Absolute Value Function is Convex/Proof 2
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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 $x, y \in \R$.
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.
$\blacksquare$