Mixing
Fundamental Theorem of Calculus for functions with one-sided derivative.
$begingroup$
Let's assume we have a continuous function $F:[0,infty)toBbb R$ such that its one-sided derivative
$$
f(t):=lim_{hsearrow 0} frac {F(t+h)-F(t)}{h}
$$
exists everywhere on $[0,infty)$.
Does the "Fundamental Theorem of Calculus" hold, i.e. for each $tin[0,infty)$ we have
$$
F(t) = F(0) + int_0^t f(s), ds?
$$
Usually we'd use the Mean Value Theorem to prove the (normal) FCT but I don't know if something similar to the MVT would be provable with only these assumptions.
If the above doesn't hold in general, what kind of condition can we impose on $F$ or $f$ to make the "FTC" holds?
Note: I also think it is possible that the assumptions that $F$ is continuous and that its one-sided derivative exists everywhere might be strong enough to deduce better property of $F$, like the existence of $F'$. If anyone know a result in this direction I'd really love to hear it.
real-analysis calculus analysis
asked Jan 17 at 16:08
BigbearZzzBigbearZzz
8,80421652
$endgroup$
add a comment |
$begingroup$
Let's assume we have a continuous function $F:[0,infty)toBbb R$ such that its one-sided derivative
$$
f(t):=lim_{hsearrow 0} frac {F(t+h)-F(t)}{h}
$$
exists everywhere on $[0,infty)$.
Does the "Fundamental Theorem of Calculus" hold, i.e. for each $tin[0,infty)$ we have
$$
F(t) = F(0) + int_0^t f(s), ds?
$$
Usually we'd use the Mean Value Theorem to prove the (normal) FCT but I don't know if something similar to the MVT would be provable with only these assumptions.
If the above doesn't hold in general, what kind of condition can we impose on $F$ or $f$ to make the "FTC" holds?
Note: I also think it is possible that the assumptions that $F$ is continuous and that its one-sided derivative exists everywhere might be strong enough to deduce better property of $F$, like the existence of $F'$. If anyone know a result in this direction I'd really love to hear it.
real-analysis calculus analysis
asked Jan 17 at 16:08
BigbearZzzBigbearZzz
8,80421652
$endgroup$
$begingroup$
the note is false -- let $F$ be the integral of the step function
$endgroup$
– gt6989b
Jan 17 at 16:12
$begingroup$
@gt6989b I'm sorry I don't get what you meant. If $F$ is the integral of a step function wouldn't that make it even classically differentiable (almost everywhere)?
$endgroup$
– BigbearZzz
Jan 17 at 16:14
$begingroup$
$F'$ exists then but is disconnected - I thought you are looking for a continuous derivative, but misread the question, sorry
$endgroup$
– gt6989b
Jan 17 at 16:16
$begingroup$
Oh I see, I thought you were talking about the main part of the question, my bad.
$endgroup$
– BigbearZzz
Jan 17 at 16:17
add a comment |
$begingroup$
Let's assume we have a continuous function $F:[0,infty)toBbb R$ such that its one-sided derivative
$$
f(t):=lim_{hsearrow 0} frac {F(t+h)-F(t)}{h}
$$
exists everywhere on $[0,infty)$.
Does the "Fundamental Theorem of Calculus" hold, i.e. for each $tin[0,infty)$ we have
$$
F(t) = F(0) + int_0^t f(s), ds?
$$
Usually we'd use the Mean Value Theorem to prove the (normal) FCT but I don't know if something similar to the MVT would be provable with only these assumptions.
If the above doesn't hold in general, what kind of condition can we impose on $F$ or $f$ to make the "FTC" holds?
Note: I also think it is possible that the assumptions that $F$ is continuous and that its one-sided derivative exists everywhere might be strong enough to deduce better property of $F$, like the existence of $F'$. If anyone know a result in this direction I'd really love to hear it.
real-analysis calculus analysis
asked Jan 17 at 16:08
BigbearZzzBigbearZzz
8,80421652
$endgroup$
Let's assume we have a continuous function $F:[0,infty)toBbb R$ such that its one-sided derivative
$$
f(t):=lim_{hsearrow 0} frac {F(t+h)-F(t)}{h}
$$
exists everywhere on $[0,infty)$.
