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Does $f'(x) = g'(x)$ for all $x in (a, b)$ imply $f(x) = g(x) + c$ for all $x in [a ,b]$?
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My question is regarding the open and closed intervals in the statement. I can see why the result holds on the open interval $(a, b)$, but does it also hold at the endpoints $a$ and $b$?
EDIT: Here is some more context on the question.
This appears in the proof that if $f$ is continuous on $[a, b]$, $f$ is integrable on $[a, b]$.
Let $U(x) = L int_a^x f$ and $U(x) = L int_a^x f$ be the lower and upper integrals evaluated on $[a, x]$, where $x in (a, b)$. It is shown that
beginequation*
L'(x) = U'(x) = f(x)
endequation*
for all $x in (a, b)$. The author then states without proof that this implies there exists a number $c$ such that
beginequation*
U(x) = L(x) + c
endequation*
for all $x in [a, b]$. The inclusion of the right endpoint $b$ is necessary in completing the proof.
In my question, I have substituted $L(x), U(x)$ with $f(x), g(x)$. To the best of my knowledge, all we know about $L(x)$ and $U(x)$ at this point is that they are differentiable on $(a, b)$.
calculus
asked Jul 20 at 18:56
tmakino
27729
add a comment |Â
up vote
1
down vote
favorite
My question is regarding the open and closed intervals in the statement. I can see why the result holds on the open interval $(a, b)$, but does it also hold at the endpoints $a$ and $b$?
EDIT: Here is some more context on the question.
This appears in the proof that if $f$ is continuous on $[a, b]$, $f$ is integrable on $[a, b]$.
Let $U(x) = L int_a^x f$ and $U(x) = L int_a^x f$ be the lower and upper integrals evaluated on $[a, x]$, where $x in (a, b)$. It is shown that
beginequation*
L'(x) = U'(x) = f(x)
endequation*
for all $x in (a, b)$. The author then states without proof that this implies there exists a number $c$ such that
beginequation*
U(x) = L(x) + c
endequation*
for all $x in [a, b]$. The inclusion of the right endpoint $b$ is necessary in completing the proof.
In my question, I have substituted $L(x), U(x)$ with $f(x), g(x)$. To the best of my knowledge, all we know about $L(x)$ and $U(x)$ at this point is that they are differentiable on $(a, b)$.
calculus
asked Jul 20 at 18:56
tmakino
27729
No: $(a,b) = (0,1)$ and $f(x) = delta(x)$ and $g(x) = delta(x-1)$.
– David G. Stork
Jul 20 at 18:58
1
You can apply Lagrange`s Mean Value Theorem on the continous funcion $h(x)=f-g$ on the closed interval [a,b] and see that it holds true. (If indeed f,g are continous on [a,b] and differentiable on (a,b))
– Sar
Jul 20 at 19:03
2
@DavidG.Stork. Is $delta$ the Dirac distribution? That's not a function.
– md2perpe
Jul 20 at 19:44
Glad to see someone studying this nice proof of integrability of continuous functions. This proof is from Spivak and not widely popular like the one via uniform continuity.
– Paramanand Singh
Jul 21 at 2:52
You should note (or prove yourself) that the functions $L, U$ are continuous on $[a, b] $ for any bounded function $f$ and your job is done by applying mean value theorem for $L-U$ on interval $[a, b]$.
– Paramanand Singh
Jul 21 at 2:55
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My question is regarding the open and closed intervals in the statement. I can see why the result holds on the open interval $(a, b)$, but does it also hold at the endpoints $a$ and $b$?
EDIT: Here is some more context on the question.
This appears in the proof that if $f$ is continuous on $[a, b]$, $f$ is integrable on $[a, b]$.
Let $U(x) = L int_a^x f$ and $U(x) = L int_a^x f$ be the lower and upper integrals evaluated on $[a, x]$, where $x in (a, b)$. It is shown that
beginequation*
L'(x) = U'(x) = f(x)
endequation*
for all $x in (a, b)$. The author then states without proof that this implies there exists a number $c$ such that
beginequation*
U(x) = L(x) + c
endequation*
for all $x in [a, b]$. The inclusion of the right endpoint $b$ is necessary in completing the proof.
