Absolute Value Function is Convex

Theorem

Let $f: \R \to \R$ be the absolute value function on the real numbers.

Then $f$ is convex.

Proof 1

Let $x_1, x_2, x_3 \in \R$ such that $x_1 < x_2 < x_3$.

Consider the expressions:

$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}$
$\dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$

The following cases are investigated:

$(1): \quad x_1, x_2, x_3 < 0$:

Then:

 $\ds \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}$ $=$ $\ds \frac {-\paren {x_2 - x_1} } {x_2 - x_1}$ Definition of Absolute Value $\ds$ $=$ $\ds -1$ $\ds \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ $=$ $\ds \frac {-\paren {x_3 - x_2} } {x_3 - x_2}$ Definition of Absolute Value $\ds$ $=$ $\ds -1$ $\ds \leadsto \ \$ $\ds \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}$ $\le$ $\ds \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ Definition of Convex Real Function

$(2): \quad x_1, x_2, x_3 > 0$:

Then:

 $\ds \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}$ $=$ $\ds \frac {x_2 - x_1} {x_2 - x_1}$ Definition of Absolute Value $\ds$ $=$ $\ds 1$ $\ds \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ $=$ $\ds \frac {x_3 - x_2} {x_3 - x_2}$ Definition of Absolute Value $\ds$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}$ $\le$ $\ds \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ Definition of Convex Real Function

$(3): \quad x_1 < 0, x_2, x_3 > 0$:
 $\ds \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}$ $=$ $\ds \frac {x_2 - \paren {-x_1} } {x_2 - x_1}$ Definition of Absolute Value $\ds$ $=$ $\ds \frac {\paren {x_2 + x_1} + \paren {x_1 - x_1} } {x_2 - x_1}$ $\ds$ $=$ $\ds \frac {x_2 - x_1} {x_2 - x_1} + \frac {2 x_1} {x_2 - x_1}$ $\ds$ $=$ $\ds 1 + \frac {2 x_1} {x_2 - x_1}$ $\ds$ $<$ $\ds 1$ as $2 x_1 < 0$ $\ds \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ $=$ $\ds \frac {x_3 - x_2} {x_3 - x_2}$ Definition of Absolute Value $\ds$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}$ $\le$ $\ds \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ Definition of Convex Real Function

$(4): \quad x_1, x_2 < 0, x_3 > 0$:
 $\ds \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}$ $=$ $\ds \frac {-\paren {x_2 - x_1} } {x_2 - x_1}$ Definition of Absolute Value $\ds$ $=$ $\ds -1$ $\ds \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ $=$ $\ds \frac {x_3 - \paren {-x_2} } {x_3 - x_3}$ Definition of Absolute Value $\ds$ $=$ $\ds \frac {x_3 + x_2 + \paren {x_3 - x_3} } {x_3 - x_2}$ $\ds$ $=$ $\ds \frac {-x_3 + x_2 + x_3 + x_3} {x_3 - x_2}$ $\ds$ $=$ $\ds \frac {-\paren {x_3 - x_2} } {x_3 - x_2} + \frac {2 x_3} {x_3 - x_2}$ $\ds$ $=$ $\ds -1 + \frac {2 x_3} {x_3 - x_2}$ $\ds$ $>$ $\ds -1$ as $2 x_3 > 0$ $\ds \leadsto \ \$ $\ds \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}$ $\le$ $\ds \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ Definition of Convex Real Function

Thus for all cases, the condition for $f$ to be convex is fulfilled.

Hence the result.

$\blacksquare$

Proof 2

Let $x, y \in \R$.

Let $\alpha, \beta \in \R_{\ge 0}$ where $\alpha + \beta = 1$.

 $\ds \map f {\alpha x + \beta y}$ $=$ $\ds \size {\alpha x + \beta y}$ Definition of $f$ $\ds$ $\le$ $\ds \size {\alpha x} + \size {\beta y}$ Triangle Inequality for Real Numbers $\ds$ $=$ $\ds \size \alpha \size x + \size \beta \size y$ Absolute Value Function is Completely Multiplicative $\ds$ $=$ $\ds \alpha \size x + \beta \size y$ Definition of Absolute Value $\ds$ $=$ $\ds \alpha \, \map f x + \beta \, \map f y$ Definition of $f$

Hence the result by definition of convex real function.

$\blacksquare$