Corrigendum: An enhanced uncertainty principle for the Vaserstein distance,Bulletin of the London Mathematical Society

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Corrigendum: An enhanced uncertainty principle for the Vaserstein distance
Bulletin of the London Mathematical Society ( IF 0.8 ) Pub Date : 2021-05-18 , DOI: 10.1112/blms.12521
Tom Carroll ₁ , Xavier Massaneda ₂ , Joaquim Ortega‐Cerdà ₂

Affiliation

One of the main results in the paper mentioned in the title is the following.

Theorem 1.Let $Q_{0} = {[0, 1]}^{d}$ be the unit cube in $R^{d}$ and let $f : Q_{0} \to R$ be a continuous function with zero mean. Let $Z (f)$ be the nodal set $Z (f) = {x \in Q : f (x) = 0}$ . Let $H^{d - 1} (Z (f))$ denote the $(d - 1)$ -dimensional Hausdorff measure of $Z (f)$ . Then

W_{1} (f^{+}, f^{-}) H^{d - 1} (Z (f)) {(\frac{{∥ f ∥}_{\infty}}{{∥ f ∥}_{1}})}^{2 - 1 / d} ≳ {∥ f ∥}_{1} .

(1)

Here $W_{1} (f^{+}, f^{-})$ indicates the Vaserstein distance between the positive and negative parts of $f$ .

The proof, which we now recall briefly, was based on a recursive selection, through a stopping time argument, of dyadic squares $Q$ where either the mass of $| f |$ is irrelevant, called empty, or $\int_{Q} f^{+}$ is much larger than $\int_{Q} f^{-}$ (or the other way around), called unbalanced.

Assume, without loss of generality, that

H^{d - 1} (Z (f))

is finite and that

f

is normalized so that

{∥ f ∥}_{1} = 1

. Consider now the standard dyadic partition of the unit cube: first split

Q_{0}

into

2^{d}

subcubes of length

1 / 2

and then, recursively, split each of the new subcubes into

2^{d}

“descendants” each with side length half that of the “parent”. A cube

Q \subset {[0, 1]}^{d}

is balanced if

\frac{1}{{100 ∥ f ∥}_{\infty}} ⩽ \frac{V_{f}^{+} (Q)}{V_{f}^{-} (Q)} ⩽ 100 {∥ f ∥}_{\infty},

where

V

indicates the volume and

V_{f}^{+} (Q) = V (Q \cap {f > 0}) and V_{f}^{-} (Q) = V (Q \cap {f < 0}) .

On the other hand,

Q

is empty if

\int_{Q} | f | < V (Q) / 10

Recursively, at each generation we set aside the cubes

\tilde{Q_{i}}

such that either:

(i) the cube $\tilde{Q_{i}}$ itself is empty;
(ii) one of the $2^{d}$ direct descendants $Q_{i}$ of $\tilde{Q_{i}}$ is full (non empty) and unbalanced. In this case, denote by $Q_{i}$ the full, unbalanced descendant of $\tilde{Q_{i}}$ for which $\int_{Q_{i}} | f |$ is maximal.

This gives a decomposition

Q_{0} = {[0, 1]}^{d} = (\cup_{i \in E} \tilde{Q_{i}}) \cup (\cup_{i \in F} \tilde{Q_{i}}) \cup R,

where

E

and

F

indicate respectively the indices of the empty and full selected cubes, and where

R

is the remaining set (whatever has not been selected in the process above).

We proved (Proposition 2) that

V (R) = 0

and that the full cubes

\tilde{Q_{i}}

carry most of the mass, i.e.

\sum_{i \in F} \int_{\tilde{Q_{i}}} | f | ⩾ 9 / 10

, and next we claimed (Remark 1) that the corresponding descendants

Q_{i}

i \in F

, still carry a significant part of that mass:

\sum_{i \in F} \int_{Q_{i}} | f | ⩾ \frac{9}{10} \times 2^{- d} .

(2)

Benjamin Jaye has kindly pointed out that this last estimate may in principle not be true in general, in that the mass in the full, unbalanced descendants $Q_{i}$ may not be comparable to the total mass of $f$ in $\tilde{Q_{i}}$ – this mass may lie predominantly in the balanced descendants of $\tilde{Q_{i}}$ . He suggested circumventing this difficulty by constructing the cubes $Q_{i}$ by means of a continuous stopping time argument together with the Besicovitch covering theorem, replacing our original dyadic-based decomposition. With these new cubes, Propositions 3 and 4 remain valid, hence also the statement of Theorem 1. The dyadic decomposition used to prove Theorem 2 requires analogous modification. The result itself, as well as all other results, remains correct as stated.

