Asymptotic behavior of gradient descent on a smooth, convex, non-negative function with no finite minimum Let $f: \mathbb{R}^d\rightarrow\mathbb{R}_{\geq 0}$ be a  function which is convex and smooth (i.e., in $C^{\infty}$). If $x^* \in \mathbb{R}^d$ is the (global) minimum of $f$, it is well known 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 perhaps there is some counter-example to this claim?
Thanks in advance!
 A: I think there are simple counterexamples of the form $f(x,y)=\phi(x)+\psi(y)$ (here $x$ and $y$ denote real variables),  where $\phi$ and $\psi$ are smooth, positive, decreasing, strictly convex functions.
The idea is that, $-\psi'(x)$ and $-\phi'(x)$ being positive decreasing functions, is not an obstruction for their ratio to oscillate between arbitrarily large and small values, on arbitrarily large intervals. This produces zig-zag orbits $z_n:=(x_n,y_n)$ for which  $z_n/\|z_n\|$ may accumulate to both $(1,0)$ and $(0,1)$.
Construction. Define some smooth, positive, decreasing functions $u(t)$ and $v(t)$ such that for all $n\in\mathbb{N}$
$$u(x):={1\over (n!)^2}\quad\text{for $n$ even, and}\; n!+1\le x\le (n+2)!$$
and
$$v(x):={1\over (n!)^2}\quad\text{for $n$ odd, and}\; n!+1\le x\le (n+2)!$$
(so one has only to extend them smoothly for $x\le2$ and  between $n!$ and $n!+1$ to have them defined everywhere). As a consequence we have
$${v(x)\over u(x)}=\begin{cases}
n^{-2}  &\text{for $n$ odd, and }\; n!+1\le x\le (n+1)! \\
n^2 &\text{for $n>0$ even, and }\; n!+1\le x\le (n+1)! \\
\end{cases}$$
while
$$n^{-2}\le {v(x)\over u(x)}\le n^2 \quad \text{for any $n>0$ and }\; n! \le x\le n!+1 \ .$$
Note also that $u$  (respectively  $v$) has finite integral on any right-unbounded interval, because the contribute to the integral on any interval $[n!+1, (n+2)!+1]$ (for $n$ even, resp. odd) is bounded by $(n+2)!/(n!)^2=(n+2)(n+1)/n!$. Therefore we can consider the smooth, decreasing, strictly convex functions
$$\phi(x):=\int_x^\infty u(t)dt\quad\text{and }\quad \psi(x):=\int_x^\infty v(t)dt\; .$$
The negative gradient iteration  $u_{n+1}=u_n-\eta\nabla f(u_n)$ writes, with $z_n:=(x_n,y_n)$
$$
\begin{cases}
x_{n+1}=x_n+\eta \,u(x_n) \\
y_{n+1}=y_n+\eta \,v(y_n)\\
\end{cases}
$$
and it follows from the above identities and inequalities  on $v/u$ that it produces sequences $(x_n,y_n)$ with $\liminf_{n\to+\infty}y_n/x_n=0$ and $\limsup_{n\to+\infty}y_n/x_n=+\infty$,
corresponding to a sequence $z_n/\|z_n\|$ that clusters both to $(1,0)$ and to $(0,1)$.
Sketch of the computation. Both sequences $x_m=x_0+\eta\sum_{k=0}^{m-1}u(x_j)$ and $y_m=x_0+\eta\sum_{k=0}^{m-1}v(y_j)$ are increasing and diverging. 
Let's define  $\mu(t)$ as  the largest integer $j$ such that $x_j\le n!$ and 
$\nu(t)$ as  the largest integer $j$ such that $y_j\le n!$. The point is to use the above identities and inequalities on $u$ and $v$ to show that, for even $n$ and $m=\mu(n!)$,  the principal part in both the above sum is given by the indices $j$ such that $\mu(n!)\le j\le \mu((n+1)!)$, so that $y_m/x_m=O(1/n^2)$. Analogously, for odd $n$ and $m=\nu(n!)$, one should get $x_m/y_m=O(1/n^2)$.
