real analysis - Continuous function with continuous one-sided derivative

Simple example of the absolute value function $x \mapsto |x|$ on $\mathbb{R}$ shows that it is possible for a continuous function to posses both the right-hand and the left-hand side derivatives and still not being differentiable on $\mathbb{R}$.

I was wondering if it is possible to assume something about one of the one-hand side derivatives to obtain differentiability.

The obvious came to my mind:

Is it true that if a continuous function $f \in C(\mathbb{R})$ has left-hand-side derivative $f_{-}^{'}$ that is continuous on $\mathbb{R}$, then the function $f$ is differentiable?

Answer

Yes.

The keystone is:

Lemma. Let $f\colon [a,b]\to\mathbb R$ be continuous and assume that $f'_+(x)$ exists and is $>0$ for all $x\in [a,b)$. Then $f$ is strictly increasing.

Assume otherwise, i.e. $f(a)\ge f(b)$.
We recursively define a map $g\colon \operatorname{Ord}\to [a,b)$ such that $g$ and $f\circ g$ are strictly inreasing. Since the class $\operatorname{Ord}$ of ordinals is a proper class and $g$ is injective, we arrive at a contradiction, thus showing the claim.

• Let $g(0)=a$.


• For a successor $\alpha=\beta+1$ assume we have already defined $g(\beta)$. For sufficently small positive $h$ we have that $g(\beta)0$. Pick one such$h$and let$g(\alpha)=g(\beta)+h$.


• If $\alpha$ is a limit ordinal, assume $g(\beta)$ is defined for all $\beta<\alpha$. Let $x=\sup_{\beta<\alpha} g(\beta)$. A priori only $x\le b$, but we need $xf(g(0))=f(a)=f(b)$. Therefore$x
  
  

$\square$

Corollary 1. (something like a one-sided Rolle theorem) Let $f\colon [a,b]\to\mathbb R$ be continuous with $f(a)=f(b)$. Assume $f_+$ exists and is continuos in $[a,b)$. Then $f'_+(x)=0$ for some $x\in[a,b)$.

Proof. Assume otherwise. Then either $f_+(x)>0$ for all $x$ or $f_+(x)<0$ for all $x$. In the first case the lemma applies and gives us a contradiction to $f(a)=f(b)$; in the other case, we consider $-f$ instead of $f$. $\square$

Corollary 2. (something like a one-sided IVT) Let $f\colon [a,b]\to\mathbb R$ be continuous. Assume $f_+$ exists and is continuos in $[a,b)$. Then $f'_+(x)=\frac{f(b)-f(a)}{b-a}$ for some $x\in[a,b)$.

Proof. Apply the previous corollary to $f(x)-\frac{f(b)-f(a)}{b-a}x$. $\square$

By symmetry, we have

Corollary 3. Let $f\colon [a,b]\to\mathbb R$ be continuous. Assume $f_-$ exists and is continuos in $(a,b]$. Then $f'_-(x)=\frac{f(b)-f(a)}{b-a}$ for some $x\in(a,b]$. $\square$

Theorem. Let $f\in C(\mathbb R)$ be a function with $f'_-$ continuous on $\mathbb R$.
Then $f\in C^1(\mathbb R)$.

Proof.

Consider aribtrary $a\in \mathbb R$.
Let $\epsilon>0$ be given.
Then by continuity of $f'_-$, for some $\delta>0$ we have $|f'_-(x)-f'_-(a)|<\epsilon$ for all $x\in(a,a+\delta)$.
Thus for $0