Kenneth Kuttler

37.1. EMBEDDING THEOREMS FOR W m,p (Rn) 1273

Proof: It is clear that h∗ is linear. It is required to show it is one to one and continuous.First suppose h∗ f = 0. Then

0 =∫

V| f (h(x))|p dx

and so f (h(x)) = 0 for a.e. x∈U. Since h is Lipschitz, it takes sets of measure zero to setsof measure zero. Therefore, f (y) = 0 a.e. This shows h∗ is one to one.

By the Meyer Serrin theorem, Theorem 37.0.8, it suffices to verify that h∗ is continuouson functions in C∞ (V ) . Let f be such a function. Then using the chain rule and productrule, (h∗ f ),i (x) = f,k (h(x))hk,i (x) ,

(h∗ f ),i j (x) =(

f,k (h(x))hk,i (x)), j

= f,kl (h(x))hl, j (x)hk,i (x)+ f,k (h(x))hk,i j (x)

etc. In general, for |α| ≤ m−1, succsessive applications of the product rule and chain ruleyield that Dα (h∗ f )(x) has the form

Dα (h∗ f )(x) = ∑|β |≤|α|

h∗(

Dβ f)(x)gβ (x)

where gβ is a bounded Lipschitz function with Lipschitz constant dependent on h and itsderivatives. It only remains to take one more derivative of the functions, Dα f for |α| =m−1. This can be done again but this time you have to use Rademacher’s theorem whichassures you that the derivative of a Lipschitz function exists a.e. in order to take the partialderivative of the gβ (x) . When this is done, the above formula remains valid for all |α| ≤m.Therefore, using the change of variables formula for multiple integrals, Corollary 35.6.14on Page 1252,∫

U|Dα (h∗ f )(x)|p dx ≤ Cm,p,h ∑

|β |≤m

∫U

∣∣∣h∗(Dβ f)(x)∣∣∣p dx

= Cm,p,h ∑|β |≤m

∫U

∣∣∣(Dβ f)(h(x))

∣∣∣p dx

= Cm,p,h ∑|β |≤m

∫V

∣∣∣(Dβ f)(y)∣∣∣p ∣∣detDh−1 (y)

∣∣dy

≤ Cm,p,h,h−1 || f ||m,p,V

This shows h∗ is continuous on C∞ (V )∩W m,p (U) and since this set is dense, this provesh∗ is continuous. The same argument applies to

(h−1)∗ and now the definitions of h∗ and(

h−1)∗ show these are inverses.

37.1 Embedding Theorems For W m,p (Rn)

Recall Theorem 35.5.1 which is listed here for convenience.

37.1. EMBEDDING THEOREMS FOR W""? (R") 1273Proof: It is clear that h* is linear. It is required to show it is one to one and continuous.First suppose h* f = 0. Then0= [ r(h(x))|"axand so f (h(x)) =0 for a.e. x € U. Since h is Lipschitz, it takes sets of measure zero to setsof measure zero. Therefore, f(y) = 0 a.e. This shows h* is one to one.By the Meyer Serrin theorem, Theorem 37.0.8, it suffices to verify that h* is continuouson functions in C* (V). Let f be such a function. Then using the chain rule and productrule, (hf) ;(x) = fx (h(x) hei (x),(h’f) ;; (x) (Fx (a(x)) hie (X))Fei (h(x) fj (&) a (X) + fx (I (X)) xi (X)etc. In general, for |a@| < m— 1, succsessive applications of the product rule and chain ruleyield that D® (h* f) (x) has the formD*(h'f)(x)= YO hY (DPF) (x) gp (x)IB|<|@|where gg is a bounded Lipschitz function with Lipschitz constant dependent on h and itsderivatives. It only remains to take one more derivative of the functions, D® f for |a| =m— 1. This can be done again but this time you have to use Rademacher’s theorem whichassures you that the derivative of a Lipschitz function exists a.e. in order to take the partialderivative of the gg (x). When this is done, the above formula remains valid for all |a@| < m.Therefore, using the change of variables formula for multiple integrals, Corollary 35.6.14on Page 1252,[\orws (hf)(x)|Pdx < Coon Df, h* (D*r) (x)|’ ax= Cops & [,\(e°2) aco] ax= Con [ (D°s) (y)|" |dewh-' (y)| ay< Cn pho! Ifllm,p.vThis shows h* is continuous on C* (V) 7W”? (U) and since this set is dense, this provesh* is continuous. The same argument applies to (h-!)* and now the definitions of h* and_ * .(h—')” show these are inverses.37.1 Embedding Theorems For W”’? (R”)Recall Theorem 35.5.1 which is listed here for convenience.