Skip to main content

Correction to: Generalized Ponce’s inequality

The Original Article was published on 14 January 2021

Introduction

We assume that \(\Omega \subset \mathbb{R}^{N}\) is an open bounded domain with Lipschitz boundary; \(( k_{\delta } ) _{\delta >0}\) is a set of radial positive functions such that \(\operatorname*{supp}k_{\delta }\subset B ( 0,\delta ) \), \(\frac{1}{C_{N}}\int _{B ( 0,\delta ) }k_{\delta } ( \vert s \vert )\,ds=1\), where \(C_{N}=\frac{1}{\operatorname*{meas} ( S^{N-1} ) }\int _{S^{N-1}} \vert \sigma \cdot \mathbf{e} \vert ^{p}\,d\mathcal{H}^{N-1} ( \sigma ) \), \(\mathcal{H}^{N-1}\) is the \(( N-1 ) \)-dimensional Hausdorff measure on the unit sphere \(S^{N-1}\), e is any unit vector in \(\mathbb{R}^{N}\), \(p>1\), and \(B(x,\delta )\) is the the ball with center x and radius δ.

In [2], under the assumptions above, the following compactness is recalled (see [2] and references therein):

Theorem 1

Assume Ω is an open bounded domain with Lipschitz boundary. Let \(( u_{\delta } ) _{\delta }\) be a sequence uniformly bounded in \(L^{p} ( \Omega )\), and let C be a positive constant such that

$$ \int _{\Omega } \int _{\Omega } \frac{k_{\delta } ( \vert x^{\prime }-x \vert ) }{ \vert x^{\prime }-x \vert ^{p}} \bigl\vert u_{\delta } \bigl( x^{\prime } \bigr) -u_{\delta } ( x ) \bigr\vert ^{p}\,dx^{\prime }\,dx\leq C $$
(1.1)

for any δ. Then, from \(( u_{\delta } ) _{\delta }\) we can extract a subsequence, still denoted by \(( u_{\delta } ) _{\delta }\), and we can find \(u\in W^{1,p} ( \Omega ) \) such that \(u_{\delta }\rightarrow u\) strongly in \(L^{p} ( \Omega ) \) as \(\delta \rightarrow 0\) and

$$ \lim_{\delta \rightarrow 0} \int _{\Omega } \int _{\Omega } \frac{k_{\delta } ( \vert x^{\prime }-x \vert ) }{ \vert x^{\prime }-x \vert ^{p}} \bigl\vert u_{\delta } \bigl( x^{\prime } \bigr) -u_{\delta } ( x ) \bigr\vert ^{p}\,dx^{\prime }\,dx\geq \int _{\Omega } \bigl\vert \nabla u ( x ) \bigr\vert ^{p}\,dx. $$
(1.2)

Even though several authors are involved in the proof, we refer to estimate (1.2) as Ponce’s inequality.

The goal of [2] is to prove the following extension of (1.2):

$$ \lim_{\delta \rightarrow 0} \int _{\Omega } \int _{\Omega }H \bigl( x^{ \prime },x \bigr) \frac{k_{\delta } ( \vert x^{\prime }-x \vert ) }{ \vert x^{\prime }-x \vert ^{p}} \bigl\vert u_{\delta } \bigl( x^{\prime } \bigr) -u_{\delta } ( x ) \bigr\vert ^{p}\,dx^{\prime } \,dx\geq \int _{\Omega }h ( x ) \bigl\vert \nabla u ( x ) \bigr\vert ^{p}\,dx, $$
(1.3)

where Ω is an open bounded set with Lipschitz boundary, \(H ( x^{\prime },x ) = \frac{h ( x^{\prime })+h(x ) }{2}\), and h is a nonnegative function from \(L^{\infty } ( \Omega ) \).

Alternatively, the goal is to check the inequality (1.2) for measurable sets, that is,

$$\begin{aligned}& \lim_{\delta \rightarrow 0} \int _{E} \int _{E} \frac{k_{\delta } ( \vert x^{\prime }-x \vert ) }{ \vert x^{\prime }-x \vert ^{p}} \bigl\vert u_{\delta } \bigl( x^{\prime } \bigr) -u_{\delta } ( x ) \bigr\vert ^{p}\,dx^{\prime }\,dx \\& \quad \geq \int _{E} \bigl\vert \nabla u ( x ) \bigr\vert ^{p}\,dx,\quad \text{for any measurable }E\subset \Omega . \end{aligned}$$
(1.4)

It must be remarked that both inequalities are true but some basis for the proofs is false. Concretely, Proposition 1 from [2, p. 3] is wrong and, consequently, those parts where it is used have to be modified. Let us go through the steps and distinguish which parts are faulty.

