Let $f$ be convex. If the right derivative of $f$ at $a$ equals left derivative of $f$ at $a$, then right derivative of $f$ is left continuous at $a$ Part of Question 9C in the Chapter 11 Appendix of Spivak's Calculus requires the reader to show that:

Let $f$ be convex. If the right derivative of $f$ at $a$ equals the left derivative of $f$ at $a$, then the right derivative of $f$ is left continuous at $a$. i.e. \begin{align}f'_+(a)=f'_-(a) \rightarrow\displaystyle \lim_{b \to a^-}f'_+(b)=f'_+(a) \end{align}

I am running into some difficulties understanding the final line of the proof that is offered by the solution manual. I will provide the solution manual's full proof, followed by some commentary on what the author is doing at each step and where, in particular, I am having trouble.

Spivak's Provided Proof

Given $\varepsilon \gt 0$, choose $c \lt a$ so that $f'_+(a)-\varepsilon=f'_-(a)-\varepsilon \lt \frac{f(a)-f(c)}{a-c}$. Then, if $c \lt b \lt a$, the secant line through $\left(b,f(b) \right)$ and $\left(a,f(a) \right)$ lies between the tangent line at $a$ and the secant line through $\left(c,f(c) \right)$ and $\left(a,f(a) \right)$, i.e., $f'_+(a)-\varepsilon\lt\frac{f(a)-f(c)}{a-c} \lt \frac{f(a)-f(b)}{a-b} \lt f'_+(a)$. This shows that $\displaystyle \lim_{b \to a^-}f'_+(b)=f'_+(a)$


The first line is justified because $f'_-(a)$ is defined as the supremum of the set $S=\left\{\frac{f(a)-f(x)}{a-x} : x \in I_{\text{convex}} \land x \lt a \right\}$. The final inequality of $f'_+(a)-\varepsilon\lt\frac{f(a)-f(c)}{a-c} \lt \frac{f(a)-f(b)}{a-b} \lt f'_+(a)$ is also correct because, and is, perhaps, made more apparent by expanding it to the following form:
$$f'_+(a)-\varepsilon\lt\frac{f(a)-f(c)}{a-c} \lt \frac{f(a)-f(b)}{a-b} \lt f'_-(a)= f'_+(a)$$
The fact that $\frac{f(a)-f(c)}{a-c} \lt \frac{f(a)-f(b)}{a-b}$ is simply a reformulation of the definition of convexity.
My confusion is why does this final inequality show that $\displaystyle \lim_{b \to a^-}f'_+(b)=f'_+(a)$?
My only guess is that the author's implicit argument is that for any $b \in (c,a)$, we have that:
$$f'_+(a)-\varepsilon \lt f'_+(b) \lt f'_+(a)$$, which necessarily implies that:
$$-\varepsilon \lt f'_+(b)-f'_+(a) \lt 0 \lt \varepsilon \implies \left|f'_+(b)-f'_+(a) \right| \lt \varepsilon$$, which is the definition of left continuity of $f'_+$ at $a$.
But why should it be the case that for any $b \in (c,a)$, $f'_+(a)-\varepsilon \lt f'_+(b) \lt f'_+(a)$?
My first guess was to demonstrate that for an arbitrary $b$, I can show that for any $x \in (b,a)$: $\frac{f(x)-f(b)}{x-b} \gt \frac{f(a)-f(c)}{a-c}$, which would necessarily imply that $f'_+(b) \geq \frac{f(a)-f(c)}{a-c}$ and therefore $f'_+(a)-\varepsilon \lt f'_+(b) \lt f'_+(a)$. However, I am having troubles demonstrating this.
Any insights would be appreciated.
 A: I too find Spivak's proof very terse. Please let me know what you think about the
following:
Lemma. For any convex $f$ the left derivative $f'_-$ is increasing and continuous from the left.
Proof. By convexity we have for all $x<y<z\,$
$$\tag{1}
\frac{f(y)-f(x)}{y-x}\le \frac{f(z)-f(x)}{z-x}\le \frac{f(z)-f(y)}{z-y}\,.
$$
The first inequality implies $f_-'(y)\le f_-'(z)$ (the left derivative is increasing). The two inequalities also imply that the quotient
$$
\frac{f(y)-f(x)}{y-x}
$$
is increasing in $y$ and in $x\,.$ Therefore,
$$
f_-'(y)=\sup_{x<y}\frac{f(y)-f(x)}{y-x}\,.
$$
Because this is increasing in $y$,
$$
\lim_{y\to z^-}f_-'(y)=\sup_{y<z}f_-'(y)=\sup_{y<z}\sup_{x<y}\frac{f(y)-f(x)}{y-x}\,.
$$
These two sups can be interchanged giving
$$
\sup_{x<z}\sup_{z>y>x}\frac{f(y)-f(x)}{y-x}\,.
$$
Because $f$ is continuous and the qotient increasing in $y$ this is
$$
\sup_{x<z}\frac{f(z)-f(x)}{z-x}=f_-'(z)\,.
$$
In other words, $f_-$ is continuous from the left.$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\square$
From (1) it also follows that
$$\tag{2}
f_-'(y)\le f'_+(y)\le f'_-(z)\le f_+'(z)\,\quad \forall y<z.
$$
This and the Lemma now implies
$$
\lim_{y\to z^-}f_-'(y)=\lim_{y\to z^-}f_+'(y)=f'_-(z)\,.
$$
Finally, if $f'_-(z)=f'_+(z)$ we get in particular Spivak's result:
$$
\lim_{y\to z^-}f_+'(y)=f'_+(z)\,.
$$
