# A function is convex if and only if

1. Nov 23, 2012

### Arian.D

1. The problem statement, all variables and given/known data
Show that a differentiable function f is convex if and only if the following inequality holds for each fixed point x0 in Rn:
f(x) ≥ f(x0) + ∇tf(x0)(x-x0) for all x in Rn, where ∇tf(x0) is the gradient vector of f at x0.

2. Relevant equations

3. The attempt at a solution

by definition of a convex function we have:

f(λx + (1-λ)x0) ≤ λf(x) + (1-λ)f(x0)

f(λ(x-x0)+x0) ≤ λ(f(x) - f(x0)) + f(x0)

f(λ(x-x0)+x0) - f(x0) ≤ λ(f(x) - f(x0))

Setting Δx = λx-x0 we'll have:

f(Δx + x0) - f(x0) ≤ λ(f(x) - f(x0))

I'm very hesitant to use this step, because vector division is not defined, but if we were dealing with real numbers (i.e, vectors of dimension 1) I could've gone further to obtain:

$$\frac{f(\Delta{x} + x_0) - f(x_0)}{\lambda} \leq f(x) - f(x_0)$$
$$\frac{f(\Delta{x} + x_0) - f(x_0)}{\lambda(x-x_0)}(x-x_0) \leq f(x) - f(x_0)$$
$$\frac{f(\Delta{x} + x_0) - f(x_0)}{\Delta{x}}(x-x_0) \leq f(x) - f(x_0)$$

taking limits from both sides as Δx goes to 0 gives us the desired result.

I don't know how to show that the converse is true, and also I don't know how to generalize what I've written to functions of several variables since vector division is not defined. any helps would be appreciated.

2. Nov 24, 2012

### Ray Vickson

Most of theses types of results generalize almost immediately from the 1-dimensional case to the multivariate case, because if $x_0, \, x \in \mathbb{R}^n$ and $p = x - x_0,$ then
$$f \text{ convex in }\mathbb{R}^n \Rightarrow \phi(t) = f(x_0 + t p)\text{ is convex in } t \in \mathbb{R},$$ and conversly. So, you need to show that $\phi(t)$ is convex if and only if
$$\phi(t) \leq \phi(t_0) + \phi'(t_0) \, (t - t_0) \, \forall \, t, t_0.$$

RGV