Convex Function

Many inequalities in analysis and probability reduce to a single geometric observation: for a “bowl-shaped” function, the straight line between two points on the graph never dips below the graph itself. Formalising that observation gives you the notion of a convex function.

The chord condition

Let $I \subseteq \mathbb{R}$ be an interval and $f : I \to \mathbb{R}$.

Definition. $f$ is convex on $I$ if for every $x, y \in I$ and every $\lambda \in [0, 1]$,

$$f\!\left(\lambda x + (1-\lambda)y\right) \;\leq\; \lambda f(x) + (1-\lambda)f(y). \tag{1}$$

The point $\lambda x + (1-\lambda)y$ is a convex combination of $x$ and $y$: as $\lambda$ ranges over $[0,1]$ it traces the segment from $y$ to $x$. The right-hand side $\lambda f(x) + (1-\lambda)f(y)$ is the corresponding point on the chord from $(x, f(x))$ to $(y, f(y))$.

In words: the graph of $f$ lies at or below every chord.

Strict convexity

Definition. $f$ is strictly convex on $I$ if inequality $(1)$ is strict whenever $x \neq y$ and $\lambda \in (0, 1)$:

$$f\!\left(\lambda x + (1-\lambda)y\right) \;<\; \lambda f(x) + (1-\lambda)f(y).$$

Strict convexity rules out any flat portion of the graph lying on a chord. Every strictly convex function is convex, but not vice versa (the identity $f(x) = x$ is convex but not strictly convex).

Concavity

$f$ is (strictly) concave if $-f$ is (strictly) convex, i.e. if the inequality in $(1)$ is reversed. Concave functions are “cap-shaped”: every chord lies at or below the graph.

Examples

Function	Domain	Convex?	Strictly?
$x^2$	$\mathbb{R}$	Yes	Yes
$e^x$	$\mathbb{R}$	Yes	Yes
$\|x\|$	$\mathbb{R}$	Yes	No (flat at $x = 0$ on the chord from $-a$ to $a$)
$\ln x$	$(0, \infty)$	No (concave)	—
$-x^2$	$\mathbb{R}$	No (concave)	—
$c$ (constant)	$\mathbb{R}$	Yes	No

Verifying $x^2$. For $\lambda \in [0,1]$ and $x, y \in \mathbb{R}$:

$$(\lambda x + (1-\lambda)y)^2 \;\leq\; \lambda x^2 + (1-\lambda)y^2$$

is equivalent to $\lambda(1-\lambda)(x-y)^2 \geq 0$, which holds for all $\lambda \in [0,1]$ and is strict when $x \neq y$ and $\lambda \in (0,1)$. So $x^2$ is strictly convex.

Equivalent two-point form

Dividing both sides of $(1)$ by setting $t = \lambda$ and rearranging, convexity can also be read as: for any three points $x < z < y$ in $I$,

$$\frac{f(z) - f(x)}{z - x} \;\leq\; \frac{f(y) - f(x)}{y - x} \;\leq\; \frac{f(y) - f(z)}{y - z}.$$

Each fraction is the slope of a secant, so this says: the secant slopes from a fixed left point are non-decreasing as the right endpoint moves right. This monotone-slope property is fully equivalent to $(1)$ and is often the quickest way to check convexity from a graph.

Summary

• $f$ is convex on $I$ if every chord from $(x, f(x))$ to $(y, f(y))$ lies at or above the graph: $f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y)$ for all $x, y \in I$, $\lambda \in [0,1]$.
• Strict convexity requires a strict inequality for $x \neq y$ and $\lambda \in (0,1)$.
• Concavity reverses the inequality; $f$ is concave iff $-f$ is convex.
• Standard examples: $x^2$ and $e^x$ are strictly convex; $\ln x$ is strictly concave.