Why does boundedness of a bilinear map imply locally Lipschitz continuity?

Question

Let $X$ be a Banach space and let $Y$ and $Z$ be normed vector spaces (over $\mathbb{R}$ or $\mathbb{C}$). Let $B:X\times Y \to Z$ be a bilinear map. Then the following are equivalent. Then if $B$ is bounded, i.e. if there is a constant $c\ge 0$ such that $$ ||B(x,y)||_Z \le c ||x|| \ ||y||_Y, $$ for all $x\in X$ and all $x\in Y$, this means that $B$ will be continuous.

The proof of this uses the fact that the bounded on $B$ implies that $B$ is locally Lipschitz continuous, and hence continuous.

But how is $B$ locally Lipschitz continuous and how does being bounded imply this fact? The notes I am using do not give a definition of locally Lipschitz continuous so I am wondering does someone know an appropriate definition, and how the map $B$ satisfies the definition?

Furthermore, why does the bounded only imply local Lipschitz continuity instead of global Lipschitz continuity?

Answer 1

To see that global Lipschitz continuity won't generally hold consider multiplication $(x,y)\mapsto xy$.

Locally Lipschitz means for each $(x_0,y_0)\in X\times Y$ there is a neighborhood $U$ of $(x_0,y_0)$ such that $B$ restricted to $U$ is Lipschitz. Presumably the norm on $X\times Y$ is equivalent with $\|(x,y)\|=\|x\|+\|y\|$, and to show the implication it suffices to take $U$ to be a ball centered at the origin with arbitrary radius $R>0$. Then for all $(x_1,y_1)$ and $(x_2,y_2)$ in $U$,

$$ \begin{align*} \|B(x_1,y_1)-B(x_2,y_2)\|&\leq \|B(x_1,y_1)-B(x_1,y_2)\|+\|B(x_1,y_2)-B(x_2,y_2)\|\\ &=\|B(x_1,y_1-y_2)\|+\|B(x_1-x_2,y_2)\|\\ &\leq c\|x_1\|\|y_1-y_2\|+c\|x_1-x_2\|\|y_2\|\\ &\leq cR(\|y_1-y_2\|+\|x_1-x_2\|)\\ &=cR\|(x_1,y_1)-(x_2,y_2)\|, \end{align*}$$

giving a local Lipschitz constant of $cR$ on $U$.