By Kontodimopoulos N., Niakas D.

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**Extra resources for A 12-year Analysis of Malmquist Total Factor Productivity in Dialysis Facilities**

**Example text**

We want to show that the set of t’s for which the above equation has solutions gt , continuous in t and with the desired properties, is both open and closed. To see that this set is open we use the implicit function theorem in a Banach space on which the Hilbert transform is bounded. For our purposes the most convenient choice is E = C 1,α (S1 ) the Banach space of functions on the circle whose ﬁrst derivatives are H¨older continuous with exponent α. Now assuming that we have solutions for 0 < t ≤ t0 deﬁne F : R × E → E by (13) F (t, u)(λ) = |Φ|(λ, gt0 (λ) + (u + ıHu)(λ)X(λ)) − t where Hu is the Hilbert transform of u (and so u + ıHu admits a holomorphic extension to ∆ which is continuous in ∆) and where X ∈ E is a holomorphic function to 36 K.

Let {An } be an increasing sequence of subsets of C and Φn a sequence of holomorphic motions of C with (28) Φn+1 |An = Φn |An . Then there is a limit holomorphic motion Φ : C → C such that Φ|An = Φn |An . Next one proves the holomorphic axiom of choice. It simply says that a holomorphic motion of a ﬁnite point set can be extended to include any other arbitrary point. 1. Let fi : ∆ → C, i = 1, 2, . . , n be holomorphic functions such that for each z ∈ ∆ we have fi (z) = fj (z), i = j. Then for each zn+1 ∈ C − {fi (0) : i = 1, 2, .

Since g(0) = (x, y, z, w) we ﬁnd (9) |(fλ (x), fλ (y), fλ (z), fλ (w))| ≤ η(M, |(x, y, z, w)|) with M = log 1+|λ| . Therefore each fλ is uniformly continuous in A and so extends 1−|λ| continuously to Fλ : A → C. Permuting the x, y, z and w entries in the equation (9) shows this extension is injective and so each Fλ (·) is a homeomorphism onto its image. For each a ∈ A − A the function F (·, a) is holomorphic since it is the local uniform limit of holomorphic functions; the joint continuity in (λ, a) follows since for every r < 1 the family {Fλ (·) : λ ∈ r∆} is equicontinuous.