# Difference of convexity and strict convexity

I know what is a function convex and what is a strict convex function. But I was wondering what is concretely the difference, haw can we differentiate both on a graph ? For example, I know that a convex function is always upper that any tangent, i.e. $f(x)\geq f(y)+f'(y)(x-y)$ for all $x$ and all $y$. I unfortunately don't thing that it make sense to says that it's strictly upper it's tangent since for $x=y$ we always have $f(x)=f(y)+f'(y)(x-y)$. So, is there a way to distinguish convexity and strict convexity just by watching the graph of a function ? And if yes, how ? What can be the specific characteristic of a strict convex function that a convex function doesn't have ? (despite the strict inequality at the definition).

• geometrically, if it's strictly convex, it cannot have line segments on its graph, so in a sense it always "bends". While, if you do away the "strict", then the function may become linear at some part of its domain. – Hayk May 14 '18 at 8:43

A function $f: \mathbb{R} \to \mathbb{R}$ is convex if for all $x,y \in \mathbb{R}$ and for all $\lambda \in (0,1)$ the following holds: $$f(\lambda x +(1-\lambda)y) \leq \lambda f(x) +(1-\lambda) f(y)$$

Geometrically this means that the line through two points $f(x)$ and $f(y)$ on the graph is always above the graph between $x$ and $y$.

We say that $f$ is strictly convex if the above inequality holds strictly, i.e. $$f(\lambda x +(1-\lambda)y) < \lambda f(x) +(1-\lambda) f(y)$$

For example, the blue function in the graph below is defined as $$f(x)=\begin{cases}-x-4, \text{ if } x\leq-4\\ 0, \text{ if } -4<x<4\\ x-4, \text{ if } x\geq 4\\ \end{cases}$$ So, $f$ is convex because the first inequality above holds. However it is not strictly convex because for $x=-2$ and $y=2$ the inequality does not hold strictly.

However, $g(x)=x^2$ is strictly convex, for example. Every strictly convex function is also convex. The opposite is not necessarily true as the above example of $f(x)$ has shown. A strictly convex function will always take a unique minimum. For a convex function which is not strictly convex the minimum needs not to be unique. For example, $f(x)$ above takes its minimum everywhere between -4 and 4. Hence, the minimum is not unique. For $g(x)=x^2$, however, there is only one unique minimum at $x=0$.