Convergence to minima for the continuous version of Backtracking Gradient Descent

The main result of this paper is: {\bf Theorem.} Let $f:\mathbb{R}^k\rightarrow \mathbb{R}$ be a $C^{1}$ function, so that $\nabla f$ is locally Lipschitz continuous. Assume moreover that $f$ is $C^2$ near its generalised saddle points. Fix real numbers $\delta0>0$ and $0<\alpha <1$. Then there is a smooth function $h:\mathbb{R}^k\rightarrow (0,\delta0]$ so that the map $H:\mathbb{R}^k\rightarrow \mathbb{R}^k$ defined by $H(x)=x-h(x)\nabla f(x)$ has the following property: (i) For all $x\in \mathbb{R}^k$, we have $f(H(x)))-f(x)\leq -\alpha h(x)||\nabla f(x)||^2$. (ii) For every $x0\in \mathbb{R}^k$, the sequence $x{n+1}=H(xn)$ either satisfies $\lim{n\rightarrow\infty}||x{n+1}-xn||=0$ or $ \lim{n\rightarrow\infty}||xn||=\infty$. Each cluster point of ${xn}$ is a critical point of $f$. If moreover $f$ has at most countably many critical points, then ${xn}$ either converges to a critical point of $f$ or $\lim{n\rightarrow\infty}||xn||=\infty$. (iii) There is a set $\mathcal{E}1\subset \mathbb{R}^k$ of Lebesgue measure $0$ so that for all $x0\in \mathbb{R}^k\backslash \mathcal{E}1$, the sequence $x{n+1}=H(xn)$, {\bf if converges}, cannot converge to a {\bf generalised} saddle point. (iv) There is a set $\mathcal{E}2\subset \mathbb{R}^k$ of Lebesgue measure $0$ so that for all $x0\in \mathbb{R}^k\backslash \mathcal{E}2$, any cluster point of the sequence $x{n+1}=H(xn)$ is not a saddle point, and more generally cannot be an isolated generalised saddle point. Some other results are proven.