In my question, I have substituted $L(x), U(x)$ with $f(x), g(x)$. To the best of my knowledge, all we know about $L(x)$ and $U(x)$ at this point is that they are differentiable on $(a, b)$.
calculus
asked Jul 20 at 18:56
tmakino
27729
My question is regarding the open and closed intervals in the statement. I can see why the result holds on the open interval $(a, b)$, but does it also hold at the endpoints $a$ and $b$?
EDIT: Here is some more context on the question.
This appears in the proof that if $f$ is continuous on $[a, b]$, $f$ is integrable on $[a, b]$.
Let $U(x) = L int_a^x f$ and $U(x) = L int_a^x f$ be the lower and upper integrals evaluated on $[a, x]$, where $x in (a, b)$. It is shown that
beginequation*
L'(x) = U'(x) = f(x)
endequation*
for all $x in (a, b)$. The author then states without proof that this implies there exists a number $c$ such that
beginequation*
U(x) = L(x) + c
endequation*
for all $x in [a, b]$. The inclusion of the right endpoint $b$ is necessary in completing the proof.
In my question, I have substituted $L(x), U(x)$ with $f(x), g(x)$. To the best of my knowledge, all we know about $L(x)$ and $U(x)$ at this point is that they are differentiable on $(a, b)$.
calculus
asked Jul 20 at 18:56
tmakino
27729
edited Jul 20 at 20:30
asked Jul 20 at 18:56
tmakino
27729
asked Jul 20 at 18:56
tmakino
27729
asked Jul 20 at 18:56
tmakino
27729
27729
No: $(a,b) = (0,1)$ and $f(x) = delta(x)$ and $g(x) = delta(x-1)$.
– David G. Stork
Jul 20 at 18:58
1
You can apply Lagrange`s Mean Value Theorem on the continous funcion $h(x)=f-g$ on the closed interval [a,b] and see that it holds true. (If indeed f,g are continous on [a,b] and differentiable on (a,b))
– Sar
Jul 20 at 19:03
2
@DavidG.Stork. Is $delta$ the Dirac distribution? That's not a function.
– md2perpe
Jul 20 at 19:44
Glad to see someone studying this nice proof of integrability of continuous functions. This proof is from Spivak and not widely popular like the one via uniform continuity.
– Paramanand Singh
Jul 21 at 2:52
You should note (or prove yourself) that the functions $L, U$ are continuous on $[a, b] $ for any bounded function $f$ and your job is done by applying mean value theorem for $L-U$ on interval $[a, b]$.
– Paramanand Singh
Jul 21 at 2:55
add a comment |Â
No: $(a,b) = (0,1)$ and $f(x) = delta(x)$ and $g(x) = delta(x-1)$.
– David G. Stork
Jul 20 at 18:58
1
You can apply Lagrange`s Mean Value Theorem on the continous funcion $h(x)=f-g$ on the closed interval [a,b] and see that it holds true. (If indeed f,g are continous on [a,b] and differentiable on (a,b))
– Sar
Jul 20 at 19:03
2
@DavidG.Stork. Is $delta$ the Dirac distribution? That's not a function.
– md2perpe
Jul 20 at 19:44
Glad to see someone studying this nice proof of integrability of continuous functions. This proof is from Spivak and not widely popular like the one via uniform continuity.
– Paramanand Singh
Jul 21 at 2:52
You should note (or prove yourself) that the functions $L, U$ are continuous on $[a, b] $ for any bounded function $f$ and your job is done by applying mean value theorem for $L-U$ on interval $[a, b]$.
– Paramanand Singh
Jul 21 at 2:55
No: $(a,b) = (0,1)$ and $f(x) = delta(x)$ and $g(x) = delta(x-1)$.