Does the "Fundamental Theorem of Calculus" hold, i.e. for each $tin[0,infty)$ we have
$$
F(t) = F(0) + int_0^t f(s), ds?
$$
Usually we'd use the Mean Value Theorem to prove the (normal) FCT but I don't know if something similar to the MVT would be provable with only these assumptions.
If the above doesn't hold in general, what kind of condition can we impose on $F$ or $f$ to make the "FTC" holds?
Note: I also think it is possible that the assumptions that $F$ is continuous and that its one-sided derivative exists everywhere might be strong enough to deduce better property of $F$, like the existence of $F'$. If anyone know a result in this direction I'd really love to hear it.
real-analysis calculus analysis
real-analysis calculus analysis
asked Jan 17 at 16:08
BigbearZzzBigbearZzz
8,80421652
asked Jan 17 at 16:08
BigbearZzzBigbearZzz
8,80421652
asked Jan 17 at 16:08
BigbearZzzBigbearZzz
8,80421652
asked Jan 17 at 16:08
BigbearZzzBigbearZzz
8,80421652
asked Jan 17 at 16:08
BigbearZzzBigbearZzz
8,80421652
8,80421652
$begingroup$
the note is false -- let $F$ be the integral of the step function
$endgroup$
– gt6989b
Jan 17 at 16:12
$begingroup$
@gt6989b I'm sorry I don't get what you meant. If $F$ is the integral of a step function wouldn't that make it even classically differentiable (almost everywhere)?
$endgroup$
– BigbearZzz
Jan 17 at 16:14
$begingroup$
$F'$ exists then but is disconnected - I thought you are looking for a continuous derivative, but misread the question, sorry
$endgroup$
– gt6989b
Jan 17 at 16:16
$begingroup$
Oh I see, I thought you were talking about the main part of the question, my bad.
$endgroup$
– BigbearZzz
Jan 17 at 16:17
add a comment |
$begingroup$
the note is false -- let $F$ be the integral of the step function
$endgroup$
– gt6989b
Jan 17 at 16:12
$begingroup$
@gt6989b I'm sorry I don't get what you meant. If $F$ is the integral of a step function wouldn't that make it even classically differentiable (almost everywhere)?
$endgroup$
– BigbearZzz
Jan 17 at 16:14
$begingroup$
$F'$ exists then but is disconnected - I thought you are looking for a continuous derivative, but misread the question, sorry
$endgroup$
– gt6989b
Jan 17 at 16:16
$begingroup$
Oh I see, I thought you were talking about the main part of the question, my bad.
$endgroup$
– BigbearZzz
Jan 17 at 16:17
$begingroup$
the note is false -- let $F$ be the integral of the step function
$endgroup$
– gt6989b
Jan 17 at 16:12
$begingroup$
the note is false -- let $F$ be the integral of the step function
$endgroup$
– gt6989b
Jan 17 at 16:12
$begingroup$
@gt6989b I'm sorry I don't get what you meant. If $F$ is the integral of a step function wouldn't that make it even classically differentiable (almost everywhere)?
$endgroup$
– BigbearZzz
Jan 17 at 16:14
$begingroup$
@gt6989b I'm sorry I don't get what you meant. If $F$ is the integral of a step function wouldn't that make it even classically differentiable (almost everywhere)?
$endgroup$
– BigbearZzz
Jan 17 at 16:14
$begingroup$
$F'$ exists then but is disconnected - I thought you are looking for a continuous derivative, but misread the question, sorry
$endgroup$
– gt6989b
Jan 17 at 16:16
$begingroup$
$F'$ exists then but is disconnected - I thought you are looking for a continuous derivative, but misread the question, sorry
$endgroup$
– gt6989b
Jan 17 at 16:16
$begingroup$
Oh I see, I thought you were talking about the main part of the question, my bad.
$endgroup$
– BigbearZzz
Jan 17 at 16:17
$begingroup$
Oh I see, I thought you were talking about the main part of the question, my bad.
$endgroup$
– BigbearZzz
Jan 17 at 16:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, that does not give an FTC.
Note that this has nothing to do with the fact that you're talking about one-sided derivatives; in fact the corresponding "normal FTC" for two-sided derivatives is also false!
