Bound on the derivative of a cut-off function Let $\rho$ be a smooth function in $\mathbb R^n$ such that $0 \leq \rho \leq 1$ and $\rho$ is supported in the unit disk and let $\rho_\epsilon(x) = \epsilon^{-n}\rho(\epsilon^{-1}\|x\|)$.
If $f$ is a $\alpha$-Holder continuous function in $\mathbb R^n$ it's not hard to show that $\| f \ast \rho_\epsilon - f \|_\infty \leq \epsilon^\alpha$. I guess that we also have  $\|f \ast \rho_\epsilon \|_{C^1} \leq \epsilon^{\alpha - 1}$.
Is this last inequality true? If not can we chose a particular $\rho$ in order to obtain it?
Thanks in advance.

Yes, this is true. The supremum part of $C^1$ was already understood, so let's look at the derivative. With  convolutions  we can choose where the derivative falls: 
$$\nabla (f*\rho_\epsilon) = f*(\nabla \rho_\epsilon)$$
Let $\eta=\nabla \rho$ and 
$\eta_\epsilon = \epsilon \nabla \rho_\epsilon$. By the chain rule, $\eta_\epsilon(x) = \epsilon^{-n}\eta(\epsilon^{-1}\|x\|)$, same relation as for $\rho$. Therefore, the   estimate you have for $f*\rho_\epsilon$ also works for $f*\eta_\epsilon$, with the modification that we don't need to subtract $f$ since $\eta$ integrates to zero:
$$\| f \ast \eta_\epsilon   \|_\infty \lesssim  \epsilon^\alpha$$
  (vector-valuedness makes no difference, just look at one component at a time). 
Hence $$\|f*(\nabla \rho_\epsilon)\|_\infty\lesssim \epsilon^{\alpha-1}$$
