English

The Gronwall inequality

Functional Analysis 2025-04-01 v1 Classical Analysis and ODEs

Abstract

We prove the following version generalization of the Gronwall inequality: Let X\mathbf X be a Banach space and UXU\subset \mathbf X an open convex set in X\mathbf X. Let f,g ⁣:[a,b]×UXf,g\colon [a,b]\times U\to \mathbf X be continuous functions and let y,z ⁣:[a,b]Uy,z\colon [a,b]\to U satisfy the initial value problems \begin{align*} y'(t)&=f(t,y(t)),\quad y(a)=y_0,\\ z'(t)&=g(t,z(t)),\quad z(a)=z_0. \end{align*} Also assume there is a constant C0C\ge 0 so that g(t,x2)g(t,x1)Cx2x1 \|g(t,x_2)-g(t,x_1)\|\le C\|x_2-x_1\| and a continuous function ϕ ⁣:[a,b][0,)\phi\colon [a,b]\to [0,\infty) so that f(t,y(t))g(t,y(t))ϕ(t). \|f(t,y(t))-g(t,y(t))\|\le \phi(t). Then for t[a,b]t\in [a,b] y(t)z(t)eCtay0z0+eCtaateCsaϕ(s)ds. \|y(t)-z(t)\| \le e^{C|t-a|}\|y_0-z_0\|+e^{C|t-a|}\int_a^te^{-C|s-a|}\phi(s)\,ds.

Keywords

Cite

@article{arxiv.2503.23639,
  title  = {The Gronwall inequality},
  author = {Ralph Howard},
  journal= {arXiv preprint arXiv:2503.23639},
  year   = {2025}
}

Comments

4 pages

R2 v1 2026-06-28T22:39:52.140Z