martin Sleziak
43.5k6113259
asked Jul 16 at 15:13
Tina
355112
closed as off-topic by user21820, Delta-u, Taroccoesbrocco, amWhy, Mostafa Ayaz Jul 25 at 11:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Delta-u, Taroccoesbrocco, amWhy, Mostafa Ayaz
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Let $f$ be a continuous function on a part $I$ of $mathbbR$. Is they exist always a function $F$ differentiable on an interval $J$ containing I such that $F'=f$ on $I$?
If $I$ is an interval , it's ok
if $I$ is an open set , it's ok
But if $I$ is only a part of $mathbbR$ ?
real-analysis integration
edited Jul 19 at 14:26
Martin Sleziak
43.5k6113259
asked Jul 16 at 15:13
Tina
355112
closed as off-topic by user21820, Delta-u, Taroccoesbrocco, amWhy, Mostafa Ayaz Jul 25 at 11:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Delta-u, Taroccoesbrocco, amWhy, Mostafa Ayaz
No, it's not true if $I$ is open. See my answer below.
– zhw.
Jul 16 at 15:39
A more interesting question might be to ask if there is an open set $J$ containing $I$ for which the conclusion is true.
– zhw.
Jul 16 at 15:52
2
@Fimpellizieri Boundedness is not enough, there are things like $sin(1/x)$ lurking about. Now if $f$ is uniformly continuous, your plan works.
– zhw.
Jul 16 at 16:01
@zhw there are some things I do not understand: every open set of R is a countable union of disjoint open intervals , so f admits antidérivative function in each interval
– Tina
Jul 16 at 16:05
@Tina True, but you told us $J$ was to be an interval.
– zhw.
Jul 16 at 16:06
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up vote
0
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up vote
0
down vote
favorite
Let $f$ be a continuous function on a part $I$ of $mathbbR$. Is they exist always a function $F$ differentiable on an interval $J$ containing I such that $F'=f$ on $I$?
If $I$ is an interval , it's ok
if $I$ is an open set , it's ok
But if $I$ is only a part of $mathbbR$ ?
real-analysis integration
edited Jul 19 at 14:26
Martin Sleziak
43.5k6113259
asked Jul 16 at 15:13
Tina
355112
Let $f$ be a continuous function on a part $I$ of $mathbbR$. Is they exist always a function $F$ differentiable on an interval $J$ containing I such that $F'=f$ on $I$?
If $I$ is an interval , it's ok
if $I$ is an open set , it's ok
But if $I$ is only a part of $mathbbR$ ?
real-analysis integration
edited Jul 19 at 14:26
Martin Sleziak
43.5k6113259
asked Jul 16 at 15:13
Tina
355112
edited Jul 19 at 14:26
Martin Sleziak
43.5k6113259
edited Jul 19 at 14:26
Martin Sleziak
43.5k6113259
edited Jul 19 at 14:26
Martin Sleziak
43.5k6113259
43.5k6113259
asked Jul 16 at 15:13
Tina
355112
asked Jul 16 at 15:13
Tina
355112
asked Jul 16 at 15:13
Tina
355112
355112
closed as off-topic by user21820, Delta-u, Taroccoesbrocco, amWhy, Mostafa Ayaz Jul 25 at 11:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Delta-u, Taroccoesbrocco, amWhy, Mostafa Ayaz
closed as off-topic by user21820, Delta-u, Taroccoesbrocco, amWhy, Mostafa Ayaz Jul 25 at 11:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Delta-u, Taroccoesbrocco, amWhy, Mostafa Ayaz
No, it's not true if $I$ is open. See my answer below.
– zhw.
Jul 16 at 15:39
A more interesting question might be to ask if there is an open set $J$ containing $I$ for which the conclusion is true.
– zhw.
Jul 16 at 15:52
2
@Fimpellizieri Boundedness is not enough, there are things like $sin(1/x)$ lurking about. Now if $f$ is uniformly continuous, your plan works.
– zhw.
Jul 16 at 16:01
@zhw there are some things I do not understand: every open set of R is a countable union of disjoint open intervals , so f admits antidérivative function in each interval
– Tina
Jul 16 at 16:05
@Tina True, but you told us $J$ was to be an interval.
– zhw.
Jul 16 at 16:06
 |Â
show 1 more comment
No, it's not true if $I$ is open. See my answer below.
– zhw.
Jul 16 at 15:39
A more interesting question might be to ask if there is an open set $J$ containing $I$ for which the conclusion is true.
– zhw.
Jul 16 at 15:52
2
@Fimpellizieri Boundedness is not enough, there are things like $sin(1/x)$ lurking about. Now if $f$ is uniformly continuous, your plan works.
– zhw.
Jul 16 at 16:01
@zhw there are some things I do not understand: every open set of R is a countable union of disjoint open intervals , so f admits antidérivative function in each interval
– Tina
Jul 16 at 16:05
@Tina True, but you told us $J$ was to be an interval.
– zhw.
Jul 16 at 16:06
No, it's not true if $I$ is open. See my answer below.
– zhw.
Jul 16 at 15:39
No, it's not true if $I$ is open. See my answer below.
– zhw.
Jul 16 at 15:39
A more interesting question might be to ask if there is an open set $J$ containing $I$ for which the conclusion is true.
– zhw.
