Let $\varphi:U\to \mathbb R^n$ a $\mathcal C^1$ application where $U\subset \mathbb R^n$ open.
Let $(\varepsilon_k)\subset \mathbb R^*_+$ a sequence of positif number s.t. $\varepsilon_k\to 0$ and denote $$C_k:=\{x\in\mathbb R^n\mid \|x-a_k\|_{\infty }\leq \frac{\varepsilon_k}{2}\},$$where $\displaystyle\|x\|_\infty :=\max_{i=1,...,n}|x_i|$
Suppose $0\in C_k$ for all $k\geq 1$ and that $\varphi(0)=0$.
Let $T:=\mathcal J_{\varphi}(0)$ the Jacobian matrix of $\varphi$ at $x=0$.
I'm trying to understand the proof of $$\liminf_{n\to \infty }\frac{m(\varphi(C_k))}{m(C_k)}\leq |\det T|,$$ where $m$ is the Lebesgue measure of $\mathbb R^n$.
So we suppose $\det T\neq 0$. Then $$\varphi(x)=Tx+\|x\|_\infty \rho(x),$$ where $\|\rho(x)\|_\infty \to 0$ when $\|x\|_\infty \to 0$. Therefore $$T^{-1}\varphi(x)=x+\|x\|_\infty T^{-1}\rho(x).$$ Observe that $\|T^{-1}\rho(x)\|_\infty \to 0$ when $\|x\|_\infty \to 0$. Recall that $0\in C_k$, i.e. $\|a_k\|_\infty \leq \frac{\varepsilon_k}{2}.$ Therefore, if $x\in C_k$, then $$\|x\|_\infty \leq \|x-a_k\|_\infty +\|a_k\|_\infty \leq \frac{\varepsilon_k}{2}+\frac{\varepsilon_k}{2} =\varepsilon_k.$$ Therefore, if $x\in C_k$, then $$T^{-1}\varphi(x)=x+y,$$ where $$\|y\|_\infty =\|x\|_\infty \|T^{-1}\rho(x)\|_\infty <\varepsilon_k\delta_k,$$ where $$2\delta_k=\sup_{0\leq \|x\|_{\infty }\leq \varepsilon_k}\|T^{-1}\rho(x)\|_\infty .$$
This is what I don't understand :
Question : For me $$\|y\|_{\infty }=\underbrace{\|x\|_\infty}_{\leq \varepsilon_k} \underbrace{\|T^{-1}\rho(x)\|_\infty}_{\leq 2\delta_k}\leq 2\varepsilon_k\delta_k,$$ So how do they get $\|y\|_\infty <\varepsilon_k\delta_k$ ? (i.e. the strict inequality and the upperbound $\varepsilon_k\delta_k$.)