ThmDex – An index of mathematical definitions, results, and conjectures.
Result R648 on D47: Lipschitz map
Proportionally bounded linear map is Lipschitz
Formulation 0
Let $R$ form a D273: Division ring.
Let $N$ and $M$ each form a D4718: Norm-metrised vector space over $R$.
Let $\Vert \cdot \Vert_N$ and $\Vert \cdot \Vert_M$ each be the D306: Vector space norm in $N$ and $M$, respectively.
Let $f : N \to M$ be a D705: Proportionally bounded linear map with respect to $N$ and $M$.
Then $f$ is a D47: Lipschitz map with respect to $N$ and $M$.
Proofs
Proof 0
Let $R$ form a D273: Division ring.
Let $N$ and $M$ each form a D4718: Norm-metrised vector space over $R$.
Let $\Vert \cdot \Vert_N$ and $\Vert \cdot \Vert_M$ each be the D306: Vector space norm in $N$ and $M$, respectively.
Let $f : N \to M$ be a D705: Proportionally bounded linear map with respect to $N$ and $M$.
Since $f$ is proportionally bounded, there exists $C > 0$ such that $\Vert f(x) \Vert_M \leq C \Vert x \Vert_N$ for every $x \in N$. If $x, y \in N$, then linearity implies \begin{equation} \Vert f(x) - f(y) \Vert_M = \Vert f(x - y) \Vert_M \leq C \Vert x - y \Vert_N \end{equation} Since $x, y \in N$ were arbitrary, the claim follows. $\square$