First proof

Proposition 2 from [2, p. 4] is true and its proof is correct. The analysis application derived that proposition establishes

$$ \liminf_{\delta \rightarrow 0} \int _{O} \int _{O}H \bigl( x^{\prime },x \bigr) \frac{k_{\delta } ( \vert x^{\prime }-x \vert ) }{ \vert x^{\prime }-x \vert ^{p}} \bigl\vert u_{\delta } \bigl( x^{\prime } \bigr) -u_{\delta } ( x ) \bigr\vert ^{p}\,dx^{ \prime } \,dx\geq \int _{O}h ( x ) \bigl\vert \nabla u ( x ) \bigr\vert ^{p}\,dx $$
(2.1)

for any symmetric nonnegative continuous function \(F\in L^{\infty } ( O\times O ) \) and any smooth open set O such that \(\vert \partial O \vert =0\). However, the proof extending (1.3) to the case where H is a measurable function of \(L^{\infty } ( \Omega ) \) is invalid because it relies on Proposition 1.

The extension to the case of measurable functions is possible because Proposition 2 from [2, p. 4] is also true for the case \(p=\infty \) and \(q=1\). Let us check it. By looking back at the original work where the idea of the proof comes from, we can check that this result is valid for all \(f\in L^{p}\) and \(\xi \in L^{q} ( \Omega ) \), with \(\frac{1}{p}+\frac{1}{q}=1\), even for the case \(p=\infty \) and \(q=1\) (see [3, p. 126]). Namely, in [3, p. 130], given \(f\in L^{p}\), we can select a family of disjoint sets \(\{ a_{kj}+\epsilon _{kj}\overline{\Omega } \} _{j}\) covering Ω such that

$$ \int _{\Omega }f ( x ) \psi ( x )\,dx\leq \sum _{i}f ( a_{ki} ) \int _{a_{ki}+\epsilon _{ki}\Omega }\psi ( x )\,dx-\frac{1}{k} \vert \Omega \vert ^{1/p} \Vert \psi \Vert _{L^{q} ( \Omega ) }$$

for any \(\psi \in L^{q}\).

Now, for simplicity, we assume \(f\in L^{\infty }\) and ξ \(L^{1}\) are nonnegative functions. Since \(\xi ^{1/q}\in L^{q}\) for any q, and \(f\in L^{p}\) for any p, the above inequality for \(\psi =\xi ^{1/q}\) reads as

$$ \int _{\Omega }f ( x ) \xi ^{1/q} ( x )\,dx\leq \sum _{i}f ( a_{ki} ) \int _{a_{ki}+\epsilon _{ki}\Omega } \xi ^{1/q} ( x )\,dx- \frac{1}{k} \vert \Omega \vert ^{1/p} \Vert \xi \Vert _{L^{1} ( \Omega ) }^{1/q}. $$

If we pass to the limit as \(p\uparrow \infty \), then \(q=\frac{p}{p-1}\downarrow 1\) and \(\xi ^{1/q} ( x ) \rightarrow \xi ( x ) \), and, consequently, by monotone and dominated convergence for series and integrals, we infer

$$ \int _{\Omega }f ( x ) \xi ( x )\,dx\leq \sum _{i}f ( a_{ki} ) \int _{a_{ki}+\epsilon _{ki}\Omega }\xi ( x )\,dx-\frac{1}{k} \Vert \xi \Vert _{L^{1} ( \Omega ) }. $$

Using this inequality and following the previous procedure, then we can conclude that (2.1) remains valid for any symmetric and nonnegative function \(F\in L^{\infty } ( O\times O ) \) and any smooth domain \(O\subset \Omega \) such that \(\vert \partial O \vert =0\).

Finally, in order to circumvent the assumption \(\vert \partial \Omega \vert =0\), we simplify as follows: for the given domain Ω, we consider Ω̃, a regular domain containing Ω whose boundary is a null set, and we extend H by zero in \(( \widetilde{\Omega }\times \widetilde{\Omega } ) \setminus ( \Omega \times \Omega ) \). We denote this extended function of H by \(H_{0}\), which is measurable, symmetric, and nonnegative. In the same way, we also appropriately extend \(u_{\delta }\) to Ω̃, so that (1.1) still holds. To do that, we first note that Ω is smooth and, therefore, we can extend u to \(\widetilde{u}\in W^{1,p} ( \widetilde{\Omega } ) \). Then, we define \(\widetilde{u}_{\delta } ( x ) =u ( x ) \) if x \(\widetilde{\Omega }\setminus \Omega \) and \(\widetilde{u}_{\delta } ( x ) =u_{\delta } ( x ) \) if \(x\in \Omega \). It is immediate to check that \(( \widetilde{u}_{\delta } ) _{\delta }\) is uniformly bounded in \(L^{p}\) and

$$ \int _{\widetilde{\Omega }} \int _{\widetilde{\Omega }}H_{0} \bigl( x^{ \prime },x \bigr) \frac{k_{\delta } ( \vert x^{\prime }-x \vert ) }{ \vert x^{\prime }-x \vert ^{p}} \bigl\vert \widetilde{u}_{\delta } \bigl( x^{\prime } \bigr) - \widetilde{u}_{\delta } ( x ) \bigr\vert ^{p}\,dx^{\prime }\,dx \leq C. $$

