Continuous Midpoint-Convex Function is Convex

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Theorem

Let $f$ be a real function which is defined on a real interval $I$.

Let $f$ be midpoint-convex and continuous on $I$.


Then $f$ is convex.


Proof

Let $x, y \in I$ and let $t \in \closedint 0 1$.

Let $\sequence {t_n}_{n \mathop \in \N}$ be a sequence of rational numbers in $\closedint 0 1$ converging to $t$.


It follows that:

\(\ds \lim \limits_{n \mathop \to\infty} \paren {t_n x + \paren {1 - t_n} y}\) \(=\) \(\ds t x + \paren {1 - t} y\)

and:

\(\ds \lim \limits_{n \mathop \to \infty} \paren {t_n \map f x + \paren {1 - t_n} \map f y}\) \(=\) \(\ds t \map f x + \paren {1 - t} \map f y\)


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From Midpoint-Convex Function is Rational Convex, $f$ is rational convex.

From Continuous Mapping is Sequentially Continuous, $f$ is sequentially continuous.


This article, or a section of it, needs explaining.
In particular: Technically, we have sequential continuity defined under the $\mathsf{Pr} \infty \mathsf{fWiki}$ link above only in terms of the general topological space. In order to connect all the pieces here, we need either to invoke the clauses which state that the real number line is a topological space under which the properties of continuity "mean the same thing" as the traditionally real-analytical definition of continuity, or we need to define sequential continuity in terms of real analysis.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.
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As $f$ is rational convex:

$\forall n \in \N: \map f {t_n x + \paren {1 - t_n} y} \le t_n \map f x + \paren {1 - t_n} \map f y$


From sequential continuity of $f$ and Inequality Rule for Real Sequences:

\(\ds \map f {t x + \paren {1 - t} y}\) \(=\) \(\ds \lim_{n \mathop \to \infty} \map f {t_n x + \paren {1 - t_n} y}\)
\(\ds \) \(\le\) \(\ds \lim_{n \mathop \to \infty} \paren {t_n \map f x + \paren {1 - t_n} \map f y}\)
\(\ds \) \(=\) \(\ds t \map f x + \paren {1 - t} \map f y\)

which completes the proof.

$\blacksquare$


Also see


Sources

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