960 CHAPTER 26. INTEGRALS AND DERIVATIVES

It remains to verify the last assertion. Note that u,i ∈ Lp (Rn) for any p > 1 because it isbounded and vanishes outside some compact set. By the first part,(∫

|uε,i−u,i|p dx)1/p

=

(∫ ∣∣∣∣∫ (u,i (x−y)−u,i (x))φ ε (y)dy∣∣∣∣p dx

)1/p

and by Minkowski’s inequality,

≤∫

φ ε (y)(∫|(u,i (x−y)−u,i (x))|p dx

)1/p

dy

=∫

B(0,ε)φ ε (y)

∣∣∣∣∣∣(u,i)y−u,i∣∣∣∣∣∣

Lp(Rn)dy

which converges to 0 from continuity of translation. This proves the lemma.Now from Corollary 26.6.2 applied to uε just described and letting y−x = v

|uε (x+v)−uε (x)−∇uε (x) ·v|

≤C(

1m(B(x,2 |v|))

∫B(x,2|v|)

|∇uε (z)−∇uε (x) |pdz)1/p

|v|.

From Lemma 26.6.4, there is a subsequence, still denoted as ε such that for each i,uε,i→ u,ipointwise a.e. and in Lp (Rn) where p > n is given. ∇u is the vector (u,1,u,2, · · · ,u,n)T .Then passing to the limit as ε → 0, for a.e. x,

|u(x+v)−u(x)−∇u(x) ·v|

≤C(

1m(B(x,2 |v|))

∫B(x,2|v|)

|∇u(z)−∇u(x) |pdz)1/p

|v|.

At every Lebesgue point x of ∇u, the above shows u(x+v)− u(x)−∇u(x) · v = o(v).Thus this has proved the following.

Lemma 26.6.5 Let u be Lipschitz continuous and vanish outside some bounded set. ThenDu(x) exists for a.e. x.

This is a good result but it is easy to give an even easier to use result. First here is atheorem which says you can extend a Lipschitz map.

Theorem 26.6.6 If h : Ω→ Rm is Lipschitz, then there exists h : Rn→ Rm which extendsh and is also Lipschitz.

Proof: It suffices to assume m = 1 because if this is shown, it may be applied to thecomponents of h to get the desired result. Suppose

|h(x)−h(y)| ≤ K |x−y|. (26.6.20)

Defineh(x)≡ inf{h(w)+K |x−w| : w ∈Ω}. (26.6.21)