This supplement contains additional materials for the main paper. Section B contains proofs of auxiliary results not included in the main paper. Section C details how the generalized virtual values coincide with traditional (ironed) virtual values in the single-good monopoly problem. Theorems, equations, and sections in the main paper are referenced using the original numbering.APPENDIX B: ADDITIONAL PROOFS PROOF OF LEMMA 4.3 (ADAPTED FROM MADARÁSZ AND PRAT (2012)): It is easy to see that statement (b) of the lemma follows from (a) by integrating over each S k in the partition; so it suffices to prove (a).As in the proof of Lemma 4.2, we can take = max x θ u(x θ) − min x θ u(x θ), and then in any mechanism, any two types' payments can differ by at most . Also, put τ = min{ε/6 1}.By Lipschitz continuity, there exists δ such that whenever θ θ are two types with d(θ θ ) < δ, then |u(x θ) − u(x θ )| < τε/6 for all x. We show this δ has the desired property.Let (x t) be any given mechanism. Let t = min θ t(θ). Let S ⊆ (X) × R be the set of values (x(θ) τt + (1 − τ)t(θ)) for θ ∈ Θ and let S be its closure, which is compact (by the above observation on payments). Then define ( x t) by simply assigning to each type θ ∈ Θ the outcome in S that maximizes its payoff, Eu(x θ) − t. This exists by compactness. This ( x t) is a mechanism: IC is satisfied by definition and IR is satisfied since the payments have only been reduced relative to those in (x t), so each type θ has the option of getting allocation x(θ) for a payment of less than t(θ), which gives nonnegative payoff. Now let d(θ θ ) < δ. We know that the outcome chosen by θ in the new mechanism can be approximated arbitrarily closely by an element of S corresponding to some type θ ; in particular, there exists θ such that Eu x θ θ − Eu x θ θ < τε 6 and t θ − τt + (1 − τ)t θ < τε 6 (B.1) Now, we know from IC for the original mechanism thatand by the definition of the new mechanism ( x t) that Eu x θ θ − t θ ≥ Eu x(θ) θ − τt + (1 − τ)t(θ)Using (twice) the fact that d(θ θ ) < δ, the latter inequality turns into Eu x θ θ − t θ ≥ Eu x(θ) θ − τt + (1 − τ)t(θ) − τε 3