The question is as in the title.
Let $f : \mathbb{R}^n \to \mathbb{R}^n$ be a smooth mapping and denote its Jacobian matrix at each $x \in \mathbb{R}^n$ as $(Df)(x)$.
By assumption, $(Df)(x)$ is a $n \times n$ square matrix and itself a smooth mapping from $\mathbb{R}^n \to \mathbb{R}^{n^2}$.
Let us further assume that the "operator norm" of $[Df](x)$ with respect to the Euclidean norm of $\mathbb{R}^n$ is bounded by $1$ for all $x \in \mathbb{R}^n$: \begin{equation} \lVert (Df)(x) \rVert_{operator} \leq 1 \text{ for all } x \in \mathbb{R}^n. \end{equation}
Then, is it true that $f$ is Lipschitz continuous with the modulus of continuity bounded by $1$? That is, \begin{equation} \lVert f(x)-f(y) \rVert \leq \lVert x-y \rVert \end{equation} for all $x,y \in \mathbb{R}^n$? Here $\lVert \cdot \rVert$ is the Euclidean norm of $\mathbb{R}^n$.
I strongly suspect that the "operator norm" is the correct matrix norm for estimating the modulus of Lipschitz continuity for vector-valued functions, but I have trouble justifying my belief rigorously..
Could anyone please help me?