Counterexamples to Calabi conjectures on minimal hypersurfaces ...

since |grad(y t ◦ ϕ)| 2 ≤ |grad Rl y t | 2 = 1. Thus,( )m − ∑ i,tc 2 it≥ (m − l)and we have letting k → +∞ in (12) thatm sup |H| ≥ (m − l)C b (r).MThis concludes the proof of the first part of Theorem 2.1.For the proof of the second part, we make use of the following characterizationof s<strong>to</strong>chastic completeness given in [13] (see [14, Theorem 3.1]): A Riemannianmanifold M is s<strong>to</strong>chastically complete if and only if for every u ∈ C 2 (M) withu ∗ = sup u < ∞ there exists a sequence {x k } such that u(x k ) > u ∗ − 1/k and∆u(x k ) < 1/k for every k ≥ 1.whereSuppose that M is s<strong>to</strong>chastically complete. Define g: N n−l × R l → R byg(z, y) = ĝ(z) = φ b (ρ N (z))⎧⎨ 1 − cos( √ bt) if b > 0, t < π/2 √ b,φ b (t) = t 2 if b = 0,⎩cosh( √ −bt) if b < 0.Then f = g ◦ ϕ is a smooth bounded function on M. Thus there exists a sequenceof points {x k } in M such thatf(x k ) > f ∗ − 1/k and ∆f(x k ) < 1/kfor k ≥ 1, where f ∗ = sup M f ≤ φ b (r) < ∞. Similar as before, we haveHess N×R lg(ϕ(x k ))(e i , e i ) = Hess N ĝ(z(x k ))(π TN e i , π TN e i )∑n−l= φ ′′b(r k )αi 2 + φ ′ b(r k ) a 2 ijHess N ρ N (z(x k ))(∂/∂θ j , ∂/∂θ j )j=2∑n−l≥ φ ′′b(r k )αi 2 + φ ′ b(r k )C b (r k )= φ ′′b(r k )α 2 i + φ ′ b(r k )C b (r k )= φ ′ b (r k)C b (r k )(81 −l∑t=1c 2 itj=2()a 2 ij1 − α 2 i −l∑t=1c 2 it)