Then, by Theorem 1, we obtain

$$ \liminf_{\delta \rightarrow 0} \int _{\widetilde{\Omega }} \int _{ \widetilde{\Omega }}H_{0} \bigl( x^{\prime },x \bigr) \frac{k_{\delta } ( \vert x^{\prime }-x \vert ) }{ \vert x^{\prime }-x \vert ^{p}} \bigl\vert \widetilde{u}_{\delta } \bigl( x^{\prime } \bigr) - \widetilde{u}_{\delta } ( x ) \bigr\vert ^{p}\,dx^{\prime }\,dx \geq \int _{\widetilde{\Omega }}H_{0} ( x,x ) \bigl\vert \nabla \widetilde{u} ( x ) \bigr\vert ^{p}\,dx. $$

Now we realize that the above inequality coincides with (2.1) for any open and bounded set Ω.

The analysis performed proving a corollary in Sect. 2.3 in [2, p. 7] is correct and therefore serves to establish that (1.4) is valid for all measurable sets \(G\subset \Omega \).

A second proof

This part of the paper deserves a stark modification because the proof given in [2] is based entirely on Proposition 1.

We first prove (1.4) and then (1.3). We assume Ω is open and \(\vert \partial \Omega \vert =0\). By hypothesis, \(( \xi _{\delta } ) _{\delta }\) is a sequence uniformly bounded in \(L^{1} ( \Omega \times \Omega )\) and, under these circumstances, we can use Chacon’s biting lemma (see [1]) to ensure the existence of a subsequence of \(\delta ^{\prime }s\), not relabeled, a decreasing sequence of measurable sets \(\mathcal{E}_{n}\subset \Omega \times \Omega \), such that \(\vert \mathcal{E}_{n} \vert \downarrow 0\), and a function \(\xi \in L^{1} ( \Omega \times \Omega ) \) such that \(\xi _{\delta }\rightharpoonup \xi \) weakly in \(L^{1} ( \Omega \times \Omega \setminus \mathcal{E}_{n} ) \) for all n. Since we are dealing with a sequence of symmetric functions, we can ensure \(( \Omega \times \Omega ) \setminus \mathcal{E}_{n}= ( \Omega \setminus E_{n} ) \times ( \Omega \setminus E_{n} ) \) where the sequence of sets \(E_{n}\subset \Omega \) is decreasing and \(\vert E_{n} \vert \downarrow 0\) if \(n\rightarrow \infty \).

Let \(O_{n}\) be any open set such that \(E_{n}\subset O_{n}\subset \Omega \), \(\vert \partial O_{n} \vert =0\), \(\vert \overline{O}_{n} \vert \downarrow 0\) if \(n\rightarrow \infty \), and \(\overline{O}_{n}\subset \Omega \) except for a null subset of \(\overline{O}_{n}\). To achieve these properties, we solely need to take \(\overline{O}_{n}\) as the infimum of the unions of open balls containing \(E_{n}\).

We apply Chacon’s biting lemma to guarantee

$$ \lim_{\delta \rightarrow 0} \iint _{A\times A}\xi _{\delta } \bigl( x^{ \prime },x \bigr)\,dx^{\prime }\,dx= \iint _{A\times A}\xi \bigl( x^{ \prime },x \bigr) \,dx^{\prime }\,dx $$
(3.1)

for any measurable \(A\times A\subset ( \Omega \setminus \overline{O}_{n} ) \times ( \Omega \setminus \overline{O}_{n} ) \). Also, inequality (1.4) for open sets provides

$$ \lim_{\delta \rightarrow 0} \iint _{A\times A}\xi _{\delta } \bigl( x^{ \prime },x \bigr)\,dx^{\prime }\,dx\geq \int _{A} \bigl\vert \nabla u ( x ) \bigr\vert ^{p}\,dx, $$
(3.2)

for any measurable set \(A\subset \Omega \setminus \overline{O}_{n}\) (here we are considering the subsequence of \(\delta ^{\prime }s\) for which (1.4) holds).

