Counterexamples to Calabi conjectures on minimal hypersurfaces ...

where G is a smooth function on [0, +∞) satisfying:(i) G(0) > 0,(ii) G ′ (t) ≥ 0 on [0, +∞),(iii) 1/ √ G(t) ∉ L 1 (0, +∞), (iv) lim sup t→+∞tG( √ t)G(t)< +∞.(6)Then, given a function u ∈ C 2 (M) with u ∗ = sup M u < +∞ there exists a sequence{x k } k∈N ⊂ M m such thatu(x k ) > u ∗ − 1/k; |gradu|(x k ) < 1/k; ∆u(x k ) < 1/k.Observe that a function G satisfying the above conditions isNow we are ready <strong>to</strong> prove Theorem 2.1.G(t) = (t + 2) 2 (log(t + 2)) 2 . (7)Proof of Theorem 2.1: Define σ : N n−l × R l → [0, +∞) byσ(z, y) = ρ R l(y),where ρ R l(y) = ‖y‖ R l is the distance function <strong>to</strong> the origin in R l . Since ϕ is properand ϕ(M) ⊂ B N (r) × R l , then the function ψ(x) = σ ◦ ϕ(x) satisfies ψ(x) → ∞ asρ M (x) = dist M (q, x) → +∞. Off a compact set, we now have|grad M ψ(x)| ≤ |grad N×Rl σ(ϕ(x))| = |grad Rl ρ R l| = 1 ≤ √ ψ(x).To compute ∆ M ψ we start with bases {∂/∂ρ N , ∂/∂θ 2 , . . .,∂/∂θ n−l } of TN and{∂/∂ρ R l, ∂/∂γ 2 , . . .,∂/∂γ l } of TR l (polar coordinates) orthonormal at x ∈ M.Then, we choose an orthonormal basis {e 1 , . . ., e m } for T x M as follows∂ ∑n−l∂ ∂e i = α i + a ij + β i∂ρ N ∂θ j ∂ρ R lj=2+l∑t=2b it∂∂γ t·Hence, we haveHess N×R l σ(ϕ(x))(e i , e i ) = Hess R l ρ R l(π R le i , π R le i ) =1σ(ϕ(x))where π R l denotes the orthogonal projection on<strong>to</strong> TR l . Here, we are usingn−l|e i | = 1 = αi 2 + ∑l∑a 2 ij + β2 i +j=25t=2b 2 itl∑t=2b 2 it ≤ 1ψ(x) ,