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I learn that the sufficient condition for Lipschitz smoothness is the function $f$ satisfys that: $$ \mid f(x+h)-f(x)-\langle\xi,h\rangle \mid \leq c\parallel h \parallel ^2 $$ where $\xi = f'$

I want to use this inequality to bound the difference between $f(x+h)$ and $f(x)+\langle\xi,h\rangle$, then what condition should be satisfied for the function $f$ in order to apply the inequality? I see someone simply assume that $f$ is sufficiently smooth before applying the inequality, is it correct?

What also confuse me is that the description above seems to have nothing to do with Lipschitz smoothness because the inequality above is the sufficient condition for Lipschitz smoothness rather than a property, basically I can't say the inequality holds because the function $f$ is Lipschitz smooth. As a result, the question is what's the property of Lipschitz smoothness, what's the application and how to use it?

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What is $\xi\,$? – Christian Blatter Aug 30 '12 at 17:48
One application of Lipschitz smoothness is the existence and uniqueness of solutions to first-order nonlinear differential equations. – process91 Aug 31 '12 at 20:23

It seems that the term Lipschitz smooth is used here to indicate that $f'$ is Lipschitz continuous (not $f$ itself). The usual notation for this property is $f\in C^{1,1}$. This property appears in obstacle problems where it is often the best regularity one can hope for. For a simple example of an obstacle problem, imagine a heavy rope hanging over a cylindrical pulley: the shape of the rope is partly a circular arc (where it touches the cylinder) and partly a line (where it hangs freely). This union of circular arc and a line is a curve of class $C^{1,1}$. It is not in class $C^2$ because the curvature has a jump discontinuity (jumps from 1/(radius of cylinder) to zero). The wikipedia article gives more sophisticated examples of this kind.

OK, let's prove something. Suppose $f\colon\mathbb R\to\mathbb R$ is differentiable and $f'$ is Lipschitz. I claim that $|f(x+h)-f(x)-hf'(x)|\le \frac{L}{2}h^2$ for all $x,h\in\mathbb R$, where $L$ is the Lipschitz constant of $f$.

Proof: $f(x+h)-f(x)=\int_x^{x+h} f'(t)\,dt = hf'(x)+\int_x^{x+h} (f'(t)-f'(x))\,dt$ where $\left|\int_x^{x+h} (f'(t)-f'(x))\,dt\right|\le \int_x^{x+h} L|t-x|\,dt = Lh^2/2$. $\quad \Box$.

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