Absolute Value Function is Convex/Proof 1

Theorem

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

Then $f$ is convex.

Proof

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

Consider the expressions:

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

The following cases are investigated:

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

Then:

 $\displaystyle \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}$ $=$ $\displaystyle \frac {-\left({x_2 - x_1}\right)} {x_2 - x_1}$ Definition of Absolute Value $\displaystyle$ $=$ $\displaystyle -1$ $\displaystyle \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$ $=$ $\displaystyle \frac {-\left({x_3 - x_2}\right)} {x_3 - x_2}$ Definition of Absolute Value $\displaystyle$ $=$ $\displaystyle -1$ $\displaystyle \implies \ \$ $\displaystyle \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}$ $\le$ $\displaystyle \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$ Definition of Convex Real Function

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

Then:

 $\displaystyle \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}$ $=$ $\displaystyle \frac {x_2 - x_1} {x_2 - x_1}$ Definition of Absolute Value $\displaystyle$ $=$ $\displaystyle 1$ $\displaystyle \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$ $=$ $\displaystyle \frac {x_3 - x_2} {x_3 - x_2}$ Definition of Absolute Value $\displaystyle$ $=$ $\displaystyle 1$ $\displaystyle \implies \ \$ $\displaystyle \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}$ $\le$ $\displaystyle \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$ Definition of Convex Real Function

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

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

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

Hence the result.

$\blacksquare$