438 CHAPTER 16. HAUSDORFF MEASURE

Letting xp≡ (x1, ...,xp−1,0) and v ≡ ep, it follows that for every xp,∂

∂xpf (xp, t) exists

for a.e. t. Thus ∂

∂xpf exists off a set of measure zero. It is similar for the other partial

derivatives, and so, taking a union of p exceptional sets of measure zero, it follows that∇ f (x) exists a.e. ■

16.3 Rademacher’s TheoremIt turns out that Lipschitz functions on Rp can be differentiated a.e. This is called Radem-acher’s theorem. It also can be shown to follow from the Lebesgue theory of differentiation.We denote Dv f (x) the directional derivative of f in the direction v. Here v is a unitvector. In the following lemma, notation is abused slightly. The symbol f (x+tv) willmean t → f (x+tv) and d

dt f (x+tv) will refer to the derivative of this function of t. It isa good idea to review Theorem 11.11.5 on integration with polar coordinates because thiswill be used in what follows. I will also denote as dx the symbol dmp (x) to save space.

Lemma 16.3.1 Let u : Rp→ R be Lipschitz with Lipschitz constant K. Let

un ≡ u∗φ n ≡∫

u(x−y)dmp (y)

where {φ n} is a mollifier,

φ n (y)≡ npφ (ny) ,

∫φ (y)dmp (y) = 1, φ (y)≥ 0, φ ∈C∞

c (B(0,1))

Then∇un (x) = ∇u∗φ n (x) (16.1)

where ∇u is defined almost everywhere according to Proposition 16.3.4. In fact,∫ b

a

∂u∂xi

(x+ tei)dt = u(x+bei)−u(x+aei) (16.2)

and∣∣∣ ∂u

∂xi

∣∣∣ ≤ K so |∇u(x)| ≤ √pK for a.e. x. Also, un (x)→ u(x) uniformly on Rp and

for a suitable subsequence, still denoted with n, ∇un (x)→ ∇u(x) for a.e. x.

Proof: To get the existence of the gradient satisfying the condition given in 16.2, applyProposition 16.3.4 to each variable. Now

un (x+hei)−un (x)

h=

∫Rp

(u(x+hei−y)−u(x−y)

h

)φ n (y)dmp (y)

=∫

B(0, 1n )

(u(x+hei−y)−u(x−y)

h

)φ n (y)dmp (y)

=∫

B(0,1)

(u(x+hei−y)−u(x−y)

h

)φ n (y)dmp (y)

Now if x−y is off a set of measure zero, the above difference quotient converges to∂u∂xi

(x−y). You just use Proposition 16.3.4 on the ith variable. If hk is any sequenceconverging to 0, you can apply the dominated conergence theorem in the above and obtain

∂un (x)

∂xi=∫

B(0,1)

∂u(x−y)∂xi

φ n (y)dmp (y) =∂u∂xi∗φ n (x)

438 CHAPTER 16. HAUSDORFF MEASURELetting xp = (x1,...,Xp—1,0) and v = ep, it follows that for every xp, a f (&p,t) existsPpfor a.e. t. Thus a f exists off a set of measure zero. It is similar for the other partialPpderivatives, and so, taking a union of p exceptional sets of measure zero, it follows thatVf (x) exists ae. Hl16.3, Rademacher’s TheoremIt turns out that Lipschitz functions on R? can be differentiated a.e. This is called Radem-acher’s theorem. It also can be shown to follow from the Lebesgue theory of differentiation.We denote D,,f (x) the directional derivative of f in the direction v. Here v is a unitvector. In the following lemma, notation is abused slightly. The symbol f (a+tv) willmean t > f(a-+tv) and 4 f (a+tv) will refer to the derivative of this function of t. It isa good idea to review Theorem 11.11.5 on integration with polar coordinates because thiswill be used in what follows. I will also denote as dx the symbol dm, (a) to save space.Lemma 16.3.1 Let u: R? > R be Lipschitz with Lipschitz constant K. Lettn = 086, = | u(@—y)dimp (y)where {@,,} is a mollifier,bay) =n?o (ny), [6 (y)amp(y) = 1. O(y) > 0, 6 EC (B(O.1))ThenVun (x) = Vuxd, (x) (16.1)where Vu is defined almost everywhere according to Proposition 16.3.4. In fact,> Ou— (a+te;)dt = u(x+be;) —u(a+ae;) (16.2)a Ox;and gu <K so |Vu(a)| < \/pK for ae. x. Also, u;(x) + u(a) uniformly on R? andfor a suitable subsequence, still denoted with n, Vu, (x) > Vu(a) for ae. x.Proof: To get the existence of the gradient satisfying the condition given in 16.2, applyProposition 16.3.4 to each variable. Nowtn (ether) — un) _ [, (oem we=Y)) 9 (y)amp(u)= oy onan)Now if z—y is off a set of measure zero, the above difference quotient converges togu (a—y). You just use Proposition 16.3.4 on the i” variable. If hy is any sequenceconverging to 0, you can apply the dominated conergence theorem in the above and obtainAun (x) =| du(x@—y) Ou/JB(0,1)Ox; Ox; , (y) dmp (y) = Ox; *Q, (x)