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The sharp quantitative isocapacitary inequality (the case of p-capacity)

  • Ekaterina Mukoseeva ORCID logo EMAIL logo
Published/Copyright: May 13, 2021

Abstract

We prove a sharp quantitative form of isocapacitary inequality in the case of a general p. This work is a generalization of the author’s paper with Guido De Philippis and Michele Marini, where we treated the case of 2-capacity.

MSC 2010: 49R05; 49Q20; 35J92

Communicated by Frank Duzaar


Funding statement: The work of the author is supported by the INDAM-grant “Geometric Variational Problems” and by the H2020-MSCA-RISE-2017 PROJECT No. 778010 IPADEGAN.

A Appendix

Here we put some inequalities that are used throughout the paper.

Lemma A.1 ([14, Lemma 2.3]).

Let p > 1 . There exists c ( p ) 0 such that if κ 0 and ξ , η R n , then

( ( κ 2 + | ξ | 2 ) p - 2 2 ξ - ( κ 2 + | η | 2 ) p - 2 2 η ) ( ξ - η ) c ( κ 2 + | ξ | 2 + | η | 2 ) p - 2 2 | ξ - η | 2 .

Moreover, there exists another constant C ( p ) 0 such that if Ω R n is an open set and for u , v W 1 , p ( Ω ) and 0 s 1 , we set u s ( x ) = s u ( x ) + ( 1 - s ) v ( x ) , then the following two inequalities hold:

  1. for p 2 ,

    (A.1) Ω | u - v | p C 0 1 1 s 𝑑 s Ω ( | u s | p - 2 u s - | v | p - 2 v ) ( u s - v ) ,

  2. for 1 < p < 2 ,

    (A.2) Ω | u - v | p C ( 0 1 1 s 𝑑 s Ω ( | u s | p - 2 u s - | v | p - 2 v ) ( u s - v ) ) p 2 ( Ω ( | u | + | v | ) p ) 1 - p 2 .

Lemma A.2.

Let x , y R N , p ( 1 , ) . Then the following inequalities hold:

  1. if p 2 , then

    | y | p | x | p + p | x | p - 2 x ( y - x ) + c | y - x | p

    for some c = c ( p ) > 0 ,

  2. if 1 < p < 2 , then

    | y | p | x | p + p | x | p - 2 x ( y - x ) + c | y - x | 2 ( | x | 2 + | y - x | 2 ) p - 2 2

    for some c = c ( p ) > 0 .

Proof.

Consider a function f : N defined as f ( x ) = | x | p . Writing Taylor expansion for f, we get

| y | p = | x | p + p | x | p - 2 x ( y - x ) + 0 1 ( 1 - t ) D 2 f ( x + t ( y - x ) ) ( y - x ) ( y - x ) 𝑑 t .

Case p = 2 . We have

| y | 2 = | x | 2 + 2 x ( y - x ) + 1 2 | y - x | 2 ,

which gives us a desired inequality. We shall consider p 2 from now on.

Case p 2 . The Hessian D 2 f ( x ) looks as follows:

D 2 f ( x ) = p | x | p - 2 Id + p ( p - 2 ) | x | p - 4 A ,

where A i , j = x i x j . We notice that

0 A ξ ξ | x | 2 | ξ | 2 for any vector  ξ N ,

yielding

D 2 f ( x ) ξ ξ c | x | p - 2 | ξ | 2 for any vector  ξ N ,

where c = c ( p ) > 0 ( c = p for p > 2 , c = p ( p - 1 ) for 1 < p < 2 ). So, we have

| y | p | x | p + p | x | p - 2 x ( y - x ) + | y - x | 2 0 1 ( 1 - t ) | x + t ( y - x ) | p - 2 𝑑 t .

Let us consider the cases of different p separately.

First, we deal with 1 < p < 2 . In this case p - 2 < 0 and so

0 1 ( 1 - t ) | x + t ( y - x ) | p - 2 𝑑 t 1 4 1 4 3 4 ( | x | + t | y - x | ) p - 2 𝑑 t c ( | x | 2 + | y - x | 2 ) p - 2 2 ,

finishing the proof of lemma in this case.

To tackle the case p > 2 , we further consider two cases. If | y - x | < 2 | x | , then

0 1 ( 1 - t ) | x + t ( y - x ) | p - 2 𝑑 t c 0 1 4 | x | p - 2 𝑑 t c | y - x | p - 2 .

If instead | y - x | 2 | x | , then

0 1 ( 1 - t ) | x + t ( y - x ) | p - 2 𝑑 t c 4 7 6 7 | y - x | p - 2 𝑑 t c | y - x | p - 2 .

Acknowledgements

The author wishes to thank Guido De Philippis for many fruitful discussions.

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Received: 2020-11-05
Revised: 2021-03-04
Accepted: 2021-03-18
Published Online: 2021-05-13
Published in Print: 2023-01-01

© 2021 Walter de Gruyter GmbH, Berlin/Boston

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