– David G. Stork
Jul 20 at 18:58
No: $(a,b) = (0,1)$ and $f(x) = delta(x)$ and $g(x) = delta(x-1)$.
– David G. Stork
Jul 20 at 18:58
1
1
You can apply Lagrange`s Mean Value Theorem on the continous funcion $h(x)=f-g$ on the closed interval [a,b] and see that it holds true. (If indeed f,g are continous on [a,b] and differentiable on (a,b))
– Sar
Jul 20 at 19:03
You can apply Lagrange`s Mean Value Theorem on the continous funcion $h(x)=f-g$ on the closed interval [a,b] and see that it holds true. (If indeed f,g are continous on [a,b] and differentiable on (a,b))
– Sar
Jul 20 at 19:03
2
2
@DavidG.Stork. Is $delta$ the Dirac distribution? That's not a function.
– md2perpe
Jul 20 at 19:44
@DavidG.Stork. Is $delta$ the Dirac distribution? That's not a function.
– md2perpe
Jul 20 at 19:44
Glad to see someone studying this nice proof of integrability of continuous functions. This proof is from Spivak and not widely popular like the one via uniform continuity.
– Paramanand Singh
Jul 21 at 2:52
Glad to see someone studying this nice proof of integrability of continuous functions. This proof is from Spivak and not widely popular like the one via uniform continuity.
– Paramanand Singh
Jul 21 at 2:52
You should note (or prove yourself) that the functions $L, U$ are continuous on $[a, b] $ for any bounded function $f$ and your job is done by applying mean value theorem for $L-U$ on interval $[a, b]$.
– Paramanand Singh
Jul 21 at 2:55
You should note (or prove yourself) that the functions $L, U$ are continuous on $[a, b] $ for any bounded function $f$ and your job is done by applying mean value theorem for $L-U$ on interval $[a, b]$.
– Paramanand Singh
Jul 21 at 2:55
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Converting my comments into an answer.
Probably you are studying the proof of integrability of continuous functions from Spivak's Calculus. This particular proof based on derivative of upper and lower Darboux integrals is not as popular as the one via uniform continuity.
You should first prove that if $f$ is any bounded function on interval $[a, b] $ then its upper and lower Darboux integrals $L, U$ given by $$L(x) =underlineint_a^x f(t) , dt,, U(x) =overlineint_a ^x f(t) , dt$$ are continuous on interval $[a, b] $.
Next one establishes that if $cin[a, b] $ and $f$ is continuous at $c$ then $L'(c)= U'(c)=f(c) $. If $c$ is an end point then the above statement refers to one sided continuity and differentiability.
Thus if $f$ is continuous on $[a, b] $ then $L, U$ are continuous on $[a, b] $ and also differentiable on $[a, b] $ (one sided derivatives at end points) and $L'(x) =U'(x) =f(x), forall xin[a, b] $. Your job is now complete via application of mean value theorem on $L-U$ on $[a, b] $.
I have given outline of the proof and if you need more details let me know.
answered Jul 21 at 3:08
Paramanand Singh
45.2k553142
Thank you, this is indeed the proof I am studying. After applying MVT to $L - U$, I ended up with $(L - U)(b) = (L - U)(a)$ since $(L - U)' = 0$ everywhere. However, I can't see how to turn this into $L = U + c$ everywhere.
– tmakino
Jul 24 at 5:00
@tmakino: When the derivative vanishes on an interval, the function vanishes on the closure of the interval, provided the function is continuous on the closure. Since $(L-U) '=0$ it follows that $L-U=c$ on $[a, b] $.
– Paramanand Singh
Jul 24 at 6:05
1
@tmakino: consider for example the function $L-U$ on interval $[a, x] $ instead of $[a, b] $ and apply mean value theorem.
– Paramanand Singh
Jul 24 at 6:13
add a comment |Â
up vote
2
down vote
Assuming that $f$ and $g$ are continuous on $[a,b]$, then, yes, it is true. For instance,$$f(a)=lim_xto af(x)=lim_xto ag(x)+c=g(a)+c.$$The same argument applies to $b$.
edited Jul 20 at 20:04
Clement C.
