I am looking at the proof given on this question given by "23rd". Continuous right derivative implies differentiability
I asked in a comment a question about 5 months ago, but didn't recieve a reply, so I am asking here instead. I have one statement in the proof I dont understand, I have marked it bold, could you please explain it to me? It is the statement "Then from $c\in E_\delta $ and $m\le g(c)\le M$ it is easy to see that $c<b$ is impossible".
Here is what the person who asked the question wanted to prove:
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. Let us denote the right derivative of $f$ by $g$.
Here is the proof by "23rd":
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 $x$. $\quad\square$
