A real function $f$ is almost convex iff it is monotonic or there exists $p \in \mathbb{R}$ such that $f$ is nonincreasing on the half-line $(-\infty, p)$ and nondecreasing on the half-line $(p,+\infty)$ Proof:

The proof is based on some simple observations about the values of an almost convex function. Suppose that $a < b$ and $f(a) \le f(b)$ Then for any $c > b$ it must be the case that $f(b) \le f(c)$ This follows from the fact that, by definition of almost convex, either $f(b) \le f(a)$ or $f(b) \le f(c)$ Since the first option is excluded by assumption, the second option must be true.

Furthermore, with $a$ $b$ as above, $f$ is nondecreasing in the half-line $[b,\infty)$ By the result of the last paragraph, it suffices to show that $f$ is non-decreasing in the open half-line $(c,\infty)$ This is tantamount to showing that, if $c < d < e$ then $f(d) \le f(e)$ From the conlusion of last paragraph, we already know that $f(c) \le f(d)$ Applying the result shown in the last paragraph to this conclusion, we further conclude that $f(d) \le f(e)$ as desired.

By replacing ``$\le$ ' by ``$\ge$ ' in the above two paragraphs suitably, we also can likewise that, if $a < b$ and $f(a) \ge f(b)$ then $f$ is nonincreasing on the half-line $(-\infty,a]$

Now assume that $f$ is almost convex but not monotonic. By the hypothesis of nonmomotonicity, there must exist $a < b < c$ such that it is the case that neither $f(a) \le f(b) \le f(c)$ nor $f(a) \ge f(b) \ge f(c)$ Furthermore, by almost-convexity, it follows that $f(b) \le f(a)$ and $f(b) \le f(c)$ This, in turn, implies that $f$ is nonincreasing on $(-\infty,a]$ and nondecreasing on $[c,+\infty)$

Let $L$ be the set of all real numbers $q$ such that $f$ is nondecreasing on the interval $(q,+\infty)$ This set is not empty because $c \in L$ It is a proper subset of the real line because, for instance, $q \notin L$ whenever $q < a$ This follows from the observation that $f$ cannot be nondecreasing on $(q,+\infty)$ because $f(a) > f(b)$ Also, $L$ must be a proper subset of the real line, because, if it were not, $f$ would be nondecreasing on the whole real line, which is contrary to assumption.

Note that, if $r < q$ and $q \notin L$ then $r \notin L$ as well. This is an expression of the fact that, if a function is not monotonic on a set, it is not monotonic on a superset, which is the contrapositive of the assertion that a the resticition of a function which is monotonic on a set to a subset is still monotonic. Since there exists a real number $r$ such that $r \notin L$ this means that $r$ is a lower bound for $L$ Since $L$ is bounded from below and not empty, it follows that $L$ has a greatest lower bound, which we shall call $p$

By construction, $f$ is non-decreasing on the half-line $(p,+\infty)$ We will now show that $f$ is nonincreasing on the half-line $(-\infty,p)$ Suppose that $q < p$ Then, by the choice of $p$ the function $f$ is not nondecreasing on the half-line $(q,+\infty)$ This means that there must exist $a,b$ such that $q < a < b$ and $f(a) > f(b)$ By the result demonstrated above, it follows that $f$ is nonincreasing on $(-\infty,a)$ hence, since $q < a$ in particular, $f$ is nononicreasing on $(-\infty,q)$ Since $f$ is nonincreasing on $(-\infty,q)$ for all $q$ it is the case that $f$ is nonincreasing on $(-\infty,p)$