Details of the modifications to the proof of Theorem 1 are as follows. The function

f

is extended by 0 from

Q_{0}

to all of

R^{d}

. The definitions of balanced and unbalanced cubes are modified so that a cube

Q

is balanced if

\frac{1}{100 \times 5^{d} {∥ f ∥}_{\infty}} ⩽ \frac{V_{f}^{+} (Q)}{V_{f}^{-} (Q)} ⩽ 100 \times 5^{d} {∥ f ∥}_{\infty} .

(3)

The choice of the factor

5^{d}

is in accordance with the constant appearing in the Besicovitch covering theorem (see below).

A similar adjustment is made to the definition of full and empty cubes, in that a cube

Q

is empty whenever

\int_{Q} | f | ⩽ \frac{1}{10 \times 5^{d}} V (Q \cap Q_{0}) .

For every $x \in Q_{0}$ such that $f (x) \neq 0$ , there exists $l (x) > 0$ such that the open cube $Q_{x}$ centred at $x$ and of side length $l (x) = l (Q_{x})$ is simultaneously balanced and unbalanced in that either $V_{f}^{+} (Q) / V_{f}^{-} (Q)$ or $V_{f}^{-} (Q) / V_{f}^{+} (Q)$ equals $100 \times 5^{d} {∥ f ∥}_{\infty}$ . This can be achieved by continuity, since for $l$ very small the cube centred at $x$ and of side length $l$ is infinitely unbalanced, while for side length $l = 2$ it is balanced. Then there must be an intermediate side length $l (x)$ for which one of the inequalities in (3) is actually an identity.

These cubes

Q_{x}

cover

Q_{0}

(up to at most a zero-measure set). According to the Besicovitch covering theorem [1, Theorem 18.1] (see also [2]), one can find

5^{d}

sequences

{(x_{i, j})}_{i ⩾ 1}

j = 1, \dots, 5^{d}

, such that for each

j

the cubes

{(Q_{x_{i, j}})}_{i ⩾ 1}

are pairwise disjoint and together still cover

Q_{0}

Q_{0} \subset ⋃_{j = 1}^{5^{d}} ⋃_{i ⩾ 1} Q_{x_{i, j}} .

Since

\int_{Q_{0}} | f | = 1

, there is at least one family of cubes

{(Q_{x_{i, j}})}_{i ⩾ 1}

such that

\sum_{i ⩾ 1} \int_{Q_{x_{i, j}}} | f | ⩾ 5^{- d} .

From this particular sequence of cubes we select those that are full, and further relabel the cubes themselves as

{(Q_{i})}_{i ⩾ 1}

. These cubes are thus disjoint and carry most of the mass, since

\sum_{i : Q_{x_{i, j}} empty} \int_{Q_{x_{i, j}}} | f | ⩽ \frac{5^{- d}}{10} \sum_{i : Q_{x_{i, j}} empty} V (Q_{x_{i, j}} \cap Q_{0}) ⩽ \frac{5^{- d}}{10} .

In conclusion, estimate (2) holds for the cubes

{(Q_{i})}_{i ⩾ 1}

, up to a change of constants. Lemma 1 follows as before with only minor modifications due to extending

f

by 0 to all of

R^{d}

With the revised choice of the side length of the cubes $Q_{i}$ , both the estimates for the size of the zero set (Proposition 3) and for the Vaserstein distance (Proposition 4) hold for the new $Q_{i}$ , with the obvious modifications. The proof of Theorem 1 then concludes as before.

The proof of Theorem 2 follows similar arguments. One option is just to take in the existing proof the cells $M_{j, l} = φ_{l} (Q_{j})$ , where the cubes $Q_{j}$ are now obtained as explained above. Another option is to use Nash's embedding theorem and consider the $d$ -dimensional compact manifold $M$ as isometrically embedded in $R^{2 d}$ . Then the standard Besicovitch theorem in $R^{2 d}$ yields a similar Besicovitch covering theorem for $M$ (with constant $5^{2 d}$ instead of $5^{d}$ ).

