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.

  • $\begingroup$ the note is false -- let $F$ be the integral of the step function $\endgroup$
    – gt6989b
    Jan 17, 2019 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, 2019 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, 2019 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, 2019 at 16:17

1 Answer 1


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.

  • $\begingroup$ 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, 2019 at 16:35
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    $\begingroup$ 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, 2019 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, 2019 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. $\endgroup$ Jan 17, 2019 at 18:28
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    $\begingroup$ 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? $\endgroup$
    – BigbearZzz
    Mar 28, 2020 at 9:18

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