We say that a subset of C n is hypoconvex if its complement is the union of complex hyperplanes. Let ∆ be the closed unit disk in C , Γ = ∂∆. We prove two conjectures of Helton and Marshall. Let ρ be a smooth function on Γ × C n whose sublevel sets have compact hypoconvex fibers over Γ . Then, with some restrictions on ρ, if Y is the set where ρ is less than or equal to 1, the polynomial convex hull of Y is the union of graphs of analytic vector valued functions with boundary in Y . Furthermore, we show that the infimum inf f ∈H ∞ (∆) n ρ(z, f (z)) ∞ is attained by a unique bounded analytic f which in fact is also smooth on Γ . We also prove that if ρ varies smoothly with respect to a parameter, so does the unique f just found.From Theorem 1 of [YK], we may conclude that if 0 < t < R, and D t z is the closed domain enclosed by the level set {w | ρ(z, w) = t}, then the complex tangent spaces to boundary points of D t z do not meet int D t z . See also [Ki]. By Corollary 4.6.9 of [Hö] applied to int D t z , we may conclude that int D t z is hypoconvex, so D t z is as well, as the intersection of D s z for s > t. D t z is also clearly strictly hypoconvex. From the same corollary, we obtain that int D t z is "C -convex", i.e., if P is a 1-dimensional complex affine subspace of C n which 680 M.A. Whittlesey meets int D t z , then the intersection is a connected and simply connected subset of P . If P ∩ int D t z = ∅, then the intersection is a smoothly bounded subset of P , because P cannot be a tangent to D t z ; hence the derivative of ρ restricted to P where ρ = t is not identically zero and P ∩ D t z is the closure of P ∩ int D t z . If P ∩ int D t z = ∅ but P does meet D t z then we claim it only meets D t z in one point. Such a P must be tangent to the boundary of D t z at any intersection point. As observed earlier, the tangent space is locally disjoint from D t z near such a point. If there are two such points, then we may perturb P slightly and obtain a P whose intersection with int D t z is not connected. Thus the intersection of any complex hyperplane with D t z is either a point or a Jordan domain with boundary as smooth as ρ. We note that D t z ⊂ K t z . The reverse inclusion also holds: Proposition 1 of [Z] shows that the interior of D s z is polynomially convex for all s, so D t z is polynomially convex as the intersection of the polynomially convex open sets D s z , s > t.Proof. It is easy to show that ( L t ) z = (L t z ) so all we must do is show thatThen w ∈ K t z since projection is analytic, so ρ(z, w ) ≤ 1. Now the polynomial w 1 v 1 + w 2 v 2 + ... + w n v n − 1 is identically 0 on L t so is 0 on L t z . Thus w = 0. Let P(). Now for |w| sufficiently small,and v ∈ T w so w / ∈ B z , so the claim holds. Consider the connected component of B z ∩ {w ρ(z, w) < 1} which contains 0. Construct a continuous path w(s), 0 ≤ s ≤ 1, from 0 to a point in the boundary of this component. Then we claim that T w(1) meets Q(L t z ) in only one point, which must be in Q(L t z ). If they met in a point in Q(L t z )...
