Absolute Value Function is Convex
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
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 12.15$ Example