958 CHAPTER 26. INTEGRALS AND DERIVATIVES

= C(∫

U|∇u(z)|p dz

)1/p(∫Sn−1

∫ r

0ρ

p′−np′ρ

n−1dρdσ

)(p−1)/p

= C(∫

U|∇u(z)|p dz

)1/p(∫

Sn−1

∫ r

0

1

ρn−1p−1

dρdσ

)(p−1)/p

= C(

p−1p−n

)(p−1)/p(∫U|∇u(z)|p dz

)1/p

r1− np

= C(

p−1p−n

)(p−1)/p(∫U|∇u(z)|p dz

)1/p

|x−y|1−np

Similarly,∫−

V|u(y)−u(z)|dz≤C

(p−1p−n

)(p−1)/p(∫V|∇u(z)|p dz

)1/p

|x−y|1−np

Therefore,

|u(x)−u(y)| ≤C(

p−1p−n

)(p−1)/p(∫B(x,2|x−y|)

|∇u(z)|p dz)1/p

|x−y|1−np

because B(x,2 |x−y|)⊇V ∪U. This proves the lemma.The following corollary is also interesting

Corollary 26.6.2 Suppose u ∈C1 (Rn) . Then

|u(y)−u(x)−∇u(x) · (y−x)|

≤C(

1m(B(x,2 |x−y|))

∫B(x,2|x−y|)

|∇u(z)−∇u(x) |pdz)1/p

| x− y|. (26.6.19)

Proof: This follows easily from letting g(y) ≡ u(y)− u(x)−∇u(x) ·(y−x) . Theng ∈C1 (Rn), g(x) = 0, and ∇g(z) = ∇u(z)−∇u(x) . From Lemma 26.6.1,

|u(y)−u(x)−∇u(x) · (y−x)|= |g(y)|= |g(y)−g(x)|

≤ C(∫

B(x,2|x−y|)|∇u(z)−∇u(x) |pdz

)1/p

|x−y|1−np

= C(

1m(B(x,2 |x−y|))

∫B(x,2|x−y|)

|∇u(z)−∇u(x) |pdz)1/p

| x− y|.

This proves the corollary.It may be interesting at this point to recall the definition of differentiability on Page

117. If you knew the above inequality held for ∇u having components in L1loc (Rn) , then at

Lebesgue points of ∇u, the above would imply Du(x) exists.