Question about left and right derivative.

Question

Let $f(x):\mathbb{R}\rightarrow\mathbb{R}$ be a function such that $\forall x\in\mathbb{R}$ there exist:

$$f'_+(x)=\lim_{\delta\rightarrow 0^+}\frac{f(x+\delta)-f(x)}{\delta}$$ $$f'_-(x)=\lim_{\delta\rightarrow 0^-}\frac{f(x+\delta)-f(x)}{\delta}$$

and $\forall x:f'_-(x)=2f'_+(x)$.

How to prove that $f$ is constant. I'm really in nowhere with this kind of problem.

EDIT: Actually I'm having an idea of using similar technique as in

https://math.stackexchange.com/a/79617/161212

but it's still getting nowhere.

Answer 1

Some reading in Saks, Theory of the integral convinced me that we can obtain this by putting together some facts from Chapter 7; this is not going to produce a very satisfying answer, though.

First of all, since upper and lower derivatives at any point are finite, it follows that $f\in VBG_*$ (Saks's notation, which is rather dated). See Theorem 10.1. (Roughly speaking, this is saying that we can partition $\mathbb R$ into countably many pieces on which $f$ is of bounded variation.)

Then Theorem 7.2 gives that $f$ is differentiable almost everywhere. Clearly, $f'=0$ at these points. The same result also implies that the Lebesgue measure of the set $N$ where $f'_+\not=0$ equals zero and, moreover, $|f(N)|=0$ as well. Now I can use Lemma 6.3 to conclude that $|f(\mathbb R)|=0$, which gives the claim since $f$ is continuous.

Answer 2

A) In fact, if a function $f$ has at each point a left and a right derivative, it has a derivative except possibly for $x\in D$, with $D$ at most countable.

See:

http://books.google.fr/books?id=rbCmt-2NxtIC&printsec=frontcover&hl=fr#v=onepage&q&f=false

Look at page 174, point 7) where you will find a proof.

B) Now you have that your continuous function $f$ has a zero derivative, except possibly for $x\in D$, with $D$ at most countable. Then $f$ is constant. The classic proof is as follows.

a) Suppose first that a continuous function $g$ has a derivative for $x\not \in D$ (supposed at most countable), and $g^{\prime}(x)>0$ for $x\not \in D$. Take $a,b$ with $b>a$. Let $d \not \in g(D)$ ($g(D)$ is at most countable) such that $d\leq g(a)$, and $E=\{x\in [a,b]; g(x)\geq d\}$. Let $c={\rm Sup}(E)$. As $g$ is continuous, we have $c\in E$. Suppose that $c<b$. Then we must have $g(c)=d$. Hence $c\not \in D$. As $g^{\prime}(c)>0$, we get that there exists $\alpha>0$ such that in $[c,c+\alpha[$ we have $g(x)\geq g(c)=d$, a contradiction. Hence $b\in E$, we have $g(b)\geq d$. As this is true for all $d\not \in g(D)$, we get $g(b)\geq g(a)$, and $g$ is increasing.

b) Take now $\varepsilon>0$, and $g(x)=f(x)+\varepsilon x$. By a) $g$ is increasing. As this is true for every $\varepsilon$, $f$ is increasing.

c) But $-f(x)$ has also a zero derivative if $x\not \in D$. Hence $-f(x)$ is increasing, and so $f$ is constant.