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Correction to: Generalized Ponce’s inequality

The Original Article was published on 14 January 2021

1 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

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

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

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

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):

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

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,

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

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.

2 First proof

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

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

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 \(G\subset \Omega \).

3 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

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

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

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

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

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

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

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

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

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

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.


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

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Correspondence to Julio Muñoz.

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Muñoz, J. Correction to: Generalized Ponce’s inequality. J Inequal Appl 2021, 80 (2021).

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