Understanding the proof of: Every convex function is continuous I am trying to understand the following proof:

Theorem 2.10. If $f$ is a convex function defined on an open interval $(a, b)$ then $f$ is continuous on $(a, b)$
Proof. Suppose $f$ is convex on $(a, b),$ and let $[c, d] \subseteq(a, b) .$ Choose $c_{1}$ and $d_{1}$ such that
$$
a<c_{1}<c<d<d_{1}<b.
$$
If $x, y \in[c, d]$ with $x<y,$ we have from Lemma 2.9 (see Figure 4$)$ that
$$
\frac{f(y)-f(x)}{y-x} \leq \frac{f(d)-f(y)}{d-y} \leq \frac{f\left(d_{1}\right)-f(d)}{d_{1}-d}
$$
and
$$
\frac{f(y)-f(x)}{y-x} \geq \frac{f(x)-f(c)}{x-c} \geq \frac{f(c)-f\left(c_{1}\right)}{c-c_{1}},
$$
showing the set
$$
\left\{\left|\frac{f(y)-f(x)}{y-x}\right|: c \leq x<y \leq d\right\}
$$
is bounded by $M>0 .$ It follows $|f(y)-f(x)| \leq M|y-x|,$ and therefore $f$ is uniformly continuous on $[c, d] .$ Recalling that uniform continuity implies continuity, we have shown that $f$ is continuous on $[c, d] .$ since the interval $[c, d]$ was arbitrary, $f$ is continuous on $(a, b)$. ${}^2$ $\square$

(transcribed from this screenshot)
My questions:

*

*Where did the modulus values in the expression $\left\{\left|\dfrac{f(y)-f(x)}{y-x}\right|\right\}$ come from?

*What about $M=0$? I think that case should also be addressed, although it is trivial. I think the idea is that if $M=0$, then $f$ is constant and hence continuous. But, how can we show that rigorously?


Since the author found to numbers $\alpha$ and $\beta$ such that you always have, when $c\leqslant x<y\leqslant d$,$$\frac{f(y)-f(x)}{y-x}\leqslant\alpha$$and$$\frac{f(y)-f(x)}{y-x}\geqslant\beta,$$then the set$$\left\{\frac{f(y)-f(x)}{y-x}\,\middle|\,c\leqslant x<y\leqslant d\right\}$$is bounded and therefore the set$$\left\{\left|\frac{f(y)-f(x)}{y-x}\right|\,\middle|\,c\leqslant x<y\leqslant d\right\}$$is bounded too. So, you can take some $M>0$ such that$$c\leqslant x<y\leqslant d\implies\left|\frac{f(y)-f(x)}{y-x}\right|<M.$$And, since you took $M>0$, there is no need to bother with the possibility that $M=0$.
