Convex functions

A function $f(x)$ is convex if for any real numbers $a<b$, each point $(c,d)$ on the line segment joining $(a,f(a))$ and $(b,f(b))$ lies above or at the point $(c,f(c))$ on the graph of $f$ with the same $x$-coordinate.

Algebraically, this condition says that

$$f((1-t)a+tb) \le (1-t)f(a) + t f(b).$$ (1)

whenever $a<b$ and for all $t \in [0,1]$. (The left hand side represents the height of the graph of the function above the $x$-value $x=(1-t)a+tb$ which is a fraction $t$ of the way from $a$ to $b$, and the right hand side represents the height of the line segment above the same $x$-value.)

Those who know what a convex set in geometry is can interpret the condition as saying that the set $S=\{(x,y):y \ge f(x)\}$ of points above the graph of $f$ is a convex set. Loosely speaking, this will hold if the graph of $f$ curves in the shape of a smile instead of a frown. For example, the function $f(x)=x^2$ is convex, as is $f(x)=x^n$ for any positive even integer.

One can also speak of a function $f(x)$ being convex on an interval $I$. This means that the condition (1) above holds at least when $a,b \in I$ (and $a<b$ and $t \in [0,1]$). For example, one can show that $f(x)=x^3$ is convex on $[0,\infty)$, and that $f(x)=\sin x$ is convex on $(-\pi,0)$.

Finally one says that a function $f(x)$ on an interval $I$ is strictly convex, if

$$f((1-t)a+tb) < (1-t)f(a) + t f(b)$$

whenever $a,b \in I$ and $a<b$ and $t \in (0,1)$. In other words, the line segment connecting two points on the graph of $f$ should lie entirely above the graph of $f$, except where it touches at its endpoints.

For convenience, here is a brief list of some convex functions. In these, $k$ represents a positive integer, $r,s$ represent real constants, and $x$ is the variable. In fact, all of these are strictly convex on the interval given, except for $x^r$ and $-x^r$ when $r$ is 0 or $1$.

$$x^{2k}, {\mathbb{R}}$$

$$x^r, [0,\infty) r \ge 1$$

$$-x^r, [0,\infty) r \in [0,1]$$

$$x^r, (0,\infty) r \le 0$$

$$-\log x, (0,\infty)$$

$$-\sin x, [0,\pi]$$

$$-\cos x, [-\pi/2,\pi/2]$$

$$\tan x, [0,\pi/2)$$

$$e^x, {\mathbb{R}}$$

$$r/(s+x) (-s,\infty) r>0$$

A sum of convex functions is convex. Adding a constant or linear function to a function does not affect convexity.

Remarks (for those who know about continuity and derivatives):
If one wants to prove rigorously that a function is convex, instead of just guessing it from the graph, it is often easier to use one of the criteria below instead of the definition of convexity.

1. Let $f(x)$ be a continuous function on an interval $I$. Then $f(x)$ is convex if and only if $(f(a)+f(b))/2 \ge f((a+b)/2)$ holds for all $a,b \in I$. Also, $f(x)$ is strictly convex if and only if $(f(a)+f(b))/2 > f((a+b)/2)$ whenever $a,b \in I$ and $a<b$.

2. Let $f(x)$ be a differentiable function on an interval $I$. Then $f(x)$ is convex if and only if $f'(x)$ is increasing on $I$. Also, $f(x)$ is strictly convex if and only if $f'(x)$ is strictly increasing on the interior of $I$.

3. Let $f(x)$ be a twice differentiable function on an interval $I$. Then $f(x)$ is convex if and only if $f''(x) \ge 0$ for all $x \in I$. Also, $f(x)$ is strictly convex if and only if $f''(x) > 0$ for all $x$ in the interior of $I$.