Lipschitz condition and differentiability

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.

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

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
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.