Let $f: \mathbb{R}^d\rightarrow\mathbb{R}_{\geq 0}$ be a smooth and convex function. If $x^* \in \mathbb{R}^d$ is the (global) minimum of $f$, it is [well known][1] that gradient descent with a sufficiently small fixed step size $\eta$,
$$ x_{n+1} = x_{n} - \eta \nabla f(x_n), $$
converges to the global minimum $x_n \rightarrow x^*$.

**Question**: What is the asymptotic behavior of gradient descent when a finite minimizer $x^*$ does not exist (*i.e.*, when $\forall x \in \mathbb{R}^d : \nabla f(x)\neq 0$)? My guess is that $x_n/||x_n||$ should converge to some finite solution, but I could not find a proof. Or is perhaps there some counter-example to this claim?

Thanks in advance!
 


  [1]: https://rkganti.wordpress.com/2015/08/21/convergence-rate-of-gradient-descent-algorithm/