47.1k33682
answered Jul 20 at 18:59
José Carlos Santos
114k1698177
1
Why assume $f$ and $g$ are continuous on $[a,b]$? That is precisely what the OP does not want to assume. Instead, they are continuous just on $(a,b)$---big and central difference.
– David G. Stork
Jul 20 at 19:07
1
@DavidG.Stork WIthout assuming that, the statement is obviously false. Let's wait and see what the OP has to say.
– José Carlos Santos
Jul 20 at 19:10
Doesn't differentiability on $[a,b]$ automatically implies continuity on $[a,b]$ or am I missing something ?
– alxchen
Jul 20 at 19:28
1
@AlexKChen You are missing the endpoints. $(a,b)=[a,b]setminusa,b$. Differentiability on $(a,b)$ implies continuity on $(a,b)$, not on $[a,b]$. (Figuring out what happens at the endpoints is the gist of the OP's question.)
– Clement C.
Jul 20 at 19:38
I've updated the question to provide more context. I tried to present the question in the most simple form possible, but I may have omitted some important information while doing so.
– tmakino
Jul 20 at 20:27
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Converting my comments into an answer.
Probably you are studying the proof of integrability of continuous functions from Spivak's Calculus. This particular proof based on derivative of upper and lower Darboux integrals is not as popular as the one via uniform continuity.
You should first prove that if $f$ is any bounded function on interval $[a, b] $ then its upper and lower Darboux integrals $L, U$ given by $$L(x) =underlineint_a^x f(t) , dt,, U(x) =overlineint_a ^x f(t) , dt$$ are continuous on interval $[a, b] $.
Next one establishes that if $cin[a, b] $ and $f$ is continuous at $c$ then $L'(c)= U'(c)=f(c) $. If $c$ is an end point then the above statement refers to one sided continuity and differentiability.
Thus if $f$ is continuous on $[a, b] $ then $L, U$ are continuous on $[a, b] $ and also differentiable on $[a, b] $ (one sided derivatives at end points) and $L'(x) =U'(x) =f(x), forall xin[a, b] $. Your job is now complete via application of mean value theorem on $L-U$ on $[a, b] $.
I have given outline of the proof and if you need more details let me know.
answered Jul 21 at 3:08
Paramanand Singh
45.2k553142
Thank you, this is indeed the proof I am studying. After applying MVT to $L - U$, I ended up with $(L - U)(b) = (L - U)(a)$ since $(L - U)' = 0$ everywhere. However, I can't see how to turn this into $L = U + c$ everywhere.
– tmakino
Jul 24 at 5:00
@tmakino: When the derivative vanishes on an interval, the function vanishes on the closure of the interval, provided the function is continuous on the closure. Since $(L-U) '=0$ it follows that $L-U=c$ on $[a, b] $.
– Paramanand Singh
Jul 24 at 6:05
1
@tmakino: consider for example the function $L-U$ on interval $[a, x] $ instead of $[a, b] $ and apply mean value theorem.
– Paramanand Singh
Jul 24 at 6:13
add a comment |Â
up vote
1
down vote
accepted
Converting my comments into an answer.
Probably you are studying the proof of integrability of continuous functions from Spivak's Calculus. This particular proof based on derivative of upper and lower Darboux integrals is not as popular as the one via uniform continuity.
You should first prove that if $f$ is any bounded function on interval $[a, b] $ then its upper and lower Darboux integrals $L, U$ given by $$L(x) =underlineint_a^x f(t) , dt,, U(x) =overlineint_a ^x f(t) , dt$$ are continuous on interval $[a, b] $.
Next one establishes that if $cin[a, b] $ and $f$ is continuous at $c$ then $L'(c)= U'(c)=f(c) $. If $c$ is an end point then the above statement refers to one sided continuity and differentiability.
