characterization of almost convex functions

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

Proof:

The proof is based on some simple observations about the values of an almost convex function. Suppose that 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 $[b,\mathrm{\infty })$. By the result of the last paragraph, it suffices to show that $f$ is non-decreasing in the open half-line $(c,\mathrm{\infty })$. This is tantamount to showing that, if , 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 and $f(a)\ge f(b)$, then $f$ is nonincreasing on the half-line $(-\mathrm{\infty },a]$.

Now assume that $f$ is almost convex but not monotonic. By the hypothesis of nonmomotonicity, there must exist 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 $(-\mathrm{\infty },a]$ and nondecreasing on $[c,+\mathrm{\infty })$.

Let $L$ be the set of all real numbers $q$ such that $f$ is nondecreasing on the interval $(q,+\mathrm{\infty })$. 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 . This follows from the observation that $f$ cannot be nondecreasing on $(q,+\mathrm{\infty })$ 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 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 $(p,+\mathrm{\infty })$. We will now show that $f$ is nonincreasing on the half-line $(-\mathrm{\infty },p)$. Suppose that . Then, by the choice of $p$, the function $f$ is not nondecreasing on the half-line $(q,+\mathrm{\infty })$. This means that there must exist $a,b$ such that and $f(a)>f(b)$. By the result demonstrated above, it follows that $f$ is nonincreasing on $(-\mathrm{\infty },a)$, hence, since , in particular, $f$ is nononicreasing on $(-\mathrm{\infty },q)$. Since $f$ is nonincreasing on $(-\mathrm{\infty },q)$ for all $q$, it is the case that $f$ is nonincreasing on $(-\mathrm{\infty },p)$.