Absolute Value Function is Convex/Proof 2

From ProofWiki
Jump to navigation Jump to search


Let $f: \R \to \R$ be the absolute value function on the real numbers.

Then $f$ is convex.


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.