中文翻译：

更正：Vaserstein 距离的增强不确定性原理

标题中提到的论文的主要结果之一如下。

定理 1.让 $问_{0} = {[0, 1]}^{d}$ 是单位立方体 ${电阻}^{d}$ 然后让 $F ：问_{0} \to 电阻$ 是一个均值为零的连续函数。让 $Z (F)$ 成为节点集 $Z (F) = {X \in 问： F (X) = 0}$ . 让 $H^{d - 1} (Z (F))$ 表示 $(d - 1)$ 维豪斯多夫测度 $Z (F)$ . 然后

宽_{1} (F^{+}, F^{-}) H^{d - 1} (Z (F)) {(\frac{{∥ F ∥}_{\infty}}{{∥ F ∥}_{1}})}^{2 - 1 / d} ≳ {∥ F ∥}_{1} .

(1)

这里 $宽_{1} (F^{+}, F^{-})$ 表示正负部分之间的 Vaserstein 距离 $F$ .

我们现在简要回顾一下这个证明，它基于通过停止时间参数对二元平方的递归选择 $问$ 其中任一质量 $| F |$ 无关紧要，称为空，或 $\int_{问} F^{+}$ 远大于 $\int_{问} F^{-}$ （或相反），称为unbalanced。

不失一般性，假设

H^{d - 1} (Z (F))

是有限的并且

F

被归一化，使得

{∥ F ∥}_{1} = 1

. 现在考虑单位立方体的标准二元划分：第一次分裂

问_{0}

进入

2^{d}

长度的子立方体

1 / 2

然后，递归地，将每个新的子立方体拆分为

2^{d}

每个“后代”的边长都是“父母”的一半。一个立方体

问 \subset {[0, 1]}^{d}

是平衡的，如果

\frac{1}{{100 ∥ F ∥}_{\infty}} ⩽ \frac{伏_{F}^{+} (问)}{伏_{F}^{-} (问)} ⩽ 100 {∥ F ∥}_{\infty},

在哪里

伏

表示音量和

伏_{F}^{+} (问) = 伏 (问 \cap {F > 0}) 和 伏_{F}^{-} (问) = 伏 (问 \cap {F < 0}) .

另一方面，

问

是空的如果

\int_{问} | F | < 伏 (问) / 10

递归地，在每一代我们都将立方体放在一边

\overset{～}{问_{一世}}

使得：

(i)立方体 $\overset{～}{问_{一世}}$ 本身是空的；
(ii)其中之一 $2^{d}$ 直系后代 $问_{一世}$ 的 $\overset{～}{问_{一世}}$ 已满（非空）且不平衡。在这种情况下，表示为 $问_{一世}$ 完整的，不平衡的后代 $\overset{～}{问_{一世}}$ 为此 $\int_{问_{一世}} | F |$ 是最大的。

这给出了一个分解

问_{0} = {[0, 1]}^{d} = (\cup_{一世 \in 乙} \overset{～}{问_{一世}}) \cup (\cup_{一世 \in F} \overset{～}{问_{一世}}) \cup 电阻,

在哪里

乙

和

F

分别表示空的和完整的选定立方体的索引，其中

电阻

是剩余的集合（在上面的过程中没有选择的任何东西）。

我们证明了（命题 2）

伏 (电阻) = 0

和完整的立方体

\overset{～}{问_{一世}}

承载大部分质量，即

\sum_{一世 \in F} \int_{\overset{～}{问_{一世}}} | F | ⩾ 9 / 10

, 接下来我们声明 (Remark 1) 相应的后代

问_{一世}

一世 \in F

，仍然承载着该质量的很大一部分：

\sum_{一世 \in F} \int_{问_{一世}} | F | ⩾ \frac{9}{10} \times 2^{- d} .

(2)

Benjamin Jaye 善意地指出，最后的估计在原则上一般来说可能不正确，因为完整的、不平衡的后代的质量 $问_{一世}$ 可能无法与总质量相比 $F$ 在 $\overset{～}{问_{一世}}$ – 这种物质可能主要存在于平衡的后代中 $\overset{～}{问_{一世}}$ . 他建议通过构建立方体来规避这个困难 $问_{一世}$ 通过连续停止时间参数和 Besicovitch 覆盖定理，取代了我们原来的基于二元的分解。有了这些新立方体，命题 3 和 4 仍然有效，因此定理 1 的陈述也是有效的。用于证明定理 2 的二进分解需要类似的修改。结果本身，以及所有其他结果，如所述保持正确。

对定理 1 证明的修改细节如下。功能

F

从 0 扩展

问_{0}

对所有

{电阻}^{d}

. 修改了平衡和非平衡立方体的定义，以便立方体

问

是平衡的，如果

\frac{1}{100 \times 5^{d} {∥ F ∥}_{\infty}} ⩽ \frac{伏_{F}^{+} (问)}{伏_{F}^{-} (问)} ⩽ 100 \times 5^{d} {∥ F ∥}_{\infty} .

