# Pretty convexity result

Where here we discover some interesting facts about continuous convex functions.

We know that a function $f:\mathbb{R}^n\to\mathbb{R}$ is convex if

$f(\tau x + (1-\tau)y) \leq \tau f(x) + (1-\tau)f(y)$

for all $\tau \in [0,1]$ and $x,y\in \mathbb{R}^n$.

We see that if $f$ is a continuous function, then an equivalent condition for convexity is that either of the following inequalities holds

\begin{aligned} f\left(\frac{x+y}{2}\right) \leq \int_0^1 f(\theta x + (1-\theta)y)\mathrm{d}\theta \leq \frac{f(x)+f(y)}{2}\end{aligned}
If $f$ is convex it is straightforward to prove the right hand side inequality. Conversely, notice that the right-hand side of that inequality is

\begin{aligned}\int_0^1 \left((1-\theta)f(x) + \theta f(y)\right)\mathrm{d}\theta = \frac{f(x)+f(y)}{2}\end{aligned}

So then, we have

\begin{aligned} &\int_0^1 f(x+\theta (y-x)) \mathrm{d}\theta \leq \int_0^1 \left((1-\theta)f(x) + \theta f(y)\right)\mathrm{d}\theta \\ \Leftrightarrow& \int_0^1 \left[ (1-\theta)f(x) + \theta f(y) - f(x+\theta (y-x)) \right] \mathrm{d}\theta \geq 0 \end{aligned}

which means that $f$ is convex. Actually, it means that $(1-\theta)f(x) + \theta f(y) \geq$ $f(x+\theta (y-x))$ for almost all $\theta \in [0,1]$, but since $f$ is assumed to be continuous, it holds for all $\theta\in[0,1]$.

There is another interesting inequality we may prove: We may use the integral form of Jensen’s inequality according to which if $f$ is convex and $g$ and $f\circ g$ are integrable on $[0,1]$ then

\begin{aligned}f\left(\int_0^1 g(\theta)\mathrm{d}\theta\right) \leq \int_0^1 f(g(\theta))\mathrm{d}\theta \end{aligned}

Here, choose $g(\theta) = x + \theta(y-x)$; we have

\begin{aligned}\int_0^1 g(\theta)\mathrm{d}\theta = \int_0^1 (x + \theta(y-x))\mathrm{d}\theta = \frac{x+y}{2}\end{aligned}

Then, we have the double-sided inequality

\begin{aligned} f\left(\frac{x+y}{2}\right) \leq \int_0^1 f(\theta x + (1-\theta)y)\mathrm{d}\theta \leq \frac{f(x)+f(y)}{2}.\end{aligned}

Assuming that $f$ is continuous, then $f$ is convex if and only if at least one of these inequalities holds.

### 2 responses

1. Interesting identities I wasn’t familiar with! With algorithms like Newton’s method (more specifically the multidimensional case) it’s the goal to find absolute extrema, and to find them a convex function is necessary – otherwise you’re not guaranteed to find that absolute minima or maxima. I’m looking forward to taking analysis the semester after next so I can get into more equations like this and how to formulate them and work with them.

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1. Indeed, there are many interesting inequalities. Maybe a good starting point is the study of those related to the first and second-order approximations of a function which is either assumed to be (i) continuously differentiable with Lipschitz gradient or (ii) twice continuously differentiable with Lipschitz Hessian. These are used in the analysis of first and second-order methods (gradient and Newton methods).

I find Nesterov’s book “Introductory Lectures on Convex Optimization” a very good reference. I uploaded a new post you might find interesting: https://mathematix.wordpress.com/2016/08/03/second-order-approximation-of-twice-differentiable-functions/

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