For example let $$F(x)=begin{cases}x^2sin(1/x^{10}),&(xne0),
\0,&(x=0).end{cases}$$
Then $F$ is differentiable everywhere but $int_{-1}^1 F'(t),dt$ does not exist (not even as a Lebesgue integral).
I suspect the result is true if you assume in addition that $f$ is continuous.
Edit: Yes, it's true if $f$ is continuous.
Lemma. If $F:Bbb RtoBbb R$ is continuous and the right-hand derivative $D_RF(x)$ exists and equals $0$ for every $x$ then $F$ is constant.
It's enough to prove this:
If $lambda>0$ then $|F(x)-F(0)|lelambda x$ for every $xge 0$.
Proof: Let $A$ be the set of $age0$ such that $|F(x)-F(0)|le lambda x$ for every $xin[0,a]$. It's clear that $$A=[0,alpha] $$for some $alphain[0,infty]$, and we need only show that $alpha=infty$. But if $alpha<infty$ then $D_RF(alpha)=0$ shows that there exists $delta>0$ with $[alpha,alpha+delta)subset A$. (Choose $delta$ so that $|F(alpha+h)-F(alpha)|<frac12lambda h$ for all $hin[0,delta)$.)
And now
Prop. Suppose $F:Bbb RtoBbb R$ is continuous and $f(x)=D_RF(x)$ exists for every $x$. If $f$ is continuous then $$F(x)=F(0)+int_0^x f$$for every $x>0$.
Proof: Define $G(x)=int_0^x f(t),dt$. Since $f$ is continuous it follows from the standard FTC that $G$ is differentiable and $G'=f$. So $D_R(F-G)=0$, hence $F-G$ is constant.
(Cor. If $f$ is continuous then $F$ is differentiable.)
Alas the question is changing. I suspect it's also true assuming just that $F$ is convex.
Edit: Yes, it's true if $F$ is convex. My version of this if anything seems simpler than the case $f$ continuous, because I saw how to use some high-powered machinery.
If $F:Bbb RtoBbb R$ is convex then $F(x)=F(0)+int_0^x D_RF$.
Proof. You say you know, and it's not hard to prove, that $F$ is locally Lipschitz. Hence it's locally absolutely continuous, so it's diferentiable almost everywhere and $F(x)-F(0)=int_0^x F'(t),t$ ((where that's a Lebesgue integral).
In case we care it follows that $F(x)-F(0)$ is actually the Riemann integral of $f=D_RF$: Since $f$ is increasing it is continuous almost everywhere, hence the Riemann integral $int_0^x f$ exists. And it equals the Lebesgue integral of $F'$, since $F'=f$ almost everywhere.
answered Jan 17 at 16:30
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
$begingroup$
Thank you for the concrete example that FCT could fail if we don't assume continuity of $f$. I should have mentioned that I am aware of this but I want to know how general $F,f$ can get. What I had in mind was something slightly more general than a convex function, whose left and right derivatives exist (but could differ) everywhere. Here $f$ need not be continuous but we know that $f$ is increasing.
$endgroup$
– BigbearZzz
Jan 17 at 16:35
$begingroup$
Don't get me wrong, I really like your answer and it was my mistake for not mentioning that I want $f$ to be at least discontinuous. Perhaps the question is just too vague and can be interpreted in many ways. The answer as it stands now has many things I didn't know before.
$endgroup$
– BigbearZzz
Jan 17 at 18:12
$begingroup$
The reason I mentioned convexity is that it was the thing that motivated this question. If $F$ is convex then it would be locally Lipschitz and hence $W^{1,1}_{text{loc}}$ so $f$ would coincide with $F'$ a.e. which means the "FTC" would work. Then it occurred to me that it would be interesting to know what can be deduced knowing about $f$ alone.
$endgroup$
– BigbearZzz
Jan 17 at 18:16
$begingroup$
@BigbearZzz Yes if $F$ is convex. My version of this is if anything simpler than the case $f$ continuous, because II saw how to use some highh-powered machinery.
$endgroup$
– David C. Ullrich
Jan 17 at 18:28
add a comment |
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$begingroup$
No, that does not give an FTC.
