1
$\begingroup$

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 $\varepsilon > 0$.

$\endgroup$
6
  • 2
    $\begingroup$ 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. $\endgroup$ – Jochen Wengenroth Jan 3 '18 at 15:52
  • $\begingroup$ Let me change the question slightly to make this a bit more reasonable. $\endgroup$ – Johnny T. Jan 3 '18 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. $\endgroup$ – Jochen Wengenroth Jan 3 '18 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? $\endgroup$ – Johnny T. Jan 3 '18 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. $\endgroup$ – Mateusz Kwaśnicki Jan 3 '18 at 19:44
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you for this. Let me change the question slightly so that the question is a bit more reasonable. $\endgroup$ – Johnny T. Jan 3 '18 at 16:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.