To be honest, I don't get the proof in Folland, "Real Analysis", p.74.
Let $\|\cdot\|$ be the max norm on $\mathbb{R}^n$.
Let $\Omega$ be open in $\mathbb{R}^n$.
Let $G:\Omega \rightarrow \mathbb{R}^n$ be a $C^1$ diffeomorphism.
Define $Q(a,\delta)=\{x\in\mathbb{R}^n : \|x-a|\|\leq \delta\}$ such that $Q(a,\delta)\subset \Omega$.
Then $\|G(x)-G(a)\|\leq \delta \sup_{y\in Q(a,\delta)} \max_{1\leq i \leq n} \sum_{j=1}^n |D_j G_i(y)|$ for all $x\in Q(a,\delta)$
I'm not sure why $\sup_{y\in Q(a,\delta)} \max_{1\leq i \leq n} \sum_{j=1}^n |D_j G_i(y)| < \infty$.
Is there any webpage or text proving this theorem precisely?
If $G$ is $C^1$, and $[x,y] \subset \Omega$, we have $G(x) -G(y) = \left( \int_0^1 DG(y+t(x-y)) dt \right) (x-y)$.
Then $\|G(x)-G(y)\| \le \| \int_0^1 DG(y+t(x-y)) dt \| \|x-y\|$, where the norm on the integral is the induced norm.
If $L = \sup_{x \in Q(a,\delta)} \| DG(x) \|$, then we have $\|G(x)-G(y)\| \le L \delta$.
To see why $L < \infty$, note that $x \mapsto \|DG(x)\|$ is continuous and $Q(a,\delta)$ is compact.
Finally, note that the induced norm is given by $\|M\|_{i,\infty} = \max_{i \in \{1,...,n\}} \sum_{j \in \{1,...,n\}} [M]_{ij}$.