Now, we first consider \(A=B ( x_{0},r ) \subset \Omega \setminus \overline{O}_{n}\) for any \(x_{0}\in \Omega \setminus \overline{O}_{n}\). Then, on the one hand, by (3.2)we have

$$ \lim_{\delta \rightarrow 0} \iint _{B ( x_{0},r ) \times B ( x_{0},r ) }\xi _{\delta } \bigl( x^{\prime },x \bigr)\,dx^{ \prime }\,dx= \iint _{B ( x_{0},r ) \times B ( x_{0},r ) } \xi \bigl( x^{\prime },x \bigr) \,dx^{\prime }\,dx. $$
(3.3)

On the other hand, since \(B ( x_{0},r ) \times B ( x_{0},r ) \) is a smooth domain, (1.4) can be applied and hence

$$ \lim_{\delta \rightarrow 0} \iint _{B ( x_{0},r ) \times B ( x_{0},r ) }\xi _{\delta } \bigl( x^{\prime },x \bigr)\,dx^{ \prime }\,dx\geq \int _{B ( x_{0},r ) } \bigl\vert \nabla u ( x ) \bigr\vert ^{p}\,dx. $$
(3.4)

By using (3.3) and (3.4), we arrive at this crucial inequality for any \(B ( x_{0},r ) \subset \Omega \setminus \overline{O}_{n}\):

$$ \iint _{B ( x_{0},r ) \times B ( x_{0},r ) } \xi \bigl( x^{\prime },x \bigr) \,dx^{\prime }\,dx\geq \int _{B ( x_{0},r ) } \bigl\vert \nabla u ( x ) \bigr\vert ^{p}\,dx\,dx^{ \prime }. $$
(3.5)

Thus, (3.2) holds for any measurable set \(A\subset \Omega \setminus \overline{O}_{n}\).

Finally, we analyze \(\lim_{\delta \rightarrow 0}\iint _{G\times G}\xi _{\delta } ( x^{ \prime },x )\,dx^{\prime }\,dx\), where \(G\subset \Omega \) is any measurable set. We note that

$$ \iint _{G\times G}\xi _{\delta } \bigl( x^{\prime },x \bigr)\,dx^{ \prime }\,dx\geq \iint _{ ( G\setminus \overline{O}_{n} ) \times ( G\setminus \overline{O}_{n} ) }\xi _{\delta } \bigl( x^{ \prime },x \bigr)\,dx^{\prime }\,dx $$

which, thanks to Chacon’s biting lemma, provides the estimate

$$ \lim_{\delta \rightarrow 0} \iint _{G\times G}\xi _{\delta } \bigl( x^{ \prime },x \bigr)\,dx^{\prime }\,dx\geq \iint _{ ( G\setminus \overline{O}_{n} ) \times ( G\setminus \overline{O}_{n} ) }\xi \bigl( x^{\prime },x \bigr) \,dx^{\prime }\,dx. $$

Since \(G\setminus \overline{O}_{n}\) is a measurable set included in \(\Omega \setminus \overline{O}_{n}\), (3.2) for measurable sets provides the estimate

$$ \iint _{ ( G\setminus \overline{O}_{n} ) \times ( G \setminus \overline{O}_{n} ) }\xi \bigl( x^{\prime },x \bigr) \,dx^{ \prime }\,dx\geq \int _{G\setminus \overline{O}_{n}} \bigl\vert \nabla u ( x ) \bigr\vert ^{p}\,dx, $$

which straightforwardly implies

$$ \lim_{\delta \rightarrow 0} \iint _{G\times G}\xi _{\delta } \bigl( x^{ \prime },x \bigr)\,dx^{\prime }\,dx\geq \int _{G\setminus \overline{O}_{n}} \bigl\vert \nabla u ( x ) \bigr\vert ^{p}\,dx. $$

By letting \(n\rightarrow \infty \), we finish the proof of (1.4).

To avoid the hypothesis \(\vert \partial \Omega \vert =0\), we proceed as in the previous section.

The analysis performed when proving a corollary in Sect. 3.1 from [2, p. 8] is correct and therefore serves to assert that (1.3) is valid for all measurable functions h.

All the changes requested are implemented in this correction.

References

  1. 1.

    Brooks, J.K., Chacon, R.V.: Continuity and compactness of measures. Adv. Math. 37, 16–26 (1980)

    MathSciNet  Article  Google Scholar 

  2. 2.

    Muñoz, J.: Generalized Ponce’s inequality. J. Inequal. Appl. 2021, 11 (2021). https://doi.org/10.1186/s13660-020-02543-1

    MathSciNet  Article  Google Scholar 

  3. 3.

    Pedregal, P.: Parametrized Measures and Variational Principles. Birkhäuser, Basel (1997)

    Book  Google Scholar 

Download references

Acknowledgements

The author would like to acknowledge most warmly Anton Egrafov’s comments.

Author information

Affiliations

Authors

Corresponding author

Correspondence to Julio Muñoz.

Rights and permissions

Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Muñoz, J. Correction to: Generalized Ponce’s inequality. J Inequal Appl 2021, 80 (2021). https://doi.org/10.1186/s13660-021-02609-8

Download citation