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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 $\varepsilon > 0$.

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    $\begingroup$ If one of the sets $U_j$ is very small but not contained in the uion of the others the corresponding $\rho_j$ takes the value $1$ and thus has to be very steep. $\endgroup$ Commented Jan 3, 2018 at 15:52
  • $\begingroup$ Let me change the question slightly to make this a bit more reasonable. $\endgroup$ Commented Jan 3, 2018 at 16:07
  • $\begingroup$ Your modification does not change the situation (as stated, you could have small components of $U_j$). For bounds on the derivatives you need more restrictions on the geometry. $\endgroup$ Commented Jan 3, 2018 at 16:24
  • $\begingroup$ Ahhhh I see. Ok, thanks! How about if I add that each $U_j$ is connected? Is it still trivially not true? $\endgroup$ Commented Jan 3, 2018 at 16:25
  • $\begingroup$ You may wish to assume that for some fixed $\epsilon > 0$ the sets $V_j = \{x : \operatorname{dist}(x, U_j^c) > \epsilon\}$ still cover your set. To see this simply follow the construction of the partition of unity: you can then convolve indicator functions with a fixed bump function, supported in a ball of radius $\epsilon$. Let me know if you like me to give more details. $\endgroup$ Commented Jan 3, 2018 at 19:44

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(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 $\|\nabla \phi\|_1$.

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Consider $B = [0,1]$. Let $U_{1,t} = (t,1-t)$ and $U_{2,t} = [0,2t) \cup (1-2t,1]$. Now use the mean value theorem to estimate the derivatives.

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  • $\begingroup$ Thank you for this. Let me change the question slightly so that the question is a bit more reasonable. $\endgroup$ Commented Jan 3, 2018 at 16:07

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