1338 CHAPTER 39. WEAK SOLUTIONS

Now (p−1) 2nn+2 ≤

( n+2n−2 −1

) 2nn+2 = 8 n

n2−4 ≤2n

n−2 and so the derivative is Lipschitz onbounded sets of H1

0 (U).

Palais Smale Conditions

Here we verify the Palais Smale conditions. Suppose then that I (uk) is bounded andI′ (uk)→ 0 in H1

0 (U). Then ∣∣∣∣12 ∥uk∥2H1

0−∫

UF (uk)dx

∣∣∣∣≤C (39.2.12)

Since I′ (uk)→ 0,uk−R−1 f (uk)→ 0 in H1

0 (U) (39.2.13)

Take inner product of the second term with uk.(R−1 f (uk) ,uk

)≡ ⟨ f (uk) ,uk⟩H−1,H1

0=∫

Uf (uk)ukdx

Then by assumption, for ε > 0, and all k large enough,

∣∣(I′ (uk) ,uk)∣∣≤ ∣∣∣∣∥uk∥2

H10 (U)−

∫U

f (uk)ukdx∣∣∣∣≤ ε ∥uk∥H1

0 (U)

Then also for large k, letting ε = 1,∣∣∣∣∫Uf (uk)ukdx

∣∣∣∣≤ ∥uk∥2H1

0 (U)+∥uk∥H10 (U)

Now from the estimates assumed and 39.2.12,

12∥uk∥2

H10≤ C+

∫U

F (uk)dx≤C+ γ

∫U

f (uk)ukdx

≤ C+ γ

(∥uk∥2

H10 (U)+∥uk∥H1

0 (U)

)and since γ < 1/2, (

12− γ

)∥uk∥2

H10≤C+∥uk∥H1

0 (U)

and so ∥uk∥H10 (U) is bounded. Hence it has a subsequence still denoted as uk which con-

verges weakly in H10 (U) to u ∈ H1

0 (U) . Since p < n+2n−2 , it follows that

p+1 <n+2n−2

+1 =2n

n−2

and so by compactness of the embedding, it follows that uk→ u strongly in Lp+1 (U). Wecan assume convergence also takes place pointwise by taking a suitable subsequence.

Now | f (u)v| ≤C (1+ |u|p) |v| . Therefore, adjusting the constants,