# Lipschitz condition and differentiability result

About lipschitz continuity of differentiable functions the following holds.

###### Theorem 1.

Let $X,Y$ be Banach spaces and let $A$ be a convex (see convex set), open subset of $X$. Let $f\colon\overline{A}\to Y$ be a function which is continuous in $\overline{A}$ and differentiable in $A$. Then $f$ is lipschitz continuous on $\overline{A}$ if and only if the derivative $Df$ is bounded on $A$ i.e.

 $\sup_{x\in A}\|Df(x)\|<+\infty.$
###### Proof.

Suppose that $f$ is lipschitz continuous:

 $\|f(x)-f(y)\|\leq L\|x-y\|.$

Then given any $x\in A$ and any $v\in X$, for all small $h\in\mathbb{R}$ we have

 $\|\frac{f(x+hv)-f(x)}{h}\|\leq L.$

Hence, passing to the limit $h\to 0$ it must hold $\|Df(x)\|\leq L$.

On the other hand suppose that $Df$ is bounded on $A$:

 $\|Df(x)\|\leq L,\qquad\forall x\in A.$

Given any two points $x,y\in\overline{A}$ and given any $\alpha\in Y^{*}$ consider the function $G:[0,1]\to\mathbb{R}$

 $G(t)=\langle\alpha,f((1-t)x+ty)\rangle.$

For $t\in(0,1)$ it holds

 $G^{\prime}(t)=\langle\alpha,Df((1-t)x+ty)[y-x]\rangle$

and hence

 $|G^{\prime}(t)|\leq L\|\alpha\|\,\|y-x\|.$

Applying Lagrange mean-value theorem to $G$ we know that there exists $\xi\in(0,1)$ such that

 $|\langle\alpha,f(y)-f(x)\rangle|=|G(1)-G(0)|=|G^{\prime}(\xi)|\leq\|\alpha\|L% \|y-x\|$

and since this is true for all $\alpha\in Y^{*}$ we get

 $\|f(y)-f(x)\|\leq L\|y-x\|$

which is the desired claim. ∎

Title Lipschitz condition and differentiability result LipschitzConditionAndDifferentiabilityResult 2013-03-22 13:32:42 2013-03-22 13:32:42 paolini (1187) paolini (1187) 5 paolini (1187) Result msc 26A16