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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

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Are you referring to Lipschitz Continuity? – Han Altae-Tran May 7 '12 at 8:32
Yes, I do. In general I know what is contuinity etc. But I cannot find an example with Lipschitz continuity in any of my math handbooks. – Misery May 7 '12 at 8:42
Are you asking for a definition of Lipschitz continuity? – Did May 7 '12 at 10:45
Yes, and how to testif my function is continous (after googling a little i know it isn't, but I need to prove it) – Misery May 7 '12 at 10:59
If your function ${\bf x}\mapsto {\bf F}({\bf x})$ is some sort of explicit expression in terms of elementary functions it is continuously differentiable wherever bona fide defined; i.e., excepted in points where $\log0$, $\sqrt{0}$ and such things occur. – Christian Blatter May 7 '12 at 12:52
up vote 2 down vote accepted

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.

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