Equivalence of Definitions of Concave Real Function

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Theorem

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

The following definitions of the concept of Concave Real Function are equivalent:

Definition 1

$f$ is concave on $I$ if and only if:

$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) \ge \alpha f \left({x}\right) + \beta f \left({y}\right)$

Definition 2

$f$ is concave on $I$ if and only if:

$\displaystyle \forall x_1, x_2, x_3 \in I: x_1 < x_2< x_3: \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \ge \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$

Definition 3

$f$ is concave on $I$ if and only if:

$\forall x_1, x_2, x_3 \in I: x_1 < x_2< x_3: \dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \ge \dfrac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}$


Proof

Let $f$ be concave real function on $I$ according to definition 1.

That is:

$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) \ge \alpha f \left({x}\right) + \beta f \left({y}\right)$

Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$.

As $\alpha + \beta = 1$, we have $x_2 = \alpha x_1 + \left({1 - \alpha}\right) x_3$.

Thus:

$\alpha = \dfrac {x_3 - x_2} {x_3 - x_1}, \beta = \dfrac {x_2 - x_1} {x_3 - x_1}$

So:

\((1):\quad\) \(\displaystyle f \left({x_2}\right)\) \(\ge\) \(\displaystyle \frac {x_3 - x_2} {x_3 - x_1} f \left({x_1}\right) + \frac {x_2 - x_1} {x_3 - x_1} f \left({x_3}\right)\) substituting for $\alpha$ and $\beta$
\(\displaystyle \iff \ \ \) \(\displaystyle \left({x_3 - x_1}\right) f \left({x_2}\right)\) \(\ge\) \(\displaystyle \left({x_3 - x_2}\right) f \left({x_1}\right) + \left({x_2 - x_1}\right) f \left({x_3}\right)\)
\(\displaystyle \iff \ \ \) \(\displaystyle \left({x_3 - x_1}\right) f \left({x_2}\right)\) \(\ge\) \(\displaystyle \left({x_3 - x_1 - x_2 + x_1}\right) f \left({x_1}\right) + \left({x_2 - x_1}\right) f \left({x_3}\right)\)
\(\displaystyle \iff \ \ \) \(\displaystyle \left({x_3 - x_1}\right) f \left({x_2}\right) - \left({x_3 - x_1}\right) f \left({x_1}\right)\) \(\ge\) \(\displaystyle \left({x_2 - x_1}\right) f \left({x_3}\right) - \left({x_2 - x_1}\right) f \left({x_1}\right)\)
\(\displaystyle \iff \ \ \) \(\displaystyle \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}\) \(\ge\) \(\displaystyle \frac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}\)


Again from $(1)$:

\((1):\quad\) \(\displaystyle f \left({x_2}\right)\) \(\ge\) \(\displaystyle \frac {x_3 - x_2} {x_3 - x_1} f \left({x_1}\right) + \frac {x_2 - x_1} {x_3 - x_1} f \left({x_3}\right)\)
\(\displaystyle \iff \ \ \) \(\displaystyle \left({x_3 - x_1}\right) f \left({x_2}\right)\) \(\ge\) \(\displaystyle \left({x_3 - x_2}\right) f \left({x_1}\right) + \left({x_2 - x_1}\right) f \left({x_3}\right)\)
\(\displaystyle \iff \ \ \) \(\displaystyle \left({x_3 - x_1}\right) f \left({x_2}\right)\) \(\ge\) \(\displaystyle \left({x_3 - x_1 - x_3 + x_2}\right) f \left({x_3}\right) + \left({x_3 - x_2}\right) f \left({x_1}\right)\)
\(\displaystyle \iff \ \ \) \(\displaystyle \left({x_3 - x_2}\right) f \left({x_3}\right) - \left({x_3 - x_2}\right) f \left({x_1}\right)\) \(\ge\) \(\displaystyle \left({x_3 - x_1}\right) f \left({x_3}\right) - \left({x_3 - x_1}\right) f \left({x_2}\right)\)
\(\displaystyle \iff \ \ \) \(\displaystyle \frac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}\) \(\ge\) \(\displaystyle \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}\)


Thus:

$\dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \ge \dfrac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1} \ge \dfrac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$


So:

$\dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \ge \dfrac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}$

demonstrating that $f$ is concave on $I$ according to definition 3, and:

$\dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \ge \dfrac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$

demonstrating that $f$ is concave on $I$ according to definition 2.


As each step is an equivalence, the argument reverses throughout.

$\blacksquare$