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P. 101
88 Direct methods
u (x)
N
1
2 N
x
1 2 3
... 1
N N N
Figure 3.1: minimizing sequence
Observe that |u | =1 a.e. and |u ν | ≤ 1/ (2ν) leading therefore to the desired
0
ν
convergence, namely
1
0 ≤ I (u ν ) ≤ 4 −→ 0,as ν →∞ .
(2ν)
Proof. We will not prove the theorem in its full generality. We refer to the
literature and in particular to Theorem 3.4.1 in [31] for a general proof; see also
the exercises below. We will prove it under the stronger following hypotheses.
¡ ¢
0
n
We will assume that f ∈ C 1 Ω × R × R , instead of C ,and
(H1+) (u, ξ) → f (x, u, ξ) is convex for every x ∈ Ω;
(H2+) there exist p> 1 and α 1 > 0, α 3 ∈ R such that
p n
f (x, u, ξ) ≥ α 1 |ξ| + α 3 , ∀ (x, u, ξ) ∈ Ω × R × R .
(H3) there exists a constant β ≥ 0 so that for every (x, u, ξ) ∈ Ω × R × R n
³ ´
p−1 p−1
|f u (x, u, ξ)| , |f ξ (x, u, ξ)| ≤ β 1+ |u| + |ξ|
¡ ¢
= ∂f/∂ξ and f u = ∂f/∂u.
where f ξ = f ξ 1 , ..., f ξ n , f ξ i i
Once these hypotheses are made, the proof is very similar to that of Theorem
2
3.1. Note also that the function f (x, u, ξ)= f (ξ)= |ξ| /2 satisfies the above
stronger hypotheses.
Part 1 (Existence). The proof is divided into three steps.
Step 1 (Compactness). Recall that by assumption on u 0 and by (H2+) we
have
−∞ <m ≤ I (u 0 ) < ∞.