Jul 16 at 15:52
A more interesting question might be to ask if there is an open set $J$ containing $I$ for which the conclusion is true.
– zhw.
Jul 16 at 15:52
2
2
@Fimpellizieri Boundedness is not enough, there are things like $sin(1/x)$ lurking about. Now if $f$ is uniformly continuous, your plan works.
– zhw.
Jul 16 at 16:01
@Fimpellizieri Boundedness is not enough, there are things like $sin(1/x)$ lurking about. Now if $f$ is uniformly continuous, your plan works.
– zhw.
Jul 16 at 16:01
@zhw there are some things I do not understand: every open set of R is a countable union of disjoint open intervals , so f admits antidérivative function in each interval
– Tina
Jul 16 at 16:05
@zhw there are some things I do not understand: every open set of R is a countable union of disjoint open intervals , so f admits antidérivative function in each interval
– Tina
Jul 16 at 16:05
@Tina True, but you told us $J$ was to be an interval.
– zhw.
Jul 16 at 16:06
@Tina True, but you told us $J$ was to be an interval.
– zhw.
Jul 16 at 16:06
 |Â
show 1 more comment
1 Answer
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Counterexample: Let $I=(-1,0)cup (0,1).$ Define $f(x) = 1/x$ for $xin I.$ Then $f$ is continuous on $I.$ Let $J$ be an interval containing $I.$ Then $J$ contains $0.$ Suppose $F$ is differentiable on $J$ and $F'=f$ on $I.$ Then for small $x>0$ we have
$$F(1/2)-F(x) = int_x^1/2 fracdtt = ln (1/2)-ln x.$$
As $xto 0^+,$ the left side $to F(1/2)-F(0),$ while the right side $to infty,$ contradiction.
answered Jul 16 at 15:38
zhw.
65.8k42870
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Counterexample: Let $I=(-1,0)cup (0,1).$ Define $f(x) = 1/x$ for $xin I.$ Then $f$ is continuous on $I.$ Let $J$ be an interval containing $I.$ Then $J$ contains $0.$ Suppose $F$ is differentiable on $J$ and $F'=f$ on $I.$ Then for small $x>0$ we have
$$F(1/2)-F(x) = int_x^1/2 fracdtt = ln (1/2)-ln x.$$
As $xto 0^+,$ the left side $to F(1/2)-F(0),$ while the right side $to infty,$ contradiction.
answered Jul 16 at 15:38
zhw.
65.8k42870
add a comment |Â
up vote
3
down vote
accepted
Counterexample: Let $I=(-1,0)cup (0,1).$ Define $f(x) = 1/x$ for $xin I.$ Then $f$ is continuous on $I.$ Let $J$ be an interval containing $I.$ Then $J$ contains $0.$ Suppose $F$ is differentiable on $J$ and $F'=f$ on $I.$ Then for small $x>0$ we have
$$F(1/2)-F(x) = int_x^1/2 fracdtt = ln (1/2)-ln x.$$
As $xto 0^+,$ the left side $to F(1/2)-F(0),$ while the right side $to infty,$ contradiction.
answered Jul 16 at 15:38
zhw.
65.8k42870
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Counterexample: Let $I=(-1,0)cup (0,1).$ Define $f(x) = 1/x$ for $xin I.$ Then $f$ is continuous on $I.$ Let $J$ be an interval containing $I.$ Then $J$ contains $0.$ Suppose $F$ is differentiable on $J$ and $F'=f$ on $I.$ Then for small $x>0$ we have
$$F(1/2)-F(x) = int_x^1/2 fracdtt = ln (1/2)-ln x.$$
As $xto 0^+,$ the left side $to F(1/2)-F(0),$ while the right side $to infty,$ contradiction.
answered Jul 16 at 15:38
zhw.
65.8k42870
Counterexample: Let $I=(-1,0)cup (0,1).$ Define $f(x) = 1/x$ for $xin I.$ Then $f$ is continuous on $I.$ Let $J$ be an interval containing $I.$ Then $J$ contains $0.$ Suppose $F$ is differentiable on $J$ and $F'=f$ on $I.$ Then for small $x>0$ we have
$$F(1/2)-F(x) = int_x^1/2 fracdtt = ln (1/2)-ln x.$$
As $xto 0^+,$ the left side $to F(1/2)-F(0),$ while the right side $to infty,$ contradiction.
answered Jul 16 at 15:38
zhw.
65.8k42870
answered Jul 16 at 15:38
zhw.
65.8k42870
answered Jul 16 at 15:38
zhw.
65.8k42870
answered Jul 16 at 15:38
zhw.
65.8k42870
65.8k42870
add a comment |Â
add a comment |Â
No, it's not true if $I$ is open. See my answer below.
– zhw.
Jul 16 at 15:39
A more interesting question might be to ask if there is an open set $J$ containing $I$ for which the conclusion is true.
– zhw.
Jul 16 at 15:52
2
@Fimpellizieri Boundedness is not enough, there are things like $sin(1/x)$ lurking about. Now if $f$ is uniformly continuous, your plan works.
– zhw.
Jul 16 at 16:01
@zhw there are some things I do not understand: every open set of R is a countable union of disjoint open intervals , so f admits antidérivative function in each interval
– Tina
Jul 16 at 16:05
@Tina True, but you told us $J$ was to be an interval.
– zhw.
Jul 16 at 16:06