Proof of Lemma 3.1.

It suffices to show that u≥0 in Ω. From (2.2), (3.4) and the integration by parts formula it follows

for all v∈W1,∞⁢(Ω) with v|∂⁡Ω=0. We denote

𝒪 := { x ∈ Ω : u ⁢ ( x ) < 0 } , u - := min ⁡ { 0 , u } , 𝒪 ′ := { x ∈ 𝒪 : | ∇ ⁡ u - ⁢ ( x ) | > 0 } .

From [2, Theorem A.1] we have u-∈W1,∞⁢(Ω) and ∇⁡u-=∇⁡u in 𝒪, ∇⁡u-=0ℝN in Ω∖𝒪. So, taking v=u- in (1) and using hypothesis (H), it follows

If meas⁡𝒪′>0, then from (2) we get the contradiction

0 > - ∫ 𝒪 ′ | ∇ ⁡ u - | 2 1 - | ∇ ⁡ u - | 2 ≥ 0 .

Consequently, meas⁡𝒪′=0 and, as ∇⁡u-=0ℝN in Ω∖𝒪, we get that ∇⁡u-=0ℝN a.e. on Ω. As Ω is a domain, we infer that u-=const., hence u-≡0 in Ω. This means u≥0 in Ω. ∎