Thus if $f$ is continuous on $[a, b] $ then $L, U$ are continuous on $[a, b] $ and also differentiable on $[a, b] $ (one sided derivatives at end points) and $L'(x) =U'(x) =f(x), forall xin[a, b] $. Your job is now complete via application of mean value theorem on $L-U$ on $[a, b] $.
I have given outline of the proof and if you need more details let me know.
answered Jul 21 at 3:08
Paramanand Singh
45.2k553142
Thank you, this is indeed the proof I am studying. After applying MVT to $L - U$, I ended up with $(L - U)(b) = (L - U)(a)$ since $(L - U)' = 0$ everywhere. However, I can't see how to turn this into $L = U + c$ everywhere.
– tmakino
Jul 24 at 5:00
@tmakino: When the derivative vanishes on an interval, the function vanishes on the closure of the interval, provided the function is continuous on the closure. Since $(L-U) '=0$ it follows that $L-U=c$ on $[a, b] $.
– Paramanand Singh
Jul 24 at 6:05
1
@tmakino: consider for example the function $L-U$ on interval $[a, x] $ instead of $[a, b] $ and apply mean value theorem.
– Paramanand Singh
Jul 24 at 6:13
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Converting my comments into an answer.
Probably you are studying the proof of integrability of continuous functions from Spivak's Calculus. This particular proof based on derivative of upper and lower Darboux integrals is not as popular as the one via uniform continuity.
You should first prove that if $f$ is any bounded function on interval $[a, b] $ then its upper and lower Darboux integrals $L, U$ given by $$L(x) =underlineint_a^x f(t) , dt,, U(x) =overlineint_a ^x f(t) , dt$$ are continuous on interval $[a, b] $.
Next one establishes that if $cin[a, b] $ and $f$ is continuous at $c$ then $L'(c)= U'(c)=f(c) $. If $c$ is an end point then the above statement refers to one sided continuity and differentiability.
Thus if $f$ is continuous on $[a, b] $ then $L, U$ are continuous on $[a, b] $ and also differentiable on $[a, b] $ (one sided derivatives at end points) and $L'(x) =U'(x) =f(x), forall xin[a, b] $. Your job is now complete via application of mean value theorem on $L-U$ on $[a, b] $.
I have given outline of the proof and if you need more details let me know.
answered Jul 21 at 3:08
Paramanand Singh
45.2k553142
Converting my comments into an answer.
Probably you are studying the proof of integrability of continuous functions from Spivak's Calculus. This particular proof based on derivative of upper and lower Darboux integrals is not as popular as the one via uniform continuity.
You should first prove that if $f$ is any bounded function on interval $[a, b] $ then its upper and lower Darboux integrals $L, U$ given by $$L(x) =underlineint_a^x f(t) , dt,, U(x) =overlineint_a ^x f(t) , dt$$ are continuous on interval $[a, b] $.
Next one establishes that if $cin[a, b] $ and $f$ is continuous at $c$ then $L'(c)= U'(c)=f(c) $. If $c$ is an end point then the above statement refers to one sided continuity and differentiability.
Thus if $f$ is continuous on $[a, b] $ then $L, U$ are continuous on $[a, b] $ and also differentiable on $[a, b] $ (one sided derivatives at end points) and $L'(x) =U'(x) =f(x), forall xin[a, b] $. Your job is now complete via application of mean value theorem on $L-U$ on $[a, b] $.
I have given outline of the proof and if you need more details let me know.
answered Jul 21 at 3:08
Paramanand Singh
45.2k553142
edited Jul 21 at 3:32
answered Jul 21 at 3:08
Paramanand Singh
45.2k553142
answered Jul 21 at 3:08
Paramanand Singh
45.2k553142
answered Jul 21 at 3:08
Paramanand Singh
45.2k553142
45.2k553142
Thank you, this is indeed the proof I am studying. After applying MVT to $L - U$, I ended up with $(L - U)(b) = (L - U)(a)$ since $(L - U)' = 0$ everywhere. However, I can't see how to turn this into $L = U + c$ everywhere.
