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?

• The necessary and sufficient condition is that $f'_-=f'_+$. Can you deduce this identity from the continuity of $f'_-$? – Siminore Oct 15 '14 at 13:37
• @Siminore: Thanks for your comment. I need to prove that the right-hand-side derivative $f'_+$ exists and is equal to $f'_{-}$. Is that what you meant? If so, then I have trouble with that. – xen Oct 15 '14 at 13:43
• See this question. – Tony Piccolo Oct 15 '14 at 14:38
• @TonyPiccolo Many thanks! I somehow overlooked that question. – xen Oct 15 '14 at 18:32

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 $g(\beta)<g(\beta)+h<b$ and $\frac{f(g(\beta)+h)-f(g(\beta))}{h}\approx f_+(g(\beta))>0$. 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 $x<b$. Because $f$ is continuous and $f\circ g$ is strictly increasing, we conclude that $f(x)=\sup_{\beta<\alpha} f(g(\beta))\ge f(g(1))>f(g(0))=f(a)=f(b)$. Therefore $x<b$ as desired and we can let $g(\alpha)=x$.

$\square$

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 $0<h<\delta$ we have $\left|\frac{f(a+h)-f(a)}{h}-f'_-(a)\right|<\epsilon$ by corollary 3. We conclude that $f'_+(a)=f'_-(a)$, i.e. $f$ is differentiable at $a$. $\square$

• Great answer! Thanks. – xen Oct 15 '14 at 18:35
• Just one remark: shouldn't it be $f(x) = \sup_{\beta < \alpha} f(g(\beta))$? – xen Oct 15 '14 at 18:44
• And one more question: what does $\mathrm{Ord}$ means here? The class of all ordinals or the class of countable ordinals? Moreover, since I only "know" that "proper class" means that this class is not a set, how the fact that $g$ is injective and $f \circ g$ is stricly increasing contradicts $f(a) \geq f(b)$? Since $g$ is not a bijection I can not see why $f$ must be increasing. – xen Oct 15 '14 at 19:30
• @xen Thanks for hinting at the typo ($f(\beta)$ isn't even defined). I mean the class of all ordinals. The countable ordinals are a set (e.g. a subset o fthe smallest uncountable ordinal). The contradiction is that by strictly increasing, the map $f$ gives a bijection between a proper class and a set (a subset of $[a,b]$); this is already a contradiction for the first ordinal of cardinality beyond the continuum. – Hagen von Eitzen Oct 16 '14 at 6:43