Lipschitz condition and differentiability

If $X$ and $Y$ are Banach spaces, e.g. $\mathbb{R}^{n}$, one can inquire about the relation between differentiability and the Lipschitz condition. If $f$ is Lipschitz, the ratio

 $\frac{\|f(q)-f(p)\|}{\|q-p\|},\quad p,q\in X$

is bounded but is not assumed to converge to a limit.

Proposition 1

Let $f:X\to Y$ be a continuously differentiable mapping (http://planetmath.org/DifferentiableMapping) between Banach spaces. If $K\subset X$ is a compact subset, then the restriction $f:K\to Y$ satisfies the Lipschitz condition.

Proof. Let $\operatorname{lin}(X,Y)$ denote the Banach space of bounded linear maps from $X$ to $Y$. Recall that the norm $\|T\|$ of a linear mapping $T\in\operatorname{lin}(X,Y)$ is defined by

 $\|T\|=\sup\{\frac{\|Tu\|}{\|u\|}:u\neq 0\}.$

Let ${\operatorname{D}f}:X\to\operatorname{lin}(X,Y)$ denote the derivative of $f$. By definition ${\operatorname{D}f}$ is continuous, which really means that $\|{\operatorname{D}f}\|:X\to\mathbb{R}$ is a continuous function. Since $K\subset X$ is compact, there exists a finite upper bound $B_{1}>0$ for $\|{\operatorname{D}f}\|$ restricted to $K$. In particular, this means that

 $\|{\operatorname{D}f}(p)u\|\leq\|{\operatorname{D}f}(p)\|\|u\|\leq B_{1}\|u\|,$

for all $p\in K,\;u\in X$.

Next, consider the secant mapping $s:X\times X\to\mathbb{R}$ defined by

 $s(p,q)=\begin{cases}\displaystyle\frac{\|f(q)-f(p)-{\operatorname{D}f}(p)(q-p)% \|}{\|q-p\|}&q\neq p\\ 0&p=q\end{cases}$

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

 $\displaystyle\|f(q)-f(p)\|$ $\displaystyle\leq\|f(q)-f(p)-{\operatorname{D}f}(p)(q-p)\|+\|{\operatorname{D}% f}(p)(q-p)\|$ $\displaystyle\leq B_{2}\|q-p\|+B_{1}\|q-p\|$ $\displaystyle=(B_{1}+B_{2})\|q-p\|$

Therefore $B_{1}+B_{2}$ is the desired Lipschitz constant. QED

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

Title Lipschitz condition and differentiability LipschitzConditionAndDifferentiability 2013-03-22 11:57:50 2013-03-22 11:57:50 Mathprof (13753) Mathprof (13753) 34 Mathprof (13753) Theorem msc 26A16 mean value inequality Derivative2