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# 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 , under the assumptions above, the following compactness is recalled (see  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  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  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 ) 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)

2. 2.

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

3. 3.

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

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## Acknowledgements

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

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Correspondence to Julio Muñoz.

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