A book of mine says the following is true, and I am having some trouble proving it. (I've considered using the Lebesgue differentiation theorem and absolute continuity, as well as elementary analysis methods.)

Let $f: [0, \infty) \rightarrow \mathbb{R}$ be continuous and have right derivatives at each point in the domain, with the right derivative function being continuous. Then $f$ is differentiable.

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    $\begingroup$ Actually, I guess I asked this prematurely because I thought of a solution and forgot to investigate it, and it turns out it worked. One needs to redo the program of Rudin Ch. 5 for right derivatives. I.e. prove the mean value theorem, and then use it to conclude the result by continuity of the right derivative. Continuity and right differentiability are used in proving the MVT. Continuity of the right derivative is used in taking the left-sided limit after using the MVT. $\endgroup$ – Jeff Jun 12 '13 at 20:12
  • $\begingroup$ You can post that as the answer to your own question :) $\endgroup$ – Ataraxia Jun 12 '13 at 20:17
  • $\begingroup$ Actually I can't because it seems like I overlooked a detail I'm working on now. The MVT requires you to have derivatives 0 at extreme points, but with only one side it is insufficient. $\endgroup$ – Jeff Jun 12 '13 at 20:20
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    $\begingroup$ Via FTC, it amounts to showing that if $f$ has null right derivative, it is constant. $\endgroup$ – Julien Jun 12 '13 at 21:13
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    $\begingroup$ @Landscape Essentially like you. For every $\epsilon>0$, the sup of $\{m\,;\,|f(x)-f(0)|\leq \epsilon|x| \forall 0\leq x\leq m\}$ must be $+\infty$. So $|f(x)-f(0)|\leq \epsilon |x|$ for all $x\geq 0$ and all $\epsilon>0$. Whence $f(x)=f(0)$ for all $x\geq 0$. $\endgroup$ – Julien Jun 13 '13 at 4:11

Let us denote the right derivative of $f$ by $g$.

Lemma: Given $a<b$ and $m\le M$, if $m\le g\le M$ on $[a,b]$, then $$m\le\frac{f(b)-f(a)}{b-a}\le M.$$

Proof: Define $$L(a)=g(a)\quad\text{and}\quad L(x)=\frac{f(x)-f(a)}{x-a},\ x\in(a,b].$$ By definition, $L$ is continuous on $[a,b]$, and it suffices to show that for every $\delta>0$, $$E_\delta:=\Big\{x\in[a,b]\,\Big|\, m-\delta\le L(y)\le M+\delta, \forall y\in[a,x] \Big\}=[a,b].$$ By definition and the continuity of $L$, we know that $E_\delta=[a,c]$ for some $c\in[a,b]$, and from $m\le g(a)\le M$ we know $c>a$. Then from $c\in E_\delta $ and $m\le g(c)\le M$ it is easy to see that $c<b$ is impossible. Therefore, $c=b$ and the lemma follows. $\quad\square$

Now let us show that $f$ is differentiable for any $x>0$. Since $g$ is continuous, given $0<h<x$, we can define $$m_h=\min_{y\in[x-h,x]}g(y),\quad M_h=\max_{y\in[x-h,x]}g(y),$$ and we know that $$\lim_{h\to 0^+}m_h=\lim_{h\to 0^+}M_h=g(y).$$ Due to the lemma, for $a=x-h$, $b=x$, $m=m_h$ and $M=M_h$, we have $$m_h\le\frac{f(x-h)-f(x)}{-h}\le M_h.$$ Let $h\to 0^+$, it follows that the left derivative of $f$ at $x$ exists and is equal to $g(x)$, i.e. $f$ is differentiable at $x$. $\quad\square$

  • $\begingroup$ Nice proof. Just a little typo: $\lim_{h\rightarrow0^+}m_h=g(x)$ rather than $g(y)$. $\endgroup$ – Fardad Pouran Jan 6 at 12:04

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