functional analysis – How to prove that support of $u_{r}$ is compact?

Let’s ${u_r}:{mathbb{R}^k}to{mathbb{R}}$, $varphi in C_{c}(mathbb{R}^n)$, $Jf$ ( jacobian of $f$, with $f: mathbb{R}^{k}to mathbb{R}^{n}$ Lipschitz ), and $chi_E$ (characteristic function on $E subset mathbb{R}^{k}$ where $E$ borel bounded subset),
begin{equation*}
u_r(w) := chi_E(z+rw)varphiBigl(frac{f(z+ rw)-f(z)}{r}Bigr) J f(z+rw).
end{equation*}

Statement: there are $ r_0> 0 $ and $ R> 0 $ such that $ supp(u_{r}) subset mathbb{B}(0,R) $ for $ r in (0, r_{0}) $.

My attempt:
In fact, since $ f $ is derivable into $z$ and $ Jf(z)> 0 $, there are $ s_{0}, lambda > 0 $ such that
begin{equation}
|{f(z’)-f(z)}| geq lambda ||{z’-z}||
end{equation}

for every $ z ‘in mathbb{B}(z,s_{0}) $. On the other hand, if $ rho> 0 $ is such that $ supp (varphi) subset mathbb{B}(0,rho) $, then
begin{equation*}
|{f(z+rw) -f(z)}|leq rrho
end{equation*}

for all $win supp(u_{r})$. Hence, if $ w in supp(u_{r}) $ with $ r < s_{0} / rho $, you have $ z + rw in mathbb{B}(z,s_{0}) $, so $ r rho geq |{f (z + rw) -f (z)}| geq lambda ||{z + rw-z}|| = lambda r ||{w}|| $, then $ ||{w}|| leq rho / lambda $, which proves statement with $ r_{0}: = s_{0} / rho $ and $ R: = rho / lambda $.

I’m not sure what $ z + rw in mathbb{B}(z,s_{0}) $ , I really appreciate it if someone could give me an idea how to improve this argument.

Another idea that I had, was to prove that the $supp(u_{0})$ is compact, which I did, in order to arrive at that the $supp(u_{r})$ is compact, which would be another way to conclude that statement in another way. But unfortunately I could not find that relationship between the $supp(u_{0})$ and the $supp(u_{r}) $I really appreciate the attention given.