Kenneth Kuttler

1269

Uz⋂

U + ta

You can imagine open sets which do not satisfy the segment condition. For example, apair of circles which are tangent at their boundaries. The condition in the above definitionbreaks down at their point of tangency.

Here is a simple lemma which will be used in the proof of the following theorem.

Lemma 37.0.10 If u ∈W m,p (U) and ψ ∈C∞c (Rn) , then uψ ∈W m,p (U) .

Proof: Let |α| ≤ m and let φ ∈C∞c (U) . Then

(Dxi (uψ))(φ) ≡ −∫

Uuψφ ,xi

= −∫

Uu((ψφ),xi

−φψ ,xi

)dx

= (Dxiu)(ψφ)+∫

Uuψ ,xi

φdx

=∫

(ψDxiu+uψ ,xi

)φdx

Therefore, Dxi (uψ)=ψDxiu+uψ ,xi∈ Lp (U) . In other words, the product rule holds. Now

considering the terms in the last expression, you can do the same argument with each ofthese as long as they all have derivatives in Lp (U) . Therefore, continuing this process thelemma is proved.

Theorem 37.0.11 Let U be an open set and suppose there exists a locally finite covering2

of U which is of the form {Ui}∞

i=1 such that each Ui is a bounded open set which satisfiesthe conditions of Definition 37.0.9. Thus there exist vectors, ai such that for all t ∈ (0,1) ,

Ui∩U + tai ⊆U.

Then C∞(U)

is dense in Xm,p (U) and so W m,p (U) = Xm,p (U) .

2This is never a problem in Rn. In fact, every open covering has a locally finite subcovering in Rn or moregenerally in any metric space due to Stone’s theorem. These are issues best left to you in case you are interested.I am usually interested in bounded sets, U, and for these, there is a finite covering.

1269aU,(\U +ta >You can imagine open sets which do not satisfy the segment condition. For example, apair of circles which are tangent at their boundaries. The condition in the above definitionbreaks down at their point of tangency.Here is a simple lemma which will be used in the proof of the following theorem.Lemma 37.0.10 [fue W™? (U) and w € Ce? (R"), thenuw ew"? (U).Proof: Let || <m and let ¢ € C? (U). Then= [wy dx— | u((y0),., 9.) ae= (Dyn) (vo) + | uy, ode(Dy; (uw)) (9)_ I (yDy,u+ uy ,,) odxTherefore, D,,; (uy) = WD,,u+uy ,, € L? (U). In other words, the product rule holds. Nowconsidering the terms in the last expression, you can do the same argument with each ofthese as long as they all have derivatives in L? (U). Therefore, continuing this process thelemma is proved.Theorem 37.0.11 Let U be an open set and suppose there exists a locally finite covering”of U which is of the form {U;};__, such that each U; is a bounded open set which satisfiesthe conditions of Definition 37.0.9. Thus there exist vectors, a; such that for allt € (0,1),U;AU +ta; CU.Then C* (U) is dense in X"? (U) and so W™? (U) = X™? (U).2This is never a problem in R”. In fact, every open covering has a locally finite subcovering in R” or moregenerally in any metric space due to Stone’s theorem. These are issues best left to you in case you are interested.Iam usually interested in bounded sets, U, and for these, there is a finite covering.