A linear transformation of $\mathbb{R}^n$ is Lipschitz

Question

Definition:

A transformation $y = Tx$ of $\mathbb{R}^n$ into itself is Lipschitz if there is a constant $c$ such that $|Tx - Tx'| \leq c|x - x'|$.

Write $y = (y_1, \ldots, y_n)$. If $y_j = f_j(x), j = 1, \ldots, n$ are the coordinate functions representing $T$. Then $T$ is Lipschitz iff each $f_j$ satisfies a Lipschitz condition $|f_j(x) - f_j(x')| \leq c_j|x - x'|$.

CLAIM: a linear transformation of $\mathbb{R}^n$ is Lipschitz

My proof is as follows:

Let $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a linear transformation. Let $x, x' \in \mathbb{R}^n$. Since $T$ is linear, we have \begin{align} (f_1(x - x'), \ldots, f_n(x - x')) = T(x - x') = Tx - Tx'= (f_1(x) - f_1(x'), \ldots, f_n(x) - f_n(x')). \end{align} Let $\displaystyle c_j = \sup_{x \neq x' \in \mathbb{R}^n} \frac{|f_j(x-x')|}{|x-x'|}$, then

\begin{align} |f_j(x) - f_j(x')| = \frac{|f_j(x - x')|}{|x - x'|} |x - x'| \leq c_j|x - x'| \end{align} for each $j$. That is, each $f_j$ satisfies a Lipschitz condition. So $T$ is Lipschitz.

Is my proof correct?

Thanks.

Answer 1

You did not prove that $c_j <\infty$. For any linear map $f: \mathbb R^{n} \to \mathbb R$ write $f(x)$as $\sum x_i f(e_i)$ where $(e_i)$ is the standard basis. Note that $|f(x)| \leq M\|x\|$ where $M=\sqrt {\sum |f(e_i)|^{2}}$ (by Cauchy Schwartz inequality). This also gives $|f(x-y)| \leq M\|x-y\|$. This $c_j <\infty$ for each $j$.