Absolute Value Function is Convex

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Absolute_Value_Function_is_Convex&oldid=342652"