1242 CHAPTER 35. WEAK DERIVATIVES

as ε → 0. By 35.5.4, there exists a subsequence ε → 0 such that for |z| < k and for eachi = 1,2, · · · ,n

(uψk)ε,i (z)→ (uψk),i (z) = u,i (z) a.e.

(uψk)ε (z)→ uψk (z) = u(z) a.e. (35.5.5)

Denoting the exceptional set by Ek, let

x,y /∈ ∪∞

k=1Ek ≡ E

and let k be so large thatB(0,k)⊇ B(x,2|x−y|).

Then by 35.4.1 and for x,y /∈ E,

|(uψk)ε (x)− (uψk)ε (y)|

≤C(∫

B(x,2|y−x|)|∇(uψk)ε |pdz

)1/p

|x−y|(1−n/p)

where C depends only on n. Similarly, by 35.4.2,∣∣(uψk)ε (x)− (uψk)ε (y)−∇(uψk)ε(x) · (y−x)

∣∣≤C(

1m(B(x,2 |x−y|))

∫B(x,2|x−y|)

|∇(uψk)ε(z)−∇(uψk)ε

(x) |pdz)1/p

| x− y|.

Now by 35.5.5 and 35.5.3 passing to the limit as ε → 0 yields

|u(x)−u(y)| ≤C(∫

B(x,2|y−x|)|∇u|pdz

)1/p

|x−y|(1−n/p) (35.5.6)

and|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|. (35.5.7)

Redefining u on the set of mesure zero, E yields 35.5.6 for all x,y. This proves the theorem.

Corollary 35.5.2 Let u,u,i ∈ Lploc (R

n) for i = 1, · · · ,n and p > n. Then the representativeof u described in Theorem 35.5.1 is differentiable a.e.

Proof: From Theorem 35.5.1

|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|. (35.5.8)