Convexity

Rudin RC p. 61

Let \phi:(a,b) \to \mathbf{R}, -\infty \le a < b \le \infty. \phi is called convex if

\phi((1 - \lambda)x + \lambda y) \le (1 - \lambda)\phi(x) + \lambda \phi(y) \qquad (1)
holds whenever x, y \in (a,b), 0 \le \lambda \le 1.

Show that this is equivalent to
\frac{\phi(t) - \phi(s)}{t - s} \le \frac{\phi(u) - \phi(t)}{u - t} \qquad (2)
whenever a < s < t < u < b.

It is easy to see that \neg (2) is equivalent to \exists x < y \in (a,b), \lambda \in (0,1), such that

\frac{ \phi((1 - \lambda) x + \lambda y) - \phi(x)}{(1 - \lambda)x + \lambda y - x} > \frac{\phi(y) - \phi((1 - \lambda)x + \lambda y)}{y - ((1 - \lambda)x + \lambda y)}

\Leftrightarrow \frac{\phi((1 - \lambda)x + \lambda y) - \phi(x)}{- \lambda x + \lambda y} > \frac{\phi(y) - \phi((1 - \lambda)x + \lambda y)}{(1 - \lambda)y - (1 - \lambda) x}

\Leftrightarrow \frac{(1- \lambda)[\phi((1 - \lambda)x + \lambda y) - \phi(x)]}{y - x} > \frac{\lambda[ \phi(y) - \phi((1 - \lambda)x + \lambda y)]}{y - x}

\Leftrightarrow (1 - \lambda)\phi((1 - \lambda)x + \lambda y) - (1 - \lambda)\phi(x) > \lambda \phi(y) - \lambda \phi((1 - \lambda)x + \lambda y)

\Leftrightarrow \phi((1 - \lambda) x + \lambda y) > \lambda \phi(y) + (1 - \lambda)\phi(x) \qquad (3),

and (3) \Rightarrow \neg (1).  Now, if \neg (1) holds, then \exists x < y \in (a,b), \lambda \in (0,1), such that (3) holds, so that we obtain \neg (2).  Thus, (1),(2) are equivalent.

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