Convex functions

Let \(I\) be an interval over \(\mathbb{R}\). It is well-known in mathematics literature that \(f : I \rightarrow \mathbb{R}\) is said to be convex if for all \(x, y \in I\) and \(t \in [0,1]\), \[f\left(tx + (1-t)y\right) \leq t f(x) + (1-t)f(y).\] Here, \([0,1]\) denotes the closed interval \(\{ t \in \mathbb{R}: 0 \leq t \leq 1\}\). The inequality is a special case of Jensen’s inequality.

We say that \(f\) is strictly convex if for all \(x, y \in I\) such that \(x \neq y\) and for all \(t \in (0,1)\), \[f\left(tx + (1-t)y\right) < t f(x) + (1-t)f(y).\] Here, \((0,1)\) denotes the open interval \(\{ t \in \mathbb{R}: 0 < t < 1\}\).

Graphically, \(f\) is strictly convex if and only if the interior of the straight line segment between any pair of points on the curve representing \(f\) is above the curve.

Recently, I encountered the following characterization¹ of a strictly convex function: \(f\) is strictly convex if and only if for all distinct \(x,y,z \in I\), \[\frac{f(x)}{(x-y)(x-z)} + \frac{f(y)}{(y-x)(y-z)} + \frac{f(z)}{(z-x)(z-y)} > 0.\]

Even though I have seen quite a bit of material on convex functions over the years, this characterization is new to me. having been unable to locate a full proof, I decided to write one out in this article.

Proof

First, assume that for all distinct \(x,y,z \in I\), \[\frac{f(x)}{(x-y)(x-z)} + \frac{f(y)}{(y-x)(y-z)} + \frac{f(z)}{(z-x)(z-y)} > 0.\]

Suppose by way of contradiction that \(f\) is not strictly convex. Then there exist \(x, y \in I\) such that \(x \neq y\) and \(t \in (0,1)\) such that \[f\left(tx + (1-t)y\right) \geq t f(x) + (1-t)f(y).\]

Setting \(z = t x + (1-t)y\), we obtain that \[\begin{align*} & \frac{f(x)}{(x-y)(x-z)} + \frac{f(y)}{(y-x)(y-z)} + \frac{f(z)}{(z-x)(z-y)} \\ =~& \frac{f(x)}{(1-t)(x-y)^2} + \frac{f(y)}{t(y-x)^2} - \frac{f(t+(1-t)y)}{t(1-t)(y-x)^2} \\ =~& \frac{tf(x) + (1-t)f(y) - f(tx+(1-t)y)}{t(1-t)(x-y)^2} \\ \leq~& 0 \end{align*}\] since \(t(1-t)(x-y)^2 > 0\), contradicting our assumption.

For the converse, assume that there exist distinct \(x,y,z \in I\) such that \[\frac{f(x)}{(x-y)(x-z)} + \frac{f(y)}{(y-x)(y-z)} + \frac{f(z)}{(z-x)(z-y)} \leq 0.\]

It is straightforward to check that the left-hand side is invariant under any permutation of \(x,y,z\). Hence, we may assume without loss of generality that \(x < z < y\).

Let \(t = \frac{y-z}{y-x}\). Then \(t \in (0,1)\). It is not difficult to verify that \(z = tx + (1-t)y\). Then, \[\begin{align*} 0 & \geq \frac{f(x)}{(x-y)(x-z)} + \frac{f(y)}{(y-x)(y-z)} + \frac{f(z)}{(z-x)(z-y)} \\ & = \frac{f(x)}{(1-t)(x-y)^2} + \frac{f(y)}{t(y-x)^2} - \frac{f(t+(1-t)y)}{t(1-t)(y-x)^2} \\ & = \frac{tf(x) + (1-t)f(y) - f(tx+(1-t)y)}{t(1-t)(x-y)^2}. \end{align*}\] Since \(t(1-t)(x-y)^2 > 0\), we have that \(0 \geq tf(x) + (1-t)f(y) - f(tx+(1-t)y)\), implying that \(f(tx+(1-t)y) \geq tf(x) + (1-t)f(y)\). Hence, \(f\) is not strictly convex.

As one can see, the steps are just the steps for the other direction in reverse. Therefore, one can potentially combine both directions into a single proof but I did not try to be quick here.