convex function

Convex function

Let $I$ be an interval and $f:I\to \mathbb{R}$. $f$ is convex on $I$ if for all $\lambda \in (0,1)$ and $x_0,x_1\in I$:

$$f((1-\lambda )x_0+\lambda x_1)\le (1-\lambda )f(x_0)+\lambda f(x_1)$$

It is concave if the inequality runs the other way ($\ge$); equivalently, if $-f$ is convex.

$(1-\lambda )x_0+\lambda x_1$ is a convex combination of the two points: as $\lambda$ runs from $0$ to $1$ it sweeps every point $x_{\lambda }$ between $x_0$ and $x_1$. The left side is the height of the graph at $x_{\lambda }$; the right side is the height of the chord joining $(x_0,f(x_0))$ and $(x_1,f(x_1))$ at that same $x_{\lambda }$. Convex means the graph stays below its chords, concave means above.

$x_{\lambda }=(1-\lambda )x_0+\lambda x_1$. The chord value (the convex combination of the two heights) sits above the curve value $f(x_{\lambda })$, at every interior point.

A function $f$ is convex iff its epigraph is a convex set.

Convex functions $x^2$ $e^x$ $|x|$

Concave functions

$\log x$
$\sqrt{x}$.