Note that this has nothing to do with the fact that you're talking about one-sided derivatives; in fact the corresponding "normal FTC" for two-sided derivatives is also false!
For example let $$F(x)=begin{cases}x^2sin(1/x^{10}),&(xne0),
\0,&(x=0).end{cases}$$
Then $F$ is differentiable everywhere but $int_{-1}^1 F'(t),dt$ does not exist (not even as a Lebesgue integral).
I suspect the result is true if you assume in addition that $f$ is continuous.
Edit: Yes, it's true if $f$ is continuous.
Lemma. If $F:Bbb RtoBbb R$ is continuous and the right-hand derivative $D_RF(x)$ exists and equals $0$ for every $x$ then $F$ is constant.
It's enough to prove this:
If $lambda>0$ then $|F(x)-F(0)|lelambda x$ for every $xge 0$.
Proof: Let $A$ be the set of $age0$ such that $|F(x)-F(0)|le lambda x$ for every $xin[0,a]$. It's clear that $$A=[0,alpha] $$for some $alphain[0,infty]$, and we need only show that $alpha=infty$. But if $alpha<infty$ then $D_RF(alpha)=0$ shows that there exists $delta>0$ with $[alpha,alpha+delta)subset A$. (Choose $delta$ so that $|F(alpha+h)-F(alpha)|<frac12lambda h$ for all $hin[0,delta)$.)
And now
Prop. Suppose $F:Bbb RtoBbb R$ is continuous and $f(x)=D_RF(x)$ exists for every $x$. If $f$ is continuous then $$F(x)=F(0)+int_0^x f$$for every $x>0$.
Proof: Define $G(x)=int_0^x f(t),dt$. Since $f$ is continuous it follows from the standard FTC that $G$ is differentiable and $G'=f$. So $D_R(F-G)=0$, hence $F-G$ is constant.
(Cor. If $f$ is continuous then $F$ is differentiable.)
Alas the question is changing. I suspect it's also true assuming just that $F$ is convex.
Edit: Yes, it's true if $F$ is convex. My version of this if anything seems simpler than the case $f$ continuous, because I saw how to use some high-powered machinery.
If $F:Bbb RtoBbb R$ is convex then $F(x)=F(0)+int_0^x D_RF$.
Proof. You say you know, and it's not hard to prove, that $F$ is locally Lipschitz. Hence it's locally absolutely continuous, so it's diferentiable almost everywhere and $F(x)-F(0)=int_0^x F'(t),t$ ((where that's a Lebesgue integral).
In case we care it follows that $F(x)-F(0)$ is actually the Riemann integral of $f=D_RF$: Since $f$ is increasing it is continuous almost everywhere, hence the Riemann integral $int_0^x f$ exists. And it equals the Lebesgue integral of $F'$, since $F'=f$ almost everywhere.
answered Jan 17 at 16:30
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
$begingroup$
Thank you for the concrete example that FCT could fail if we don't assume continuity of $f$. I should have mentioned that I am aware of this but I want to know how general $F,f$ can get. What I had in mind was something slightly more general than a convex function, whose left and right derivatives exist (but could differ) everywhere. Here $f$ need not be continuous but we know that $f$ is increasing.
$endgroup$
– BigbearZzz
Jan 17 at 16:35
$begingroup$
Don't get me wrong, I really like your answer and it was my mistake for not mentioning that I want $f$ to be at least discontinuous. Perhaps the question is just too vague and can be interpreted in many ways. The answer as it stands now has many things I didn't know before.
$endgroup$
– BigbearZzz
Jan 17 at 18:12
$begingroup$
The reason I mentioned convexity is that it was the thing that motivated this question. If $F$ is convex then it would be locally Lipschitz and hence $W^{1,1}_{text{loc}}$ so $f$ would coincide with $F'$ a.e. which means the "FTC" would work. Then it occurred to me that it would be interesting to know what can be deduced knowing about $f$ alone.
$endgroup$
– BigbearZzz
Jan 17 at 18:16
$begingroup$
@BigbearZzz Yes if $F$ is convex. My version of this is if anything simpler than the case $f$ continuous, because II saw how to use some highh-powered machinery.
