1
$\begingroup$

I do have a problem with achieving convergence in Newton method (using Armijo rule) for system of algebraic non-linear equations. I suspect that my function is not continuously differentiable, however I'd like to be sure if that is so. How do I test it if my F( x ) is Lipschitz continuously differentiable?

Thanks in advance, Regards

  • 0
    Then see [here](http://en.wikipedia.org/wiki/Lipschitz_continuity).2012-05-16

1 Answers 1

2

Two conditions are being confused here, because the second one is less known and is inconsistently named.

  1. A function $f$ (either scalar or vector valued) is Lipschitz if there is a constant $L$ such that $\|f(x)-f(y)\|\le L\|x-y\|$ for all $x,y$. This is denoted by $f\in\mathrm{Lip}(U)$ or $f\in C^{0,1}(U)$ where $U$ is the domain of $f$.
  2. A function $f$ is Lipschitz smooth or Lipschitz continuously differentiable if there is a constant $L$ such that $\|\nabla f(x)-\nabla f(y)\|\le L\|x-y\|$ for all $x,y$. This is denoted by $\nabla f\in\mathrm{Lip}(U)$ or $f\in C^{1,1}(U)$ where $U$ is the domain of $f$.

That is, the second condition is the Lipschitz continuity of first-order derivatives. A sufficient condition for $f\in C^{1,1}$ is that the second derivative of $f$ is bounded: see $C^{1,1}$ regularity.