Directional derivative zero, minimum and convexity

Question

Let $f:\mathbb R^n \to \mathbb R$ be convex, let $u\in \mathbb R^n$, $v\in \mathbb R^n\setminus \{0\}$ and assume that the directional derivative of $f$ at $u$ in direction $v$ is $0$.

I'm wondering if this implies that $f$ has a global minimum at $x$. The function $g:\mathbb R\to \mathbb R, t\mapsto f(x+tv)$ is convex and its right-derivative at $0$ is zero, hence $g$ has a global minimum at $0$, thus $f$ is minimized at $x$ when restricted along direction $v$.

I don't see a reason why $x$ should be a global minimizer. Can someone provide a counterexample ?

Answer 1

A counterexample: $f(x, y) = x$ is convex, and the directional derivative in $y$-direction is zero everywhere. But $f$ has no (local or global) minimum at all.

If $f:\mathbb R^n \to \mathbb R$ is differentiable then a necessary condition for a (local or global) minimum at a point $x_0$ is that $\nabla f(x_0) = 0$. For a convex function this condition is also sufficient for a global minimum.

Answer 2

As pointed out in the comments with $f(x, y) = x$ and $v$ going in the $y$-direction, there are plenty of very accessible counterexamples.

In fact, if $g:\Bbb R^m\to \Bbb R^n$ with $m>n$ is differentiable, then there has to be, at every single point of $\Bbb R^m$, directions where the directional derivative of $g$ is zero. And yet clearly not every function has local minima or maxima (the above counterexample generalizes neatly to any dimension).

We can probably see this fact most easily if we consider the derivative of $g$ at a point $u$ as a linear transformation $J$. (This is the so-called Jacobian of $g$ at $u$, and its matrix form, assuming a Cartesian coordinate system on $\Bbb R^m$ and $\Bbb R^n$, has as columns all the partial derivatives of $g$. In the special case of $n = 1, m>1$ the Jacobian is often called the gradient, and in the special case $m = 1, n>1$ it is called the velocity vector.) Then the directional derivative at $u$ in the direction of $v$ is $Jv$. And since $J$ is a linear transformation from $m$-dimensional space to $n$-dimensional space, if $m>n$ then $J$ must have a non-trivial kernel.