$endgroup$
– David C. Ullrich
Jan 17 at 18:28
add a comment |
$begingroup$
No, that does not give an FTC.
Note that this has nothing to do with the fact that you're talking about one-sided derivatives; in fact the corresponding "normal FTC" for two-sided derivatives is also false!
For example let $$F(x)=begin{cases}x^2sin(1/x^{10}),&(xne0),
\0,&(x=0).end{cases}$$
Then $F$ is differentiable everywhere but $int_{-1}^1 F'(t),dt$ does not exist (not even as a Lebesgue integral).
I suspect the result is true if you assume in addition that $f$ is continuous.
Edit: Yes, it's true if $f$ is continuous.
Lemma. If $F:Bbb RtoBbb R$ is continuous and the right-hand derivative $D_RF(x)$ exists and equals $0$ for every $x$ then $F$ is constant.
It's enough to prove this:
If $lambda>0$ then $|F(x)-F(0)|lelambda x$ for every $xge 0$.
Proof: Let $A$ be the set of $age0$ such that $|F(x)-F(0)|le lambda x$ for every $xin[0,a]$. It's clear that $$A=[0,alpha] $$for some $alphain[0,infty]$, and we need only show that $alpha=infty$. But if $alpha<infty$ then $D_RF(alpha)=0$ shows that there exists $delta>0$ with $[alpha,alpha+delta)subset A$. (Choose $delta$ so that $|F(alpha+h)-F(alpha)|<frac12lambda h$ for all $hin[0,delta)$.)
And now
Prop. Suppose $F:Bbb RtoBbb R$ is continuous and $f(x)=D_RF(x)$ exists for every $x$. If $f$ is continuous then $$F(x)=F(0)+int_0^x f$$for every $x>0$.
Proof: Define $G(x)=int_0^x f(t),dt$. Since $f$ is continuous it follows from the standard FTC that $G$ is differentiable and $G'=f$. So $D_R(F-G)=0$, hence $F-G$ is constant.
(Cor. If $f$ is continuous then $F$ is differentiable.)
Alas the question is changing. I suspect it's also true assuming just that $F$ is convex.
Edit: Yes, it's true if $F$ is convex. My version of this if anything seems simpler than the case $f$ continuous, because I saw how to use some high-powered machinery.
If $F:Bbb RtoBbb R$ is convex then $F(x)=F(0)+int_0^x D_RF$.
Proof. You say you know, and it's not hard to prove, that $F$ is locally Lipschitz. Hence it's locally absolutely continuous, so it's diferentiable almost everywhere and $F(x)-F(0)=int_0^x F'(t),t$ ((where that's a Lebesgue integral).
In case we care it follows that $F(x)-F(0)$ is actually the Riemann integral of $f=D_RF$: Since $f$ is increasing it is continuous almost everywhere, hence the Riemann integral $int_0^x f$ exists. And it equals the Lebesgue integral of $F'$, since $F'=f$ almost everywhere.
answered Jan 17 at 16:30
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
$begingroup$
Thank you for the concrete example that FCT could fail if we don't assume continuity of $f$. I should have mentioned that I am aware of this but I want to know how general $F,f$ can get. What I had in mind was something slightly more general than a convex function, whose left and right derivatives exist (but could differ) everywhere. Here $f$ need not be continuous but we know that $f$ is increasing.
$endgroup$
– BigbearZzz
Jan 17 at 16:35
$begingroup$
Don't get me wrong, I really like your answer and it was my mistake for not mentioning that I want $f$ to be at least discontinuous. Perhaps the question is just too vague and can be interpreted in many ways. The answer as it stands now has many things I didn't know before.
$endgroup$
– BigbearZzz
Jan 17 at 18:12
$begingroup$
The reason I mentioned convexity is that it was the thing that motivated this question. If $F$ is convex then it would be locally Lipschitz and hence $W^{1,1}_{text{loc}}$ so $f$ would coincide with $F'$ a.e. which means the "FTC" would work. Then it occurred to me that it would be interesting to know what can be deduced knowing about $f$ alone.
$endgroup$
– BigbearZzz
Jan 17 at 18:16
$begingroup$
@BigbearZzz Yes if $F$ is convex. My version of this is if anything simpler than the case $f$ continuous, because II saw how to use some highh-powered machinery.
