This is a contradiction. Thus the liminf condition must hold.
Next consider another operator. Let Σ_{1} be ∂Ω\ Σ_{0} and has positive surface
measure. Let r → a
(r,t)
be lower semicontinuous and r → b
(r,t)
be upper
semicontinuous. Let 0 < δ
(t)
≤ a
(r,t)
≤ b
(r,t)
≤
-1-
δ(t)
. Also let both of these functions
be measurable in t. Now γ : V → L^{2}
(Σ1)
and so γ^{∗} : L^{2}
(Σ1)
→ V^{′} defined
in the usual way. Then z ∈ B
(u,t)
will mean z = γ^{∗}w for some w ∈ L^{2}
(Σ1)
with
w (x ) ∈ [a(γu (x ),t),b(γu(x),t)]
for a.e. x such that
∫
〈z,v〉 = w (x)γv(x)
Σ1
Using Sobolev embedding theorems, if u_{n}→ u weakly in V , then from the Sobolev
embedding theorem u_{n}→ u strongly in a suitable Sobolev space of fractional order such
that the embedding of V into this space is compact and the trace map is still continuous.
Thus there is a subsequence such that γu_{n}
(x)
→ γu
(x)
pointwise a.e. and w_{n}→ w in
L^{2}
(Σ1 )
. Then by the semicontinuity properties of a,b we obtain from routine
considerations that w
= 0. Then taking the union of the exceptional sets
for all k, it follows that w
(x)
≤ b
(γu(x),t)
a.e. The other side of the inequality can be
shown similarly. Letting z_{n}∈ B
(un,t)
and v ∈ V, is it true that
lim ni→nf∞ 〈zn,un − v〉 ≥ 〈z(v),un − v〉
for some z
(v)
∈ B
(u,t)
? Suppose not. Then from the above, there is a subsequence such
that the limit equals the liminf but which has the inequality turned around for some v
and all z ∈ B
(u,t)
. Then from what was just shown, letting w_{n} go with z_{n}, there is a
further subsequence such that w_{n}→ w weakly in L^{2}
Then a.e. t_{n} is a Lebesgue point of u,_{xn} for F_{}
(t1,⋅⋅⋅,tn−1)
where F_{(t1,⋅⋅⋅,tn−1)
} consists of
those points of F where
(t1,⋅⋅⋅,tn− 1)
is fixed. Also let t_{n} be a point of density
of F_{(t1,⋅⋅⋅,tn−1)
}. Of course m_{1} a.e. points of F_{(t1,⋅⋅⋅,tn−1)
} are points of density.
Therefore, there exists a sequence h_{k}→ 0+ such that t_{n}− h_{k}→ t_{n} as k →∞ and
(tn − hk)
∈ F_{(t1,⋅⋅⋅,tn−1)
}. Otherwise there would be some open set about t_{n} which
excludes points of F_{(t1,⋅⋅⋅,tn−1)
} which would imply that t_{n} is not actually a point of
density. Then using the fundamental theorem of calculus, we get for such points which
are points of F_{(t1,⋅⋅⋅,tn− 1)
} the fact that u,_{xn}
(t1,⋅⋅⋅,tn−1,tn)
= 0. Thus for a.e.
t_{n}∈ F_{(t1,⋅⋅⋅,tn−1)
},u,_{xn}
(t1,⋅⋅⋅,tn−1,tn)
= 0. Thus u,_{xn}
(t1,⋅⋅⋅,tn−1,tn)
= 0 for a.e. t_{n} in
F_{(t1,⋅⋅⋅,tn−1)
}. Similar reasoning holds for differentiation with respect to the other
variables. Thus ∇u = 0 a.e. on F. ■