– tmakino
Jul 24 at 5:00
@tmakino: When the derivative vanishes on an interval, the function vanishes on the closure of the interval, provided the function is continuous on the closure. Since $(L-U) '=0$ it follows that $L-U=c$ on $[a, b] $.
– Paramanand Singh
Jul 24 at 6:05
1
@tmakino: consider for example the function $L-U$ on interval $[a, x] $ instead of $[a, b] $ and apply mean value theorem.
– Paramanand Singh
Jul 24 at 6:13
add a comment |Â
Thank you, this is indeed the proof I am studying. After applying MVT to $L - U$, I ended up with $(L - U)(b) = (L - U)(a)$ since $(L - U)' = 0$ everywhere. However, I can't see how to turn this into $L = U + c$ everywhere.
– tmakino
Jul 24 at 5:00
@tmakino: When the derivative vanishes on an interval, the function vanishes on the closure of the interval, provided the function is continuous on the closure. Since $(L-U) '=0$ it follows that $L-U=c$ on $[a, b] $.
– Paramanand Singh
Jul 24 at 6:05
1
@tmakino: consider for example the function $L-U$ on interval $[a, x] $ instead of $[a, b] $ and apply mean value theorem.
– Paramanand Singh
Jul 24 at 6:13
Thank you, this is indeed the proof I am studying. After applying MVT to $L - U$, I ended up with $(L - U)(b) = (L - U)(a)$ since $(L - U)' = 0$ everywhere. However, I can't see how to turn this into $L = U + c$ everywhere.
– tmakino
Jul 24 at 5:00
Thank you, this is indeed the proof I am studying. After applying MVT to $L - U$, I ended up with $(L - U)(b) = (L - U)(a)$ since $(L - U)' = 0$ everywhere. However, I can't see how to turn this into $L = U + c$ everywhere.
– tmakino
Jul 24 at 5:00
@tmakino: When the derivative vanishes on an interval, the function vanishes on the closure of the interval, provided the function is continuous on the closure. Since $(L-U) '=0$ it follows that $L-U=c$ on $[a, b] $.
– Paramanand Singh
Jul 24 at 6:05
@tmakino: When the derivative vanishes on an interval, the function vanishes on the closure of the interval, provided the function is continuous on the closure. Since $(L-U) '=0$ it follows that $L-U=c$ on $[a, b] $.
– Paramanand Singh
Jul 24 at 6:05
1
1
@tmakino: consider for example the function $L-U$ on interval $[a, x] $ instead of $[a, b] $ and apply mean value theorem.
– Paramanand Singh
Jul 24 at 6:13
@tmakino: consider for example the function $L-U$ on interval $[a, x] $ instead of $[a, b] $ and apply mean value theorem.
– Paramanand Singh
Jul 24 at 6:13
add a comment |Â
up vote
2
down vote
Assuming that $f$ and $g$ are continuous on $[a,b]$, then, yes, it is true. For instance,$$f(a)=lim_xto af(x)=lim_xto ag(x)+c=g(a)+c.$$The same argument applies to $b$.
edited Jul 20 at 20:04
Clement C.
47.1k33682
answered Jul 20 at 18:59
José Carlos Santos
114k1698177
1
Why assume $f$ and $g$ are continuous on $[a,b]$? That is precisely what the OP does not want to assume. Instead, they are continuous just on $(a,b)$---big and central difference.
– David G. Stork
Jul 20 at 19:07
1
@DavidG.Stork WIthout assuming that, the statement is obviously false. Let's wait and see what the OP has to say.
– José Carlos Santos
Jul 20 at 19:10
Doesn't differentiability on $[a,b]$ automatically implies continuity on $[a,b]$ or am I missing something ?
– alxchen
Jul 20 at 19:28
1
@AlexKChen You are missing the endpoints. $(a,b)=[a,b]setminusa,b$. Differentiability on $(a,b)$ implies continuity on $(a,b)$, not on $[a,b]$. (Figuring out what happens at the endpoints is the gist of the OP's question.)
– Clement C.
