Kenneth Kuttler

1252 CHAPTER 35. WEAK DERIVATIVES

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|. (35.6.26)

Defineh(x)≡ inf{h(w)+K |x−w| : w ∈Ω}. (35.6.27)

If x ∈Ω, then for all w ∈Ω,

h(w)+K |x−w| ≥ h(x)

by 35.6.26. This shows h(x) ≤ h(x). But also you could take w = x in 35.6.27 whichyields h(x)≤ h(x). Therefore h(x) = h(x) if x ∈Ω.

Now suppose x,y ∈ Rn and consider∣∣h(x)−h(y)

∣∣. Without loss of generality assumeh(x)≥ h(y) . (If not, repeat the following argument with x and y interchanged.) Pick w∈Ω

such thath(w)+K |y−w|− ε < h(y).

Then ∣∣h(x)−h(y)∣∣= h(x)−h(y)≤ h(w)+K |x−w|−

[h(w)+K |y−w|− ε]≤ K |x−y|+ ε.

Since ε is arbitrary, ∣∣h(x)−h(y)∣∣≤ K |x−y|

and this proves the theorem.This yields a simple corollary to Theorem 35.6.12.

Corollary 35.6.14 Let h : Ω→ Rn be Lipschitz continuous and one to one where Ω is aLebesgue measurable set. Then if f ≥ 0 is Lebesgue measurable,∫

h(Ω)f (y)dmn =

∫Ω

f (h(x))∣∣detDh(x)

∣∣dmn. (35.6.28)

where h denotes a Lipschitz extension of h.

1252 CHAPTER 35. WEAK DERIVATIVESProof: It suffices to assume m = | because if this is shown, it may be applied to thecomponents of h to get the desired result. Suppose|n(x) —h(y)| < K|x—y|. (35.6.26)Define 7h(x) = inf{h(w) + K|x—w|: we Q}. (35.6.27)If x € Q, then for all w € Q,h(w)+K |x—w| > A(x)by 35.6.26. This shows h(x) < h(x). But also you could take w =x in 35.6.27 whichyields h(x) < h(x). Therefore h(x) = h(x) ifx €Q.Now suppose x, y € R” and consider |h (x) —h(y) |. Without loss of generality assumeh(x) >h(y). (If not, repeat the following argument with x and y interchanged.) Pick w € Qsuch thath(w) +Kly—w|—e </h(y).Then _ _ _ _(x) —h(y)| =A(x) —A(y) <h(w) + K |x—w|—[h(w) +K ly —w|—€] <K|x—y| +e.Since € is arbitrary,|h(x) —h(y)| <K|x—y|and this proves the theorem.This yields a simple corollary to Theorem 35.6.12.Corollary 35.6.14 Leth: Q — R" be Lipschitz continuous and one to one where Q is aLebesgue measurable set. Then if f > 0 is Lebesgue measurable,| f (y)dmy = | Ff (lh(x)) |det D(x) | dm. (35.6.28)h(Q) Qwhere h denotes a Lipschitz extension of h.