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

analysiscalculusreal-analysis

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.

Best Answer

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.

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