Question about Lipschitz function Suppose $A = (a_{ij})$, $1 \leq i \leq m$, $1 \leq j \leq n$ is an $m \times n$ matrix. then $A: \mathbb{R}^n \to \mathbb{R}^m$ given by
$$ A(x) = A(x_1,\dots,x_n) = \left( \sum_{j=1}^n a_{1j}x_j,\dots, \sum_{j=1}^n a_{mj}x_j \right) $$ then we have 
$$ ||A(x)||_{\infty} \leq C ||x||_{\infty}.$$
Using this fact, can we show that any $A$ is Lipschitz function?
$$|| A(x) - A(y) || = || A(x-y)|| \leq C ||x-y|| \;\;\;\;\; \forall x,y \in \mathbb{R}^n .$$
Do we need to show indeed that $A$ is a linear map?
 A: Let $A = (a_{i,j})_{\substack{1 \leq i \leq m \\ 1 \leq j \leq n}}$ a real $m \times n$ matrix. On $\mathbb{R}^{m}$, we define the norm $\Vert \cdot \Vert_{\infty, \mathbb{R}^{m}}$ by :
$$ \forall x \in \mathbb{R}^{m}, \; \Vert x \Vert_{\infty , \mathbb{R}^{m}} = \max \limits_{1 \leq i \leq m} \vert x_{i} \vert $$
and we could give a similar definition for $\Vert \cdot \Vert_{\infty,\mathbb{R}^{n}}$ on $\mathbb{R}^{n}$. We have :
$$
\begin{align*}
\Vert Ax \Vert_{\infty,\mathbb{R}^{m}} & = {} \max \limits_{1 \leq i \leq m} \Big\vert \sum_{j=1}^{n} a_{i,j}x_{j} \Big\vert \\
 &\leq \max \limits_{1 \leq i \leq m} \sum_{j=1}^{n} \vert a_{i,j} \vert \vert x_{j} \vert \\
 &\leq \Big( \max \limits_{1 \leq i \leq m} \sum_{j=1}^{n} \vert a_{i,j} \vert \Big) \Vert x \Vert_{\infty,\mathbb{R}^{n}}
\end{align*}
$$
By letting $\displaystyle C=\max \limits_{1 \leq i \leq m} \sum_{j=1}^{n} \vert a_{i,j} \vert$, we have : 
$$\forall x \in \mathbb{R}^{n}, \; \Vert Ax \Vert_{\infty} \leq C \Vert x \Vert_{\infty} \tag{$\star$}$$ 
By the way, $C$ can also be written $\Vert A \Vert$ as it defines a norm on $\mathrm{Mat}_{m \times n}(\mathbb{R})$. Now, let $\Vert \cdot \Vert_{\mathbb{R}^{m}}$ (respectively $\Vert \cdot \Vert_{\mathbb{R}^{n}}$) be any norm on $\mathbb{R}^{m}$ (resp. $\mathbb{R}^{n}$). Since these two spaces are finite-dimensional vector spaces, it follows from the norm equivalence theorem that $\Vert \cdot \Vert_{\mathbb{R}^{m}}$ and $\Vert \cdot \Vert_{\infty,\mathbb{R}^{m}}$ (resp. $\Vert \cdot \Vert_{\mathbb{R}^{n}}$ and $\Vert \cdot \Vert_{\infty,\mathbb{R}^{n}}$ are equivalent. By definition, it means that there exist $(c_{1},c_{2}) \in (0,+\infty)^{2}$, $(d_{1},d_{2}) \in (0,+\infty)^{2}$ such that :
$$ \forall x \in \mathbb{R}^{m}, \; c_{1} \Vert x \Vert_{\infty,\mathbb{R}^{m}} \leq \Vert x \Vert_{\mathbb{R}^{m}} \leq c_{2} \Vert x \Vert_{\infty,\mathbb{R}^{m}} $$
and :
$$ \forall x \in \mathbb{R}^{m}, \; d_{1} \Vert x \Vert_{\infty,\mathbb{R}^{n}} \leq \Vert x \Vert_{\mathbb{R}^{n}} \leq d_{2} \Vert x \Vert_{\infty,\mathbb{R}^{n}} $$
Using these norm equivalences and $(\star)$, you can easily prove that there exists a constant $C'>0$ such that :
$$ \forall x \in \mathbb{R}^{n}, \; \Vert Ax \Vert_{\mathbb{R}^{m}} \leq C' \Vert x \Vert_{\mathbb{R}^{n}} $$
And now, using the linearity of $A$, you get the conclusion :
$$ \forall (x,y) \in \big( \mathbb{R}^{n} \big)^{2}, \; \Vert Ax - Ay \Vert_{\mathbb{R}^{m}} \leq C' \Vert x - y \Vert_{\mathbb{R}^{n}} $$
