Theorem: Absolute Value Function is Convex

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:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}$
:$\dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
The following cases are investigated:
:$(1): \quad x_1, x_2, x_3 < 0$:
Then:
{{begin-eqn}}
{{eqn | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
      | r = \frac {-\paren {x_2 - x_1} } {x_2 - x_1}
      | c = {{Defof|Absolute Value}}
}}
{{eqn | r = -1
      | c = 
}}
{{eqn | l = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
      | r = \frac {-\paren {x_3 - x_2} } {x_3 - x_2}
      | c = {{Defof|Absolute Value}}
}}
{{eqn | r = -1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
      | o = \le
      | r = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
      | c = {{Defof|Convex Real Function}}
}}
{{end-eqn}}
:$(2): \quad x_1, x_2, x_3 > 0$:
Then:
{{begin-eqn}}
{{eqn | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
      | r = \frac {x_2 - x_1} {x_2 - x_1}
      | c = {{Defof|Absolute Value}}
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | l = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
      | r = \frac {x_3 - x_2} {x_3 - x_2}
      | c = {{Defof|Absolute Value}}
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
      | o = \le
      | r = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
      | c = {{Defof|Convex Real Function}}
}}
{{end-eqn}}
:$(3): \quad x_1 < 0, x_2, x_3 > 0$:
{{begin-eqn}}
{{eqn | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
      | r = \frac {x_2 - \paren {-x_1} } {x_2 - x_1}
      | c = {{Defof|Absolute Value}}
}}
{{eqn | r = \frac {\paren {x_2 + x_1} + \paren {x_1 - x_1} } {x_2 - x_1} 
      | c = 
}}
{{eqn | r = \frac {x_2 - x_1} {x_2 - x_1} + \frac {2 x_1} {x_2 - x_1}
      | c = 
}}
{{eqn | r = 1 + \frac {2 x_1} {x_2 - x_1}
      | c = 
}}
{{eqn | o = <
      | r = 1
      | c = as $2 x_1 < 0$
}}
{{eqn | l = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
      | r = \frac {x_3 - x_2} {x_3 - x_2}
      | c = {{Defof|Absolute Value}}
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
      | o = \le
      | r = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
      | c = {{Defof|Convex Real Function}}
}}
{{end-eqn}}
:$(4): \quad x_1, x_2 < 0, x_3 > 0$:
{{begin-eqn}}
{{eqn | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
      | r = \frac {-\paren {x_2 - x_1} } {x_2 - x_1}
      | c = {{Defof|Absolute Value}}
}}
{{eqn | r = -1
      | c = 
}}
{{eqn | l = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
      | r = \frac {x_3 - \paren {-x_2} } {x_3 - x_3}
      | c = {{Defof|Absolute Value}}
}}
{{eqn | r = \frac {x_3 + x_2 + \paren {x_3 - x_3} } {x_3 - x_2} 
      | c = 
}}
{{eqn | r = \frac {-x_3 + x_2 + x_3 + x_3} {x_3 - x_2} 
      | c = 
}}
{{eqn | r = \frac {-\paren {x_3 - x_2} } {x_3 - x_2} + \frac {2 x_3} {x_3 - x_2}
      | c = 
}}
{{eqn | r = -1 + \frac {2 x_3} {x_3 - x_2}
      | c = 
}}
{{eqn | o = >
      | r = -1
      | c = as $2 x_3 > 0$
}}
{{eqn | ll= \leadsto
      | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
      | o = \le
      | r = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
      | c = {{Defof|Convex Real Function}}
}}
{{end-eqn}}
Thus for all cases, the condition for $f$ to be convex is fulfilled.
Hence the result.
{{qed}}