Kenneth Kuttler

442 CHAPTER 16. HAUSDORFF MEASURE

Proof: This follows easily from letting g(y) ≡ u(y)− u(x)−∇u(x) · (y−x) . Asexplained above, |∇u(x)| ≤ √pK at every point where ∇u exists, the exceptional pointsbeing in a set of measure zero. Then g(x) = 0, and ∇g(y) = ∇u(y)−∇u(x) at the pointsy where the gradient of g exists. From Corollary 16.3.3,

|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|

It follows that at Lebesgue points of ∇u, the above expression is o(|x−y|) and so at allsuch points u is differentiable. As to 16.7, this follows from an application of Lemma16.1.2 to f (t) = u(x+tv). ■

Note that for a.e. x,Dvu(x) = ∇u(x) ·v. If you have a line with direction vector v,does it follow that Du(x+ tv) exists for a.e. t? We know the directional derivative existsa.e. t but it might not be clear that it is ∇u(x) ·v.

For |w|= 1, denote the measure of Section 11.11 defined on the unit sphere Sp−1 as σ .Let Nw be defined as those t ∈ [0,∞) for which Dwu(x+ tw) ̸= ∇u(x+ tw) ·w.

B≡{w ∈ Sp−1 : Nw has positive measure

}This is contained in the set of points of Rp where the derivative of v(·) ≡ u(x+ ·) failsto exist. Thus from Section 11.11 the measure of this set is

∫B∫

Nwρn−1dρdσ (w) This

must equal zero from what was just shown about the derivative of the Lipschitz function vexisting a.e. and so σ (B) = 0. The claimed formula follows from this. Thus we obtain thefollowing corollary.

Corollary 16.3.5 Let u be Lipschitz. Then for any x and v ∈ Sp−1 \Bx where σ (Bx) =0, it follows that for all t,

u(x+tv)−u(x) =∫ t

0Dvu(x+ sv)ds =

∫ t

0∇u(x+ sv) ·vds

In all of the above, the function u is defined on all of Rp. However, it is always thecase that Lipschitz functions can be extended off a given set. Thus if a Lipschitz function isdefined on some set Ω, then it can always be considered the restriction to Ω of a Lipschitzmap defined on all of Rp.

Theorem 16.3.6 If h : Ω→ Rm is Lipschitz, then there exists h : Rp→ Rm whichextends h and is also Lipschitz.

442 CHAPTER 16. HAUSDORFF MEASUREProof: This follows easily from letting g(y) = u(y) —u(a) — Vu(a)-(y—a). Asexplained above, |Vu(a)| < \/pK at every point where Vu exists, the exceptional pointsbeing in a set of measure zero. Then g(a) = 0, and Vg(y) = Vu(y) — Vu (a) at the pointsy where the gradient of g exists. From Corollary 16.3.3,\u(y) —u(x) —Vu(ax) -(y—2)| =|¢(y)| =|2(y) —8(a)|1/q ie(. yl) |Vu(z) — Vu (a) Mas) ja—y| 4c(/., ajay MH) vu(e)|ide) ten1 1/q¢(sa@apco recaie-y [Yu(=)—Yula) Pas) le ulNow this is no larger thanIACc1 7 _ I/q<c(5 Fede ayy Iw nay (PH) — VMN EVPRY as) le ylIt follows that at Lebesgue points of Vu, the above expression is o(|a— y|) and so at allsuch points u is differentiable. As to 16.7, this follows from an application of Lemma16.1.2 to f (t) =u(a-+tv).Note that for a.e. x, D,u(x) = Vu(a)-v. If you have a line with direction vector v,does it follow that Du(a-+tv) exists for a.e. t? We know the directional derivative existsa.e. t but it might not be clear that it is Vu(a)-v.For |w| = 1, denote the measure of Section 11.11 defined on the unit sphere 5 SPlaso.Let Ny be defined as those t € [0, °°) for which Dyu(a+tw) A Vu(a+tw)-wB= {w € §?-!: Ny, has positive measure }This is contained in the set of points of R? where the derivative of v(-) = u(a-+-) failsto exist. Thus from Section 11.11 the measure of this set is Jp Jy,, p”!dpdo (w) Thismust equal zero from what was just shown about the derivative of the Lipschitz function vexisting a.e. and so o (B) = 0. The claimed formula follows from this. Thus we obtain thefollowing corollary.Corollary 16.3.5 Let u be Lipschitz. Then for any x and v € S?~!\ By where 0 (Bz) =0, it follows that for all t,u(a+tv) —u(a) = [ Dou(e+sv)ds= | Vu(e+s0)-vdsIn all of the above, the function u is defined on all of R’. However, it is always thecase that Lipschitz functions can be extended off a given set. Thus if a Lipschitz function isdefined on some set Q, then it can always be considered the restriction to Q of a Lipschitzmap defined on all of R?.Theorem 16.3.6 If hh: Q— R"” is Lipschitz, then there exists h : R? — R” whichextends h and is also Lipschitz.