Correction to: Generalized Ponce’s inequality

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

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

(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.

2 First proof

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

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

(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 $G\subset \Omega$.

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

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

(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.