# 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, 2018 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 $x=0$.

"A strictly convex function will always take a unique minimum." The function $$x\mapsto e^x$$ on the real numbers is strictly convex but has no local or global minimizer. In fact, a strictly convex function defined (of course) on a convex set has at most one global minimizer, and the only local minimizer or stationary point (if the function is differentiable) is in fact the global minimizer if it exists.

• Why the function need to be defined on a convex set? Dec 18, 2022 at 11:32
• The definition of a convex function involves points that are convex combinations of other points. Therefore, the definition of a convex function requires that the domain be convex. Any book or paper that omits this assumption is perhaps relying on the reader to infer that the domain is necessarily convex. In careful writing of mathematics, it is often helpful to the reader to explicitly state this assumption rather than require that the reader infer it. I hope this is helpful. Apr 1 at 19:20