Jul 20 at 19:38
I've updated the question to provide more context. I tried to present the question in the most simple form possible, but I may have omitted some important information while doing so.
– tmakino
Jul 20 at 20:27
add a comment |Â
up vote
2
down vote
Assuming that $f$ and $g$ are continuous on $[a,b]$, then, yes, it is true. For instance,$$f(a)=lim_xto af(x)=lim_xto ag(x)+c=g(a)+c.$$The same argument applies to $b$.
edited Jul 20 at 20:04
Clement C.
47.1k33682
answered Jul 20 at 18:59
José Carlos Santos
114k1698177
1
Why assume $f$ and $g$ are continuous on $[a,b]$? That is precisely what the OP does not want to assume. Instead, they are continuous just on $(a,b)$---big and central difference.
– David G. Stork
Jul 20 at 19:07
1
@DavidG.Stork WIthout assuming that, the statement is obviously false. Let's wait and see what the OP has to say.
– José Carlos Santos
Jul 20 at 19:10
Doesn't differentiability on $[a,b]$ automatically implies continuity on $[a,b]$ or am I missing something ?
– alxchen
Jul 20 at 19:28
1
@AlexKChen You are missing the endpoints. $(a,b)=[a,b]setminusa,b$. Differentiability on $(a,b)$ implies continuity on $(a,b)$, not on $[a,b]$. (Figuring out what happens at the endpoints is the gist of the OP's question.)
– Clement C.
Jul 20 at 19:38
I've updated the question to provide more context. I tried to present the question in the most simple form possible, but I may have omitted some important information while doing so.
– tmakino
Jul 20 at 20:27
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Assuming that $f$ and $g$ are continuous on $[a,b]$, then, yes, it is true. For instance,$$f(a)=lim_xto af(x)=lim_xto ag(x)+c=g(a)+c.$$The same argument applies to $b$.
edited Jul 20 at 20:04
Clement C.
47.1k33682
answered Jul 20 at 18:59
José Carlos Santos
114k1698177
Assuming that $f$ and $g$ are continuous on $[a,b]$, then, yes, it is true. For instance,$$f(a)=lim_xto af(x)=lim_xto ag(x)+c=g(a)+c.$$The same argument applies to $b$.
edited Jul 20 at 20:04
Clement C.
47.1k33682
answered Jul 20 at 18:59
José Carlos Santos
114k1698177
edited Jul 20 at 20:04
Clement C.
47.1k33682
edited Jul 20 at 20:04
Clement C.
47.1k33682
edited Jul 20 at 20:04
Clement C.
47.1k33682
47.1k33682
answered Jul 20 at 18:59
José Carlos Santos
114k1698177
answered Jul 20 at 18:59
José Carlos Santos
114k1698177
answered Jul 20 at 18:59
José Carlos Santos
114k1698177
114k1698177
1
Why assume $f$ and $g$ are continuous on $[a,b]$? That is precisely what the OP does not want to assume. Instead, they are continuous just on $(a,b)$---big and central difference.
– David G. Stork
Jul 20 at 19:07
1
@DavidG.Stork WIthout assuming that, the statement is obviously false. Let's wait and see what the OP has to say.
– José Carlos Santos
Jul 20 at 19:10
Doesn't differentiability on $[a,b]$ automatically implies continuity on $[a,b]$ or am I missing something ?
– alxchen
Jul 20 at 19:28
1
@AlexKChen You are missing the endpoints. $(a,b)=[a,b]setminusa,b$. Differentiability on $(a,b)$ implies continuity on $(a,b)$, not on $[a,b]$. (Figuring out what happens at the endpoints is the gist of the OP's question.)
– Clement C.
Jul 20 at 19:38
I've updated the question to provide more context. I tried to present the question in the most simple form possible, but I may have omitted some important information while doing so.
– tmakino
Jul 20 at 20:27
add a comment |Â
1
Why assume $f$ and $g$ are continuous on $[a,b]$? That is precisely what the OP does not want to assume. Instead, they are continuous just on $(a,b)$---big and central difference.
– David G. Stork
Jul 20 at 19:07
1
@DavidG.Stork WIthout assuming that, the statement is obviously false. Let's wait and see what the OP has to say.
– José Carlos Santos
Jul 20 at 19:10
Doesn't differentiability on $[a,b]$ automatically implies continuity on $[a,b]$ or am I missing something ?
– alxchen
Jul 20 at 19:28
1
@AlexKChen You are missing the endpoints. $(a,b)=[a,b]setminusa,b$. Differentiability on $(a,b)$ implies continuity on $(a,b)$, not on $[a,b]$. (Figuring out what happens at the endpoints is the gist of the OP's question.)
– Clement C.
Jul 20 at 19:38
I've updated the question to provide more context. I tried to present the question in the most simple form possible, but I may have omitted some important information while doing so.
– tmakino
Jul 20 at 20:27
1
1
Why assume $f$ and $g$ are continuous on $[a,b]$? That is precisely what the OP does not want to assume. Instead, they are continuous just on $(a,b)$---big and central difference.
– David G. Stork
Jul 20 at 19:07
Why assume $f$ and $g$ are continuous on $[a,b]$? That is precisely what the OP does not want to assume. Instead, they are continuous just on $(a,b)$---big and central difference.
– David G. Stork
Jul 20 at 19:07
1
1
@DavidG.Stork WIthout assuming that, the statement is obviously false. Let's wait and see what the OP has to say.
– José Carlos Santos
Jul 20 at 19:10
@DavidG.Stork WIthout assuming that, the statement is obviously false. Let's wait and see what the OP has to say.
– José Carlos Santos
Jul 20 at 19:10
Doesn't differentiability on $[a,b]$ automatically implies continuity on $[a,b]$ or am I missing something ?
– alxchen
Jul 20 at 19:28
Doesn't differentiability on $[a,b]$ automatically implies continuity on $[a,b]$ or am I missing something ?
– alxchen
Jul 20 at 19:28
1
1
@AlexKChen You are missing the endpoints. $(a,b)=[a,b]setminusa,b$. Differentiability on $(a,b)$ implies continuity on $(a,b)$, not on $[a,b]$. (Figuring out what happens at the endpoints is the gist of the OP's question.)
– Clement C.
Jul 20 at 19:38
@AlexKChen You are missing the endpoints. $(a,b)=[a,b]setminusa,b$. Differentiability on $(a,b)$ implies continuity on $(a,b)$, not on $[a,b]$. (Figuring out what happens at the endpoints is the gist of the OP's question.)
– Clement C.
Jul 20 at 19:38
I've updated the question to provide more context. I tried to present the question in the most simple form possible, but I may have omitted some important information while doing so.
– tmakino
Jul 20 at 20:27
I've updated the question to provide more context. I tried to present the question in the most simple form possible, but I may have omitted some important information while doing so.
– tmakino
Jul 20 at 20:27
add a comment |Â
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No: $(a,b) = (0,1)$ and $f(x) = delta(x)$ and $g(x) = delta(x-1)$.
– David G. Stork
Jul 20 at 18:58
1
You can apply Lagrange`s Mean Value Theorem on the continous funcion $h(x)=f-g$ on the closed interval [a,b] and see that it holds true. (If indeed f,g are continous on [a,b] and differentiable on (a,b))
– Sar
Jul 20 at 19:03
2
@DavidG.Stork. Is $delta$ the Dirac distribution? That's not a function.
– md2perpe
Jul 20 at 19:44
Glad to see someone studying this nice proof of integrability of continuous functions. This proof is from Spivak and not widely popular like the one via uniform continuity.
– Paramanand Singh
Jul 21 at 2:52
You should note (or prove yourself) that the functions $L, U$ are continuous on $[a, b] $ for any bounded function $f$ and your job is done by applying mean value theorem for $L-U$ on interval $[a, b]$.
– Paramanand Singh
Jul 21 at 2:55