(3)

因子的选择

5^{d}

符合 Besicovitch 覆盖定理中出现的常数（见下文）。

对满立方体和空立方体的定义进行了类似的调整，因为立方体

问

什么时候都是空的

\int_{问} | F | ⩽ \frac{1}{10 \times 5^{d}} 伏 (问 \cap 问_{0}) .

对于每 $X \in 问_{0}$ 以至于 $F (X) \neq 0$ ，那里存在 $升 (X) > 0$ 使得开放的立方体 $问_{X}$ 以 $X$ 和边长 $升 (X) = 升 (问_{X})$ 同时平衡和不平衡，因为要么 $伏_{F}^{+} (问) / 伏_{F}^{-} (问)$ 或者 $伏_{F}^{-} (问) / 伏_{F}^{+} (问)$ 等于 $100 \times 5^{d} {∥ F ∥}_{\infty}$ . 这可以通过连续性来实现，因为对于 $升$ 以非常小的立方体为中心 $X$ 和边长 $升$ 是无限不平衡的，而对于边长 $升 = 2$ 它是平衡的。那么一定有一个中间边长 $升 (X)$ 其中（3）中的一个不等式实际上是一个恒等式。

这些立方体

问_{X}

覆盖

问_{0}

（最多一个零测量集）。根据 Besicovitch 覆盖定理 [ 1 , Theorem 18.1]（另见 [ 2 ]），可以发现

5^{d}

序列

{(X_{一世, j})}_{一世 ⩾ 1}

j = 1, \dots, 5^{d}

，这样对于每个

j

立方体

{(问_{X_{一世, j}})}_{一世 ⩾ 1}

成对不相交并且在一起仍然覆盖

问_{0}

：

问_{0} \subset ⋃_{j = 1}^{5^{d}} ⋃_{一世 ⩾ 1} 问_{X_{一世, j}} .

自从

\int_{问_{0}} | F | = 1

，至少有一个立方体族

{(问_{X_{一世, j}})}_{一世 ⩾ 1}

以至于

\sum_{一世 ⩾ 1} \int_{问_{X_{一世, j}}} | F | ⩾ 5^{- d} .

从这个特定的立方体序列中，我们选择那些已满的立方体，并进一步将立方体本身重新标记为

{(问_{一世})}_{一世 ⩾ 1}

. 这些立方体因此是不相交的并承载了大部分质量，因为

\sum_{一世 ： 问_{X_{一世, j}} 空的} \int_{问_{X_{一世, j}}} | F | ⩽ \frac{5^{- d}}{10} \sum_{一世 ： 问_{X_{一世, j}} 空的} 伏 (问_{X_{一世, j}} \cap 问_{0}) ⩽ \frac{5^{- d}}{10} .

总之，估计（2）适用于立方体

{(问_{一世})}_{一世 ⩾ 1}

，直到常数的变化。引理 1 与以前一样，只是由于扩展而进行了微小的修改

F

由 0 到所有

{电阻}^{d}

随着立方体边长的修正选择 $问_{一世}$ ，零集大小（命题 3）和 Vaserstein 距离（命题 4）的估计值都适用于新的 $问_{一世}$ ，有明显的修改。定理 1 的证明然后如前结束。

定理 2 的证明遵循类似的论证。一种选择是采用现有的证明细胞 $米_{j, 升} = φ_{升} (问_{j})$ ，其中立方体 $问_{j}$ 现在获得如上所述。另一种选择是使用纳什嵌入定理并考虑 $d$ 维紧凑流形 $米$ 作为等距嵌入 ${电阻}^{2 d}$ . 那么标准的贝西科维奇定理 ${电阻}^{2 d}$ 产生类似的贝西科维奇覆盖定理 $米$ （与常数 $5^{2 d}$ 代替 $5^{d}$ ）。

更新日期：2021-05-18

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