Equivalence of Definitions of Strictly Convex Real Function
Theorem
Let $f$ be a real function which is defined on a real interval $I$.
The following definitions of the concept of Strictly Convex Real Function are equivalent:
Definition 1
$f$ is strictly convex on $I$ if and only if:
- $\forall x, y \in I, x \ne y: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} < \alpha \map f x + \beta \map f y$
Definition 2
$f$ is strictly convex on $I$ if and only if:
- $\forall x_1, x_2, x_3 \in I: x_1 < x_2 < x_3: \dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} < \dfrac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$
Definition 3
$f$ is strictly convex on $I$ if and only if:
- $\forall x_1, x_2, x_3 \in I: x_1 < x_2 < x_3: \dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} < \dfrac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}$
Proof
Let $f$ be strictly convex real function on $I$ according to definition 1.
That is:
- $\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} < \alpha \, \map f x + \beta \, \map f y$
Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$.
As $\alpha + \beta = 1$, we have:
- $x_2 = \alpha x_1 + \paren {1 - \alpha} x_3$
Thus:
- $\alpha = \dfrac {x_3 - x_2} {x_3 - x_1}$
- $\beta = \dfrac {x_2 - x_1} {x_3 - x_1}$
So:
| \(\text {(1)}: \quad\) | \(\ds \map f {x_2}\) | \(<\) | \(\ds \frac {x_3 - x_2} {x_3 - x_1} \, \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}\) | substituting for $\alpha$ and $\beta$ | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \paren {x_3 - x_1} \map f {x_2}\) | \(<\) | \(\ds \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \paren {x_3 - x_1} \map f {x_2}\) | \(<\) | \(\ds \paren {x_3 - x_1 - x_2 + x_1} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \paren {x_3 - x_1} \map f {x_2} - \paren {x_3 - x_1} \map f {x_1}\) | \(<\) | \(\ds \paren {x_2 - x_1} \map f {x_3} - \paren {x_2 - x_1} \map f {x_1}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}\) | \(<\) | \(\ds \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}\) |
Again from $(1)$:
| \(\text {(1)}: \quad\) | \(\ds \map f {x_2}\) | \(<\) | \(\ds \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \paren {x_3 - x_1} \map f {x_2}\) | \(<\) | \(\ds \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \paren {x_3 - x_1} \map f {x_2}\) | \(<\) | \(\ds \paren {x_3 - x_1 - x_3 + x_2} \map f {x_3} + \paren {x_3 - x_2} \map f {x_1}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \paren {x_3 - x_2} \map f {x_3} - \paren {x_3 - x_2} \map f {x_1}\) | \(<\) | \(\ds \paren {x_3 - x_1} \map f {x_3} - \paren {x_3 - x_1} \map f {x_2}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}\) | \(<\) | \(\ds \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}\) |
Thus:
- $\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1} < \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
So:
- $\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$
demonstrating that $f$ is strictly convex on $I$ according to definition 3, and:
- $\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
demonstrating that $f$ is strictly convex on $I$ according to definition 2.
As each step is an equivalence, the argument reverses throughout.
$\blacksquare$