Lipschitz condition and differentiability

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

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

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

Proposition 1

Let $f\mathrm{:}X\mathrm{\to }Y$ be a continuously differentiable mapping (http://planetmath.org/DifferentiableMapping) betweenBanach spaces. If $K\mathrm{\subset }X$ is a compactsubset, then the restriction $f\mathrm{:}K\mathrm{\to }Y$ satisfies the Lipschitzcondition.

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

$$\parallel T\parallel =sup\{\frac{\parallel Tu\parallel }{\parallel u\parallel }:u\ne 0\}.$$

Let $\mathrm{D}f:X\to \mathrm{lin}(X,Y)$ denote the derivative of $f$. By definition$\mathrm{D}f$ is continuous, which really means that$\parallel \mathrm{D}f\parallel :X\to ℝ$is a continuous function. Since$K\subset X$ is compact, there exists a finite upper bound $B_1>0$ for$\parallel \mathrm{D}f\parallel$ restricted to $K$. In particular, this means that

$$\parallel \mathrm{D}f(p)u\parallel \le \parallel \mathrm{D}f(p)\parallel \parallel u\parallel \le B_1\parallel u\parallel ,$$

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

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

This mapping is continuous, because $f$ is assumed to be continuouslydifferentiable. Hence, there is a finiteupper bound $B_2>0$ for $s$ restricted to the compact set $K\times K$. Itfollows that for all $p,q\in K$ we have

$\parallel f(q)-f(p)\parallel$$\le \parallel f(q)-f(p)-\mathrm{D}f(p)(q-p)\parallel +\parallel \mathrm{D}f(p)(q-p)\parallel$

$\le B_2\parallel q-p\parallel +B_1\parallel q-p\parallel$

$=(B_1+B_2)\parallel q-p\parallel$

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

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