Let $B \subseteq \mathbb{R}^n$ be a product of closed bounded intervals in $\mathbb{R}$. Fix $N>0$. Suppose I want to cover $B$ with $N$ open sets, $U\_1, \ldots, U\_N$, and get a smooth partition of unity $\rho\_1, \ldots, \rho\_N$ with respect to these sets. I was wondering is it possible to do this in a way that the derivatives of the $\rho\_j$'s are bounded independent of the choice of the $U\_j$'s? 


I was wondering maybe I am asking for too much here and that this is not possible, or maybe it's possible? I have no idea... I would appreciate any comments or suggestions. Thank you. 


PS I would like to change the question slightly. I would like to assume that each $U\_j$ is not too small in that each $U\_j$ contains an open set of the form $(x\_1 - \varepsilon, x\_1+ \varepsilon) \times \cdots \times (x\_n - \varepsilon, x\_n+ \varepsilon)$ for some $\varepsilon > 0$. 



Can we get smooth parition of unity with uniformity?
(This is an extended version of my comment above).


Suppose that $U\_j$, $j = 1, \ldots, N$, is an open cover of $B$, $\epsilon > 0$ and that the sets $$V\_j = \{x \in U\_j : \operatorname{dist}(x, U\_j^c) > \epsilon\}, \; j = 1, \ldots, N,$$ also form an open cover of the $\epsilon$-neighbourhood of $B$. Then one can find the partition of unity $\rho\_j$, $j = 1, \ldots, N$, such that $\rho\_j$ is zero outside of $U\_j$ and $\nabla \rho\_j$ is bounded by a constant that depends only on $\epsilon$ and the dimension. (Actually, the same statement is true for derivatives of $\rho\_j$ of arbitrary order).


To prove the above claim, define $$W\_j = V\_j \setminus (V\_1 \cup \ldots \cup V\_{j-1}) , \; j = 1, \ldots, N ,$$ and let $\phi$ be a bump function supported in $B(0, \epsilon)$ (that is, $\phi$ is infinitely smooth, non-negative, with total mass $1$). Then $\rho\_j = \mathbb{1}\_{V\_j} \* \phi$, $j = 1, \ldots, N$, form a smooth partition of unity on $B$: all these functions take values in $[0, 1]$, they are infinitely smooth and $$ \rho\_1 + \ldots + \rho\_N = \mathbb{1}\_{V\_1 \cup \ldots \cup V\_N} \* \phi $$ is equal to $1$ on $B$, because $V\_1 \cup \ldots \cup V\_N$ contains the $\epsilon$-neighbourhood of $B$. Furthermore, $\rho\_j$ is clearly equal to zero in the complement of $U\_j$. Finally, $\nabla \rho\_j = \mathbb{1}\_{V\_j} \* \nabla \phi$ is bounded by $\|\nabla \phi\|\_1$.


