Is the sequence $(x_n)$ in Gradient Descent algorithm always convergent?

Question

Let $f \in \mathcal C^1(\mathbb R^n,\mathbb R)$ be convex and $\nabla f$ be $L$-Lipschitz continuous. The sequence $(x_n)$ in Gradient Descent algorithm is defined as $$x_{n+1} = x_n -\gamma_n \nabla f(x_n)$$ where $\gamma_n>0$ is the step size.

It's is well-known that $(f(x_n))$ will converge to the minimum of $f$. I've searched through my lecture note and many other materials on the Internet, but it seems that they do not mention about the convergence of $(x_n)$. As such,

I would like to ask if the sequence $(x_n)$ converges.

Thank you so much for your clarification!

Answer 1

Thank you @copper.hat for providing a counter-example here.

Let $f(x)=x$. Then $x_{n+1} = x_n -\gamma$ is divergent for all $\gamma> 0$.

Answer 2

Yes. However it can happen that there is no unique minimizer and in that case the limit depends on your starting point. For example for $f(x,y)=x^2$ the second component of the limit is simply the second component of the initial point.