TITLE: Gradient descent for differentiable convex functions
QUESTION [9 upvotes]: Suppose $f\colon\mathbb{R}^n\to\mathbb{R}$ is convex and differentiable, and assume that $f$ has a minimizer.  

If $(x_k)$ is the sequence generated by exact gradient descent, must it converge to a minimizer? 
Here "exact gradient descent" means that $x_{k+1} = x_k-t_k\nabla f(x_k)$ where it is assumed that $t_k$ is a minimizer of the function $t\mapsto f(x_k-t\nabla f(x_k))$ for $t\geq 0$ (the existence of $t_k$ is assumed for all $k$). 

Reference or counterexample would be great. 
(I am aware of Wolfe's example demonstrating the importance of differentiability. I am also aware that this works when $f$ is strictly convex and coercive.)

REPLY [2 votes]: We have assumed that $f : \mathbb{R}^N \rightarrow \mathbb{R}$ is convex, differentiable and has a minimiser. Let the line search function be denoted by,
$$
\begin{align}
  g_k : \mathbb{R} \rightarrow& \mathbb{R} \\
  t \mapsto& f(x_k - t \nabla f(x_k))
\end{align}
$$
Lemma 1. If $\frac{dg_k}{dt}(0) = 0$ then $x_k$ is the minimum of $f$.
Proof. The derivative of $g_k$ is $\frac{dg_k}{dt} = \nabla f(x_k - t\nabla f(x_k)) \cdot \nabla f(x_k)$. Suppose that $\frac{dg_k}{dt}(0) = 0$, then $\nabla f(x_k)\cdot\nabla f(x_k) =0$, i.e. $\|\nabla f(x_k)\|^2 = 0$, and so $x_k$ is at the minimum. $\square$
The function $g_k$ is,

Convex as an affine transformation of $f$, which is convex.
Non constant. If it were then $\frac{dg_k}{dt} = 0$ for every $t \in \mathbb{R}$ and so $x_k$ is the minima by Lemma 1.

Lemma 2. If $x_{k+1} = x_k$ then $x_{k+1}$ is the minima. 
Proof. The value $t_k$ is the minimiser of the function $g_k$. If $t_k = 0$ then, $\frac{dg}{dt}(0) = 0$, and so, by Lemma 1 we are at the minimum. Instead assume that $t_k \neq 0$. If $x_{k+1} = x_k$ then $t_k \nabla f(x_k) = 0$ and so $x_k$ is the minimum. $\square$
Lemma 3. The sequence defined by $z_k = f(x_k)$ is strictly decreasing unless $x_k$ is the global minima of $f$, in which case $x_n = x_k$, $\forall n\geq k$.
Proof. It suffices to prove that $g_k(t_k) < g_k(t)$ for every $t \in \mathbb{R}$. We know that $g_k(t_k) \leq g_k(t) $ by assumption. If $\exists t^* \neq t_k$ such that  $g_k(t_k) = g_k(t^*) $, then $g_k$ is constant. By Lemma 1 it follows that $x_k$ is the minima. If it is the minima then $\nabla f(x_k) = 0$ and so $x_{k+1} = x_k$. $\square$
Theorem The sequence $z_k = f(x_k)$ converges to the minimum $x^*$. 
Proof. The sequence $z_k$ is a real and strictly decreasing for all $z_k > f(x^*)$. It is bounded below by $f(x^*)$ and so it converges its infimum $ \ell = \inf_k{z_k} $ by the monotone convergence theorem. But $f(x_k) \to \ell$ which means that $\nabla f(x_{k}) \to 0$ as $k \to \infty$. $\nabla f(x_{k}) = 0$ only at the minimum of $f$, and so $\ell$ is the minimum.