1336 CHAPTER 39. WEAK SOLUTIONS

and I (u)≥ a> 0 for all ∥u∥= r. Suppose also that there exists v,∥v∥> r such that I (v)≤ 0.Then define

Γ≡ {g ∈C ([0,1] ;H) : g(0) = 0,g(1) = v}

Letc≡ inf

g∈Γmax

0≤t≤1I (g(t))

Then c is a critical value of I meaning that there exists u such that I (u) = c and I′ (u) = 0.In particular, there is u ̸= 0 such that I′ (u) = 0.

This nice example is in Evans [49]. Let the Hilbert space be H10 (U) where U is a

bounded open set. To avoid cases, assume U is in R3 or higher. The main results will workin general but it would involve cases. Consider the functional

12∥u∥2

H10−∫

UF (u)dx≡ I1 (u)− I2 (u)

where F ′ (u) = f (u) , f (0) = 0. Here it is assumed that

| f (u)| ≤C (1+ |u|p) ,∣∣ f ′ (u)∣∣≤C

(1+ |u|p−1

), 1 < p <

n+2n−2

(39.2.8)

Also suppose that0≤ F (u)≤ γ f (u)u where 0 < γ < 1/2 (39.2.9)

and finally thatα |u|p+1 ≤ F (u)≤ A |u|p+1 , α,A > 0 (39.2.10)

Let R : H10 (U)→ H−1 (U) be the Riesz map.

Showing Functional is C1,1

Then it is not hard to verify that (I1 (u) ,v) = (u,v) and so it is clearly the case that I′ (u)exists and is a continuous function of u. In addition to this, it is Lipschitz.

Next consider I2.

I2 (u+ v)− I2 (u) =∫

UF (u+ v)−F (u)dx

=∫

Uf (u)v+

12

f ′ (û)v2dx, û ∈ [u,u+ v]

Now H10 (U) embeds continuously into L2n/(n−2) (U) . Because of the estimate for f (u) ,

we can regard f (u) as being in H−1 (U) as follows.∣∣∣∣∫Uf (u)vdx

∣∣∣∣≤ (∫U| f (u)|2n/(n+2) dx

)(n+2)/2n(∫U|v|2n/(n−2)

)(n−2)/2n

≤(∫

UC(

1+ |u|2n/(n−2))

dx)(n+2)/2n

∥v∥H10