This contribution deals with the dominance relation on the class of conjunctors, containing as particular cases the subclasses of quasi-copulas, copulas and t-norms. The main results pertain to the summand-wise nature of the dominance relation, when applied to ordinal sum conjunctors, and to the relationship between the idempotent elements of two conjunctors involved in a dominance relationship. The results are illustrated on some well-known parametric families of t-norms and copulas.

Necessary and sufficient conditions for the existence of compactly supported $L^p$-solutions for the two-dimensional two-scale dilation equations are given.

Exponential polynomials are the building bricks of spectral synthesis. In some cases it happens that exponential polynomials should be extended from subgroups to whole groups. To achieve this aim we prove an extension theorem for exponential polynomials which is based on a classical theorem on the extension of homomorphisms.

Let x be an indeterminate over ℂ. We investigate solutions $$\begin{array}{c}\hfill \alpha (s,x)=\sum _{n\ge 0}{\alpha }_n\left(s\right)x^n,\end{array}$$αn : ℂ → ℂ, n ≥ 0, of the first cocycle equation $$\begin{array}{c}\hfill \alpha (s+t,x)=\alpha (s,x)\alpha \left(t,F(s,x)\right),s,t\in ,\left(\mathrm{Co}1\right)\end{array}$$in ℂ [[x]], the ring of formal power series over ℂ, where (F(s,x))s ∈ ℂ is an iteration group of type II, i.e. it is a solution of the translation equation $$\begin{array}{c}\hfill F(s+t,x)=F(s,F(t,x\left)\right),s,t\in ,\left(\mathrm{T}\right)\end{array}$$of the form F(s,x) ≡ x + ck(s)xk mod xk+1, where k ≥ 2 and ck ≠ 0 is necessarily an additive function. It is easy to prove that the coefficient functions αn(s) of $$\begin{array}{c}\hfill \alpha (s,x)=1+\sum _{n\ge 1}{\alpha }_n\left(s\right)x^n\end{array}$$are polynomials in ck(s).It is possible to replace...

Let E be a real normed space and a complex Banach algebra with unit. We characterize the continuous solutions f: E → of the functional equation $f(x+y)=li{m}_{n\to \infty }\left(f(x/n)f(y/n)\right)ⁿ$.