Papers
Topics
Authors
Recent
Detailed Answer
Quick Answer
Concise responses based on abstracts only
Detailed Answer
Well-researched responses based on abstracts and relevant paper content.
Custom Instructions Pro
Preferences or requirements that you'd like Emergent Mind to consider when generating responses
Gemini 2.5 Flash
Gemini 2.5 Flash 43 tok/s
Gemini 2.5 Pro 49 tok/s Pro
GPT-5 Medium 17 tok/s Pro
GPT-5 High 19 tok/s Pro
GPT-4o 96 tok/s Pro
Kimi K2 197 tok/s Pro
GPT OSS 120B 455 tok/s Pro
Claude Sonnet 4 36 tok/s Pro
2000 character limit reached

Convergence to minima for the continuous version of Backtracking Gradient Descent (1911.04221v2)

Published 11 Nov 2019 in math.OC, cs.LG, cs.NA, math.NA, and stat.ML

Abstract: 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 $C2$ near its generalised saddle points. Fix real numbers $\delta_0>0$ and $0<\alpha <1$. Then there is a smooth function $h:\mathbb{R}k\rightarrow (0,\delta_0]$ 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 $x_0\in \mathbb{R}k$, the sequence $x_{n+1}=H(x_n)$ either satisfies $\lim_{n\rightarrow\infty}||x_{n+1}-x_n||=0$ or $ \lim_{n\rightarrow\infty}||x_n||=\infty$. Each cluster point of ${x_n}$ is a critical point of $f$. If moreover $f$ has at most countably many critical points, then ${x_n}$ either converges to a critical point of $f$ or $\lim_{n\rightarrow\infty}||x_n||=\infty$. (iii) There is a set $\mathcal{E}1\subset \mathbb{R}k$ of Lebesgue measure $0$ so that for all $x_0\in \mathbb{R}k\backslash \mathcal{E}_1$, the sequence $x{n+1}=H(x_n)$, {\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 $x_0\in \mathbb{R}k\backslash \mathcal{E}_2$, any cluster point of the sequence $x{n+1}=H(x_n)$ is not a saddle point, and more generally cannot be an isolated generalised saddle point. Some other results are proven.

Citations (18)
List To Do Tasks Checklist Streamline Icon: https://streamlinehq.com

Collections

Sign up for free to add this paper to one or more collections.

Summary

We haven't generated a summary for this paper yet.

Dice Question Streamline Icon: https://streamlinehq.com

Follow-Up Questions

We haven't generated follow-up questions for this paper yet.

Authors (1)

Don't miss out on important new AI/ML research

See which papers are being discussed right now on X, Reddit, and more:

“Emergent Mind helps me see which AI papers have caught fire online.”

Philip

Philip

Creator, AI Explained on YouTube