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[parent] Lipschitz condition and differentiability (Theorem)

If $X$ and $Y$ are Banach spaces, e.g. $\reals^n$ , one can inquire about the relation between differentiability and the Lipschitz condition. If $f$ is Lipschitz, the ratio $$\frac{ \Vert f(q)-f(p)\Vert}{\Vert q-p \Vert},\quad p,q\in X$$ is bounded but is not assumed to converge to a limit.

Proposition 1   Let $f:X\to Y$ be a continuously differentiable mapping between Banach spaces. If $K\subset X$ is a compact subset, then the restriction $f:K\to Y$ satisfies the Lipschitz condition.
Proof. Let $\lin(X,Y)$ denote the Banach space of bounded linear maps from $X$ to $Y$ . Recall that the norm $\Vert T\Vert$ of a linear mapping $T\in\lin(X,Y)$ is defined by $$\Vert T \Vert = \sup \{ \frac{\Vert Tu \Vert}{\Vert u\Vert} : u\neq 0\}.$$

Let $\Df:X\to \lin(X,Y)$ denote the derivative of $f$ . By definition $\Df$ is continuous, which really means that $\Vert \Df \Vert: X\to \reals$ is a continuous function. Since $K\subset X$ is compact, there exists a finite upper bound $B_1>0$ for $\Vert \Df\Vert$ restricted to $K$ . In particular, this means that $$\Vert \Df(p) u \Vert \leq \Vert \Df(p)\Vert \Vert u\Vert \leq B_1 \Vert u\Vert,$$ for all $p\in K,\; u\in X$ .

Next, consider the secant mapping $s:X\times X\to\reals$ defined by

\begin{displaymath}s(p,q) = \begin{cases} \displaystyle \frac{\Vert f(q) - f(p) ... ...p)(q-p)\Vert}{\Vert q-p\Vert} & q\neq p \ 0 & p=q \end{cases}\end{displaymath}
This mapping is continuous, because $f$ is assumed to be continuously differentiable. Hence, there is a finite upper bound $B_2>0$ for $s$ restricted to the compact set $K\times K$ . It follows that for all $p,q\in K$ we have
$\displaystyle \Vert f(q) - f(p) \Vert$ $\displaystyle \leq \Vert f(q) - f(p) - {\operatorname{D}f}(p)(q-p)\Vert + \Vert {\operatorname{D}f}(p)(q-p)\Vert$    
  $\displaystyle \leq B_2 \Vert q-p\Vert + B_1 \Vert q-p\Vert$    
  $\displaystyle = (B_1+B_2)\Vert q-p\Vert$    

Therefore $B_1+ B_2$ is the desired Lipschitz constant. QED

Neither condition is stronger. For example, the function $f:\reals \to \reals$ given by $f(x) = x^2$ is differentiable but not Lipschitz.




"Lipschitz condition and differentiability" is owned by Mathprof. [ full author list (4) | owner history (2) ]
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See Also: derivative

Other names:  mean value inequality

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Lipschitz condition and differentiability result (Result) by paolini
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Cross-references: differentiable, function, stronger, QED, Lipschitz constant, compact set, continuously differentiable, mapping, secant, upper bound, finite, compact, continuous, derivative, linear mapping, norm, bounded linear maps, proof, restriction, compact subset, limit, converge, bounded, ratio, Lipschitz condition, relation, Banach spaces
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This is version 30 of Lipschitz condition and differentiability, born on 2001-11-12, modified 2006-09-13.
Object id is 776, canonical name is CalculatingLipschitzRatios.
Accessed 12552 times total.

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AMS MSC26A16 (Real functions :: Functions of one variable :: Lipschitz classes)

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Proof contained by Koro on 2003-07-27 19:04:18
Since this entry contains its own proofs, it should be marked as such.
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