$endgroup$
– David C. Ullrich
Jan 17 at 18:28
add a comment |
$begingroup$
No, that does not give an FTC.
Note that this has nothing to do with the fact that you're talking about one-sided derivatives; in fact the corresponding "normal FTC" for two-sided derivatives is also false!
For example let $$F(x)=begin{cases}x^2sin(1/x^{10}),&(xne0),
\0,&(x=0).end{cases}$$
Then $F$ is differentiable everywhere but $int_{-1}^1 F'(t),dt$ does not exist (not even as a Lebesgue integral).
I suspect the result is true if you assume in addition that $f$ is continuous.
Edit: Yes, it's true if $f$ is continuous.
Lemma. If $F:Bbb RtoBbb R$ is continuous and the right-hand derivative $D_RF(x)$ exists and equals $0$ for every $x$ then $F$ is constant.
It's enough to prove this:
If $lambda>0$ then $|F(x)-F(0)|lelambda x$ for every $xge 0$.
Proof: Let $A$ be the set of $age0$ such that $|F(x)-F(0)|le lambda x$ for every $xin[0,a]$. It's clear that $$A=[0,alpha] $$for some $alphain[0,infty]$, and we need only show that $alpha=infty$. But if $alpha<infty$ then $D_RF(alpha)=0$ shows that there exists $delta>0$ with $[alpha,alpha+delta)subset A$. (Choose $delta$ so that $|F(alpha+h)-F(alpha)|<frac12lambda h$ for all $hin[0,delta)$.)
And now
Prop. Suppose $F:Bbb RtoBbb R$ is continuous and $f(x)=D_RF(x)$ exists for every $x$. If $f$ is continuous then $$F(x)=F(0)+int_0^x f$$for every $x>0$.
Proof: Define $G(x)=int_0^x f(t),dt$. Since $f$ is continuous it follows from the standard FTC that $G$ is differentiable and $G'=f$. So $D_R(F-G)=0$, hence $F-G$ is constant.
(Cor. If $f$ is continuous then $F$ is differentiable.)
Alas the question is changing. I suspect it's also true assuming just that $F$ is convex.
Edit: Yes, it's true if $F$ is convex. My version of this if anything seems simpler than the case $f$ continuous, because I saw how to use some high-powered machinery.
If $F:Bbb RtoBbb R$ is convex then $F(x)=F(0)+int_0^x D_RF$.
Proof. You say you know, and it's not hard to prove, that $F$ is locally Lipschitz. Hence it's locally absolutely continuous, so it's diferentiable almost everywhere and $F(x)-F(0)=int_0^x F'(t),t$ ((where that's a Lebesgue integral).
In case we care it follows that $F(x)-F(0)$ is actually the Riemann integral of $f=D_RF$: Since $f$ is increasing it is continuous almost everywhere, hence the Riemann integral $int_0^x f$ exists. And it equals the Lebesgue integral of $F'$, since $F'=f$ almost everywhere.
answered Jan 17 at 16:30
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
No, that does not give an FTC.
Note that this has nothing to do with the fact that you're talking about one-sided derivatives; in fact the corresponding "normal FTC" for two-sided derivatives is also false!
For example let $$F(x)=begin{cases}x^2sin(1/x^{10}),&(xne0),
\0,&(x=0).end{cases}$$
Then $F$ is differentiable everywhere but $int_{-1}^1 F'(t),dt$ does not exist (not even as a Lebesgue integral).
I suspect the result is true if you assume in addition that $f$ is continuous.
Edit: Yes, it's true if $f$ is continuous.
Lemma. If $F:Bbb RtoBbb R$ is continuous and the right-hand derivative $D_RF(x)$ exists and equals $0$ for every $x$ then $F$ is constant.
It's enough to prove this:
If $lambda>0$ then $|F(x)-F(0)|lelambda x$ for every $xge 0$.
Proof: Let $A$ be the set of $age0$ such that $|F(x)-F(0)|le lambda x$ for every $xin[0,a]$. It's clear that $$A=[0,alpha] $$for some $alphain[0,infty]$, and we need only show that $alpha=infty$. But if $alpha<infty$ then $D_RF(alpha)=0$ shows that there exists $delta>0$ with $[alpha,alpha+delta)subset A$. (Choose $delta$ so that $|F(alpha+h)-F(alpha)|<frac12lambda h$ for all $hin[0,delta)$.)
