CORRIGENDUM TO “BACKWARD ITERATION IN STRONGLY CONVEX DOMAINS" ADV. IN MATH., 228, PP. 2837–2854. MARCO ABATE AND JASMIN RAISSY* Abstract. We correct a gap in two lemmas in [2], providing a new proof of the main results of that paper for hyperbolic and strongly elliptic self-maps of a bounded strongly convex domain with C 2 boundary. We have found a gap in the proofs of Lemmas 2.2 and 2.5 of our paper [2]. In this note we fill these gaps, giving a proof of the main results using different arguments. More precisely we prove the following version of [2, Theorem 0.1]: Theorem 1. Let D ⋐ C n be a bounded strongly convex domain with C 2 boundary. Let f ∈ Hol(D,D) be either hyperbolic or strongly elliptic, with Wolff point τ ∈ D. Let {z k }⊂ D be a backward orbit for f with bounded Kobayashi step. Then: (i) the sequence {z k } converges to a boundary fixed point σ ∈ ∂D; (ii) if σ = τ then σ is repelling; (iii) σ = τ if and only if {z k } goes to σ inside a K-region, that is, there exists M> 0 so that z k ∈ K p (σ, M ) eventually, where p is any point in D. Remark 2. If f is strongly elliptic then clearly σ = τ . We conjecture that σ = τ in the hyperbolic case too. Remark 3. The following proof does not work in the parabolic case, considered in the original version of [2, Theorem 0.1]. Thus the behavior of backward orbits for a parabolic self-map is still not understood, even (as far as we know) in the unit ball of C n (see [4]). Proof. The proof is divided into two cases according to whether f is hyperbolic or strongly elliptic. We will freely use the notations introduced in [2]. Hyperbolic case. We begin by proving part (i) following the approach already indicated in [2, Remark 2.1]. Lemma 4. Let D ⋐ C n be a bounded strongly convex domain with C 2 boundary. Let f ∈ Hol(D,D) be hyperbolic with Wolff point τ ∈ ∂D and let {z k }⊂ D be a backward orbit for f with bounded Kobayashi step a> 0. Then {z k } converges to a boundary fixed point σ ∈ ∂D. Proof. First of all, recall that [3, Lemma 2.4 and Remark 3] yields a constant C 1 > 0 such that (1) ‖z k − z k+1 ‖ 2 + |〈z k − z k+1 ,z k 〉| ≤ C 2 1 1 − ˆ a 2 d(z k ,∂D), and so (2) ‖z k − z k+1 ‖≤ C 1 √ 1 − ˆ a 2  d(z k ,∂D) ≤ C 1 1 − ˆ a  d(z k ,∂D) , * Partially supported by ANR project LAMBDA, ANR-13-BS01-0002. 1