# Fundamental Theorem of Calculus for functions with one-sided derivative.

Let's assume we have a continuous function $$F:[0,\infty)\to\Bbb R$$ such that its one-sided derivative $$f(t):=\lim_{h\searrow 0} \frac {F(t+h)-F(t)}{h}$$ exists everywhere on $$[0,\infty)$$.

Does the "Fundamental Theorem of Calculus" hold, i.e. for each $$t\in[0,\infty)$$ 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.

• the note is false -- let $F$ be the integral of the step function Jan 17, 2019 at 16:12
• @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)? Jan 17, 2019 at 16:14
• $F'$ exists then but is disconnected - I thought you are looking for a continuous derivative, but misread the question, sorry Jan 17, 2019 at 16:16
• Oh I see, I thought you were talking about the main part of the question, my bad. Jan 17, 2019 at 16:17

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 $$F(x)=\begin{cases}x^2\sin(1/x^{10}),&(x\ne0), \\0,&(x=0).\end{cases}$$

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 $$F:\Bbb R\to\Bbb R$$ 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 $$|F(x)-F(0)|\le\lambda x$$ for every $$x\ge 0$$.

Proof: Let $$A$$ be the set of $$a\ge0$$ such that $$|F(x)-F(0)|\le \lambda x$$ for every $$x\in[0,a]$$. It's clear that $$A=[0,\alpha]$$for some $$\alpha\in[0,\infty]$$, 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 $$|F(\alpha+h)-F(\alpha)|<\frac12\lambda h$$ for all $$h\in[0,\delta)$$.)

And now

Prop. Suppose $$F:\Bbb R\to\Bbb R$$ 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 $$F:\Bbb R\to\Bbb R$$ 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 $$F(x)-F(0)=\int_0^x F'(t)\,dt$$ (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.

• 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. Jan 17, 2019 at 16:35
• 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. Jan 17, 2019 at 18:12
• 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. Jan 17, 2019 at 18:16
• @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. Jan 17, 2019 at 18:28
• It's been over a year now, sorry for reviving the comment to your answer. I've been reading through some old posts including this one and was curious if there exists a differentiable function with positive derivative whose derivative is not (Lebesgue) integrable? Mar 28, 2020 at 9:18