And now
Prop. Suppose $F:Bbb RtoBbb R$ is continuous and $f(x)=D_RF(x)$ exists for every $x$. If $f$ is continuous then $$F(x)=F(0)+int_0^x f$$for every $x>0$.
Proof: Define $G(x)=int_0^x f(t),dt$. Since $f$ is continuous it follows from the standard FTC that $G$ is differentiable and $G'=f$. So $D_R(F-G)=0$, hence $F-G$ is constant.
(Cor. If $f$ is continuous then $F$ is differentiable.)
Alas the question is changing. I suspect it's also true assuming just that $F$ is convex.
Edit: Yes, it's true if $F$ is convex. My version of this if anything seems simpler than the case $f$ continuous, because I saw how to use some high-powered machinery.
If $F:Bbb RtoBbb R$ is convex then $F(x)=F(0)+int_0^x D_RF$.
Proof. You say you know, and it's not hard to prove, that $F$ is locally Lipschitz. Hence it's locally absolutely continuous, so it's diferentiable almost everywhere and $F(x)-F(0)=int_0^x F'(t),t$ ((where that's a Lebesgue integral).
In case we care it follows that $F(x)-F(0)$ is actually the Riemann integral of $f=D_RF$: Since $f$ is increasing it is continuous almost everywhere, hence the Riemann integral $int_0^x f$ exists. And it equals the Lebesgue integral of $F'$, since $F'=f$ almost everywhere.
answered Jan 17 at 16:30
David C. UllrichDavid C. Ullrich
61k43994
edited Jan 17 at 18:26
answered Jan 17 at 16:30
David C. UllrichDavid C. Ullrich
61k43994
answered Jan 17 at 16:30
David C. UllrichDavid C. Ullrich
61k43994
answered Jan 17 at 16:30
David C. UllrichDavid C. Ullrich
61k43994
61k43994
$begingroup$
Thank you for the concrete example that FCT could fail if we don't assume continuity of $f$. I should have mentioned that I am aware of this but I want to know how general $F,f$ can get. What I had in mind was something slightly more general than a convex function, whose left and right derivatives exist (but could differ) everywhere. Here $f$ need not be continuous but we know that $f$ is increasing.
$endgroup$
– BigbearZzz
Jan 17 at 16:35
$begingroup$
Don't get me wrong, I really like your answer and it was my mistake for not mentioning that I want $f$ to be at least discontinuous. Perhaps the question is just too vague and can be interpreted in many ways. The answer as it stands now has many things I didn't know before.
$endgroup$
– BigbearZzz
Jan 17 at 18:12
$begingroup$
The reason I mentioned convexity is that it was the thing that motivated this question. If $F$ is convex then it would be locally Lipschitz and hence $W^{1,1}_{text{loc}}$ so $f$ would coincide with $F'$ a.e. which means the "FTC" would work. Then it occurred to me that it would be interesting to know what can be deduced knowing about $f$ alone.
$endgroup$
– BigbearZzz
Jan 17 at 18:16
$begingroup$
@BigbearZzz Yes if $F$ is convex. My version of this is if anything simpler than the case $f$ continuous, because II saw how to use some highh-powered machinery.
$endgroup$
– David C. Ullrich
Jan 17 at 18:28
add a comment |
$begingroup$
Thank you for the concrete example that FCT could fail if we don't assume continuity of $f$. I should have mentioned that I am aware of this but I want to know how general $F,f$ can get. What I had in mind was something slightly more general than a convex function, whose left and right derivatives exist (but could differ) everywhere. Here $f$ need not be continuous but we know that $f$ is increasing.
$endgroup$
– BigbearZzz
Jan 17 at 16:35
$begingroup$
Don't get me wrong, I really like your answer and it was my mistake for not mentioning that I want $f$ to be at least discontinuous. Perhaps the question is just too vague and can be interpreted in many ways. The answer as it stands now has many things I didn't know before.
