Sobolev Inequalities

# Sobolev Inequalities - A NALYSIS TOOLS W ITH APPLICATIONS...

This preview shows pages 1–4. Sign up to view the full content.

ANALYSIS TOOLS WITH APPLICATIONS 493 27. Sobolev Inequalities 27.1. Morrey’s Inequality. Notation 27.1. Let S d 1 be the sphere of radius one centered at zero inside R d . For a set Γ S d 1 ,x R d , and r (0 , ) , let Γ x,r { x + : ω Γ such that 0 s r } . So Γ x,r = x + Γ 0 ,r where Γ 0 ,r is a cone based on Γ , seeF igure49be low . Γ Γ Figure 49. The cone Γ 0 ,r . Notation 27.2. If Γ S d 1 is a measurable set let | Γ | = σ ( Γ ) be the surface “area” of Γ . Notation 27.3. If R d is a measurable set and f : R d C is a measurable function let f := Z f ( x ) dx := 1 m ( ) Z f ( x ) dx. By Theorem 8.35, (27.1) Z Γ x,r f ( y ) dy = Z Γ 0 ,r f ( x + y ) dy = Z r 0 dt t d 1 Z Γ f ( x + ) ( ω ) and letting f =1 in this equation implies (27.2) m ( Γ x,r )= | Γ | r d /d. Lemma 27.4. Let Γ S d 1 be a measurable set such that | Γ | > 0 . For u C 1 ( Γ x,r ) , (27.3) Z Γ x,r | u ( y ) u ( x ) | dy 1 | Γ | Z Γ x,r | u ( y ) | | x y | d

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
494 BRUCE K. DRIVER Proof. Write y = x + with ω S d 1 , then by the fundamental theorem of calculus, u ( x + ) u ( x )= Z s 0 u ( x + ) · ωdt and therefore, Z Γ | u ( x + ) u ( x ) | ( ω ) Z s 0 Z Γ | u ( x + ) | ( ω ) dt = Z s 0 t d 1 dt Z Γ | u ( x + ) | | x + x | d 1 ( ω ) = Z Γ x,s | u ( y ) | | y x | d 1 dy Z Γ x,r | u ( y ) | | x y | d 1 dy, wherein the second equality we have used Eq. (27.1). Multiplying this inequality by s d 1 and integrating on s [0 ,r ] gives Z Γ x,r | u ( y ) u ( x ) | dy r d d Z Γ x,r | u ( y ) | | x y | d 1 dy = m ( Γ x,r ) | Γ | Z Γ x,r | u ( y ) | | x y | d 1 dy which proves Eq. (27.3). Corollary 27.5. Suppose d<p ≤∞ , Γ B S d 1 such that | Γ | > 0 ,r (0 , ) and u C 1 ( Γ x,r ) . Then (27.4) | u ( x ) | C ( | Γ | ,r,d,p ) k u k W 1 ,p ( Γ x,r ) where C ( | Γ | ,r,d,p ):= 1 | Γ | 1 /p max Ã d 1 /p r , μ p 1 p d 1 1 /p ! · r 1 d/p . Proof. For y Γ x,r , | u ( x ) | | u ( y ) | + | u ( y ) u ( x ) | and hence using Eq. (27.3) and Hölder’s inequality, | u ( x ) | Z Γ x,r | u ( y ) | dy + 1 | Γ | Z Γ x,r | u ( y ) | | x y | d 1 dy 1 m ( Γ x,r ) k u k L p ( Γ x,r ) k 1 k L p ( Γ x,r ) + 1 | Γ | k u k L p ( Γ x,r ) k 1 | x ·| d 1 k L q ( Γ x,r ) (27.5) where q = p p 1 as before. Now k 1 |·| d 1 k q L q ( Γ 0 ,r ) = Z r 0 dt t d 1 Z Γ ¡ t d 1 ¢ q ( ω ) = | Γ | Z r 0 dt ¡ t d 1 ¢ 1 p p 1 = | Γ | Z r 0 dt t d 1 p 1 and since 1 d 1 p 1 = p d p 1
ANALYSIS TOOLS WITH APPLICATIONS 495 we f nd (27.6) k 1 |·| d 1 k L q ( Γ 0 ,r ) = μ p 1 p d | Γ | r p d p 1 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 10/11/2010 for the course MATH 11 taught by Professor Smith during the Three '10 term at ADFA.

### Page1 / 19

Sobolev Inequalities - A NALYSIS TOOLS W ITH APPLICATIONS...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online