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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.

MSC:
 39B12 Iteration theory, iterative and composite equations 26E25 Set-valued functions
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References:
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