954 CHAPTER 26. INTEGRALS AND DERIVATIVES

Now |∇un| = |∇u∗φ n| by Lemma 26.5.2 and this last is bounded. Also, by this lemma,∇un (z)→ ∇u(z) a.e. and un (x)→ u(x) for all x. Therefore, we can pass to a limit in theabove and obtain 26.5.13.

Note you can write 26.5.13 in the form

|u(x)−u(y)| ≤ C(

1|x−y|p

∫B(x,2|x−y|)

|∇u(z) |qdz)1/q

|x−y|

= Ĉ(

1mp (B(x,2 |x−y|))

∫B(x,2|x−y|)

|∇u(z) |qdz)1/q

|x−y|

Before leaving this remarkable formula, note that if you are in any situation where theabove formula holds and ∇u exists in some sense and is in Lq,q > p, then u would need tobe continuous. This is the basis for the Sobolev embedding theorem.

Here is Rademacher’s theorem.

Theorem 26.5.5 Suppose u is Lipschitz with constant K then if x is a point where ∇u(x)exists,

|u(y)−u(x)−∇u(x) · (y−x)|

≤C(

1m(B(x,2 |x−y|))

∫B(x,2|x−y|)

|∇u(z)−∇u(x) |qdz)1/q

| x− y|. (26.5.14)

Also u is differentiable at a.e. x and also

u(x+tv)−u(x) =∫ t

0Dvu(x+ sv)ds (26.5.15)

Proof: This follows easily from letting g(y) ≡ u(y)− u(x)−∇u(x) ·(y−x) . As ex-plained above, |∇u(x)| ≤√pK at every point where ∇u exists, the exceptional points beingin a set of measure zero. Then g(x) = 0, and ∇g(y) =∇u(y)−∇u(x) at the points y wherethe gradient of g exists. From Corollary 26.5.4,

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

≤ C(∫

B(x,2|x−y|)|∇u(z)−∇u(x) |qdz

)1/q

|x−y|1−pq

= C(∫

B(x,2|x−y|)|∇u(z)−∇u(x) |qdz

)1/q 1|x−y|p

1q|x−y|

= C(

1m(B(x,2 |x−y|))

∫B(x,2|x−y|)

|∇u(z)−∇u(x) |qdz)1/q

|x− y|.

Now this is no larger than

≤C(

1m(B(x,2 |x−y|))

∫B(x,2|x−y|)

|∇u(z)−∇u(x)|(2√pK)q−1 dz)1/q

|x− y|

954 CHAPTER 26. INTEGRALS AND DERIVATIVESNow |Vu,| = |Vu*@,,| by Lemma 26.5.2 and this last is bounded. Also, by this lemma,Vuln (z) + Vu(z) a.e. and uy, (x) > u(x) for all x. Therefore, we can pass to a limit in theabove and obtain 26.5.13. JjNote you can write 26.5.13 in the form(= IVu(e) tae) Ixy!Cc — | Vu(z ac) x-yIx—yl? Ja(x,2|x-y))- 1 1/q= (sa BRI ban (ule) ae) RyBefore leaving this remarkable formula, note that if you are in any situation where theabove formula holds and Vu exists in some sense and is in L1,q > p, then u would need tobe continuous. This is the basis for the Sobolev embedding theorem.Here is Rademacher’s theorem.lu (x) —u(y)|Theorem 26.5.5 Suppose u is Lipschitz with constant K then if x is a point where Vu(x)exists,| (y) — u(x) — Vu(x) -(y—x)|1 q 1/q<C( petra Iran gp Me PH ac) Ix— yl. (265.14)Also u is differentiable at a.e. x and alsotu(x+tv) — u(x) =| Dyu(x+sv)ds (26.5.15)0Proof: This follows easily from letting g(y) = u(y) — u(x) — Vu(x)-(y —x). As ex-plained above, |Vu (x)| < \/pK at every point where Vu exists, the exceptional points beingin a set of measure zero. Then g (x) =0, and Vg (y) = Vu(y) — Vu (x) at the points y wherethe gradient of g exists. From Corollary 26.5.4,u(x) — Vu(x) -(y—x)|u(yIg(y)| =lg(y)—g (x))-(y)1/q 1 ;= c(f,e Vu Vu(x tas) ——{ |x—og yp IW Ce) Vas) de)" sy1 1/47 C(srprm airy Iran yy IM) — YH Maz) |Ix— y|.Now this is no larger than1/q< c(f, [Yu (a) ~ Yul) |) Ix—y|'~4B(x,2|x—y|)1 4 I/q<e(,, BR IRI Jasean-»y MH) — VC] (2VPRI" i) ed