Kenneth Kuttler

1334 CHAPTER 39. WEAK SOLUTIONS

and so uδ ∈C∞(Uδ 0

)and

∇uδ (x) =∫

Uu(t)∇φ δ (x− t)dt =

∫B(0,δ )

∇u(x−y)φ δ (y)dy = 0.

Therefore, uδ equals a constant on Uδ 0 because Uδ 0 is a connected open set and uδ is asmooth function defined on this set which has its gradient equal to 0. By Minkowski’sinequality, (∫

Uδ0

|u(x)−uδ (x)|2 dx

)1/2

≤∫

B(0,δ )φ δ (y)

(∫Uδ0

|u(x)−u(x−y)|2 dx

)1/2

and this converges to 0 as δ → 0 by continuity of translation in L2. It follows there exists asequence of constants, cδ ≡ uδ (x) such that {cδ} converges to u in L2

(Uδ 0

). Consequently,

a subsequence, still denoted by uδ , converges to u a.e. By Eggoroff’s theorem there existsa set, Nk having measure no more than 3−kmn

(Uδ 0

)such that uδ converges to u uniformly

on NCk . Thus u is constant on NC

k . Now ∑k mn (Nk) ≤ 12 mn

(Uδ 0

)and so there exists x0 ∈

Uδ 0 \∪∞k=1Nk. Therefore, if x /∈Nk it follows u(x)= u(x0) and so, if u(x) ̸= u(x0) it must be

the case that x ∈ ∩∞k=1Nk, a set of measure zero. This shows that u equals a constant a.e. on

Uδ 0 =Uηk . Since k is arbitrary, 39.1.3 shows u is a.e. equal to a constant on U. Therefore,u equals the restriction of a function of S to U and so γu equals this constant in L2 (∂Ω) .Since the surface measure of Γ is positive, the constant must equal zero. Therefore, |||·||| isa norm.

It remains to verify that it is equivalent to the usual norm. It is clear that |||u||| ≤ ||u||1,2 .What about the other direction? Suppose it is not true that for some constant, K, ||u||1,2 ≤K |||u||| . Then for every k ∈ N, there exists uk ∈V such that

||uk||1,2 > k |||uk||| .

Replacing uk with uk/ ||uk||1,2 , it can be assumed that ||uk||1,2 = 1 for all k. Therefore, usingthe compactness of the embedding of H1 (U) into L2 (U) , there exists a subsequence, stilldenoted by uk such that

uk → u weakly in V, (39.1.4)uk → u strongly in L2 (U) , (39.1.5)

|||uk||| → 0, (39.1.6)uk → u weakly in (V, |||·|||) . (39.1.7)

From 39.1.6 and 39.1.7, it follows u = 0. Therefore, |uk|L2(U)→ 0. This with 39.1.6 con-tradicts the fact that ||uk||1,2 = 1 and this proves the equivalence of the two norms.

The proof of the above theorem yields the following interesting corollary.

1334 CHAPTER 39. WEAK SOLUTIONSand so us € C* (Us,) andVus(x)= [u(t)Voa(x—t)dr= | Vulx—y)Galv)dy=0.Therefore, us equals a constant on Us, because Us, is a connected open set and ug is asmooth function defined on this set which has its gradient equal to 0. By Minkowski’sinequality,U,1/2< Doan?) [/, ws) —a00-v) Pa) dyand this converges to 0 as 5 — 0 by continuity of translation in L”. It follows there exists asequence of constants, cs = ug (x) such that {cs } converges to u in L? (Us, ) . Consequently,a subsequence, still denoted by us, converges to u a.e. By Eggoroff’s theorem there existsa set, N, having measure no more than 3~*m, (Us,) such that us converges to u uniformly1/2|u (x) — ug s) Pa)on NE. Thus u is constant on NE. Now Yi; 711 (Nx) < 5Mn (Us,) and so there exists xg €Us, \Ug1Ne- Therefore, if x ¢ Nj it follows u (x) =u (Xo) and so, if u(x) Au (Xo) it must bethe case that x € M;-_, Nx, a set of measure zero. This shows that u equals a constant a.e. onUs, = Un, Since k is arbitrary, 39.1.3 shows u is a.e. equal to a constant on U. Therefore,u equals the restriction of a function of G to U and so yu equals this constant in L? (dQ).Since the surface measure of T is positive, the constant must equal zero. Therefore, |||-||| isa norm.It remains to verify that it is equivalent to the usual norm. It is clear that |||w||| < ||u||, 9-What about the other direction? Suppose it is not true that for some constant, K, ||u||, 9 <K |||u|||. Then for every k € N, there exists uz, € V such thatH|orelli2 > Allee llReplacing ux with ux/ ||ug||; 2, it can be assumed that ||1,||; » = 1 for all k. Therefore, usingthe compactness of the embedding of H! (U) into L? (U), there exists a subsequence, stilldenoted by u, such thatur — uweakly in V, (39.1.4)uz — ustrongly in L?(U), (39.1.5)I||ux||| —> 0, (39.1.6)uy —> u weakly in (V,|||-|]|). (39.1.7)From 39.1.6 and 39.1.7, it follows u = 0. Therefore, lux|22(v) — 0. This with 39.1.6 con-tradicts the fact that ||u,||, . = 1 and this proves the equivalence of the two norms.The proof of the above theorem yields the following interesting corollary.