Is $f(x) = e^{−∥x∥^2}$ with $x∈R^n$ concave?

Question

Is $f(x) = e^{−∥x∥^2}$ with $x∈R^n$ concave? According to my calculations the $e^{-x}$ is convex, the norm is also convex. Then the whole function is convex.

However, when I plot a visual graph in 2D, this function is concave (looks like a multivariate Gaussian).

And here https://math.stackexchange.com/a/1972485/539984 they say that it is nor neither not convex, however, I do not understand the proof. The Hessian is $\geq0$ then why the conclusion is that it is not convex?

Thank you for clarifying!

Answer 1

Composition of convex functions is not necessarily convex and $e^{-x^2}$ is a counterexample to this, being the composition of $x^2$ and $e^{-x}$. However, if $f: \mathbb{R}\rightarrow \mathbb{R}$ is convex non-decreasing and $g: \mathbb{R}^n\rightarrow \mathbb{R}$ is convex, then $f\circ g$ is also convex.

The criterion for the convexity of $C^2$ functions $\mathbb{R}^n\rightarrow \mathbb{R}$ goes as follows:

1. If $f$ is convex, then it has everywhere positive semi-definite Hessian.
2. If $f$ has everywhere positive definite Hessian, then $f$ is strictly convex.

An analogous principle is true for concave functions.

One can easily check that $f=e^{-\|x\|^2}$ has positive semi-definite Hessian at some points (far from the origin) and negative semi-definite Hessian at some other points (close to the origin). Thus $f$ is neither convex everywhere nor concave everywhere.