Convex Functions

Definition

$g$ is convex on the interval $(a, b)$ if $\forall x_{1}, x_{2} \in (a, b), 0 < k < 1$
We have $g (k x_{1} + (1 - k) x_{2}) \leq k g (x_{1}) + (1 - k) g (x_{2})$

When we change $k$ , we just change the point in between $x_{1}, x_{2}$ .
$g (k x_{1} + (1 - k) x_{2})$
$k g (x_{1}) + (1 - k) g (x_{2})$
Visually:
This is very time consuming to prove a function is convex via this definition.
Quickly show it's convex via second derivative test.
- $g (x)$ is convex if $g^{″} (x) > 0$
Example:
- $g (x) = \ln (x)$ not convex
- $g (x) = - \ln (x)$ is convex
- $g^{'} (x) = - \frac{1}{x}$
- $g^{″} (x) = - \frac{1}{x^{2}} > 0$
- $g (x) = x^{2}$ is convex
- $g^{'} (x) = 2 x$
- $g^{″} (x) = 2 > 0$ so it's convex