Absolute Value Function is Convex/Proof 1

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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:

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


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.

$\blacksquare$

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