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Also set-valued functions do not like iterative roots. (English) Zbl 1144.39017
Let \(X\) be a nonempty set, \(c \in X\), \(n \in \mathbb N\) and \(f: X \to 2^{X}\). A set-valued function \(g: X \to 2^{X}\) is an iterative square root of \(f\) if
\[ f(x) = g^{2}(x) = \bigcup_{u \in g(x)}g(u) \]
for \(x \in X\). Let \(\#A\) denote the cardinality of a subset \(A\subset X\).
Theorem 1. If (i) \(\#f(x) = 1\) for every \(x \in X \setminus \{c\}\), (ii) \(f(x_{o}) = \{c\}\) for an \(x_{o} \in X\), (iii) \(\#f(c) > n\) and (iv) \(\#\{x \in X: f(x) = \{y\}\} \leq n\) for every \( y \in X\), then \(f\) has no iterative square roots.
Theorem 2. If (i), (ii) hold, (v) \(\#f(c) > 1\) and (vi) \(c \in f(c)\), then \(f\) has no iterative square roots.
Some examples of multifunctions which have no iterative square roots are given. Appropriate examples show that Theorem 1 does not follow from Theorem 2 nor conversely.

39B12 Iteration theory, iterative and composite equations
26E25 Set-valued functions
Full Text: DOI
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