## 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.