$endgroup$
– BigbearZzz
Jan 17 at 18:12
$begingroup$
The reason I mentioned convexity is that it was the thing that motivated this question. If $F$ is convex then it would be locally Lipschitz and hence $W^{1,1}_{text{loc}}$ so $f$ would coincide with $F'$ a.e. which means the "FTC" would work. Then it occurred to me that it would be interesting to know what can be deduced knowing about $f$ alone.
$endgroup$
– BigbearZzz
Jan 17 at 18:16
$begingroup$
@BigbearZzz Yes if $F$ is convex. My version of this is if anything simpler than the case $f$ continuous, because II saw how to use some highh-powered machinery.
$endgroup$
– David C. Ullrich
Jan 17 at 18:28
$begingroup$
Thank you for the concrete example that FCT could fail if we don't assume continuity of $f$. I should have mentioned that I am aware of this but I want to know how general $F,f$ can get. What I had in mind was something slightly more general than a convex function, whose left and right derivatives exist (but could differ) everywhere. Here $f$ need not be continuous but we know that $f$ is increasing.
$endgroup$
– BigbearZzz
Jan 17 at 16:35
$begingroup$
Thank you for the concrete example that FCT could fail if we don't assume continuity of $f$. I should have mentioned that I am aware of this but I want to know how general $F,f$ can get. What I had in mind was something slightly more general than a convex function, whose left and right derivatives exist (but could differ) everywhere. Here $f$ need not be continuous but we know that $f$ is increasing.
$endgroup$
– BigbearZzz
Jan 17 at 16:35
$begingroup$
Don't get me wrong, I really like your answer and it was my mistake for not mentioning that I want $f$ to be at least discontinuous. Perhaps the question is just too vague and can be interpreted in many ways. The answer as it stands now has many things I didn't know before.
$endgroup$
– BigbearZzz
Jan 17 at 18:12
$begingroup$
Don't get me wrong, I really like your answer and it was my mistake for not mentioning that I want $f$ to be at least discontinuous. Perhaps the question is just too vague and can be interpreted in many ways. The answer as it stands now has many things I didn't know before.
$endgroup$
– BigbearZzz
Jan 17 at 18:12
$begingroup$
The reason I mentioned convexity is that it was the thing that motivated this question. If $F$ is convex then it would be locally Lipschitz and hence $W^{1,1}_{text{loc}}$ so $f$ would coincide with $F'$ a.e. which means the "FTC" would work. Then it occurred to me that it would be interesting to know what can be deduced knowing about $f$ alone.
$endgroup$
– BigbearZzz
Jan 17 at 18:16
$begingroup$
The reason I mentioned convexity is that it was the thing that motivated this question. If $F$ is convex then it would be locally Lipschitz and hence $W^{1,1}_{text{loc}}$ so $f$ would coincide with $F'$ a.e. which means the "FTC" would work. Then it occurred to me that it would be interesting to know what can be deduced knowing about $f$ alone.
$endgroup$
– BigbearZzz
Jan 17 at 18:16
$begingroup$
@BigbearZzz Yes if $F$ is convex. My version of this is if anything simpler than the case $f$ continuous, because II saw how to use some highh-powered machinery.
$endgroup$
– David C. Ullrich
Jan 17 at 18:28
$begingroup$
@BigbearZzz Yes if $F$ is convex. My version of this is if anything simpler than the case $f$ continuous, because II saw how to use some highh-powered machinery.
$endgroup$
– David C. Ullrich
Jan 17 at 18:28
add a comment |
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$begingroup$
the note is false -- let $F$ be the integral of the step function
$endgroup$
– gt6989b
Jan 17 at 16:12
$begingroup$
@gt6989b I'm sorry I don't get what you meant. If $F$ is the integral of a step function wouldn't that make it even classically differentiable (almost everywhere)?
$endgroup$
– BigbearZzz
Jan 17 at 16:14
$begingroup$
$F'$ exists then but is disconnected - I thought you are looking for a continuous derivative, but misread the question, sorry
$endgroup$
– gt6989b
Jan 17 at 16:16
$begingroup$
Oh I see, I thought you were talking about the main part of the question, my bad.
$endgroup$
– BigbearZzz
Jan 17 at 16:17