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**Example text**

Assume that the generalized translation operators L p satisfy the finiteness condi- tion: (F): For any A, BE tBo(Q), there is a compact set F so large that (Lpf)(q) for almost all pEA and q E B provided that suppfn F = 0. =0 It is obvious that if L p satisfy (F), then right generalized translations also satisfy (F). By using condition (F), one can define the action of generalized translation operators L p Section 2 47 Hypercomplex Systems and Related Objects f from C(Q). 8o CQ) of arbitrary points p, q E Q, we take a set F according to condition (F) andput (LpJ)(q) = (LpfFXq), where the function fF(r) is equal to fCr) for rE F upon all functions and equal to zero, otherwise (