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# characterization of almost convex functions

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 $f(a)\leq f(b)$. Then for any $c>b$, it must be the case that $f(b)\leq f(c)$. This follows from the fact that, by definition of almost convex, either $f(b)\leq f(a)$ or $f(b)\leq f(c)$. 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 $f(d)\leq f(e)$. From the conlusion of last paragraph, we already know that $f(c)\leq f(d)$. Applying the result shown in the last paragraph to this conclusion, we further conclude that $f(d)\leq f(e)$, as desired.

By replacing “$\leq$” by “$\geq$” in the above two paragraphs suitably, we also can likewise that, if $a<b$ and $f(a)\geq f(b)$, 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 $f(a)\leq f(b)\leq f(c)$ nor $f(a)\geq f(b)\geq f(c)$. Furthermore, by almost-convexity, it follows that $f(b)\leq f(a)$ and $f(b)\leq f(c)$. 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 $(-\infty,p)$.

## Mathematics Subject Classification

26A51*no label found*

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