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Sobolev Inequalities

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

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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 , see Figure 49 below. 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 1 dy.

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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 + ) · ω dt 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 fi 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 /q = μ p 1 p d | Γ | p 1 p r 1 d p . Combining Eqs. (27.5), Eq. (27.6) along with the identity, (27.7) 1 m ( Γ x,r ) k 1 k L q ( Γ x,r ) = 1 m ( Γ x,r ) m ( Γ x,r ) 1 /q = ¡ | Γ | r d /d ¢ 1 /p , shows | u ( x ) | k u k L p ( Γ x,r ) ¡ | Γ | r d /d ¢ 1 /p + 1 | Γ | k u k L p ( Γ x,r ) μ p 1 p d | Γ | 1 1 /p r 1 d/p = 1 | Γ | 1 /p " k u k L p ( Γ x,r ) d 1 /p r + k u k L p ( Γ x,r ) μ p 1 p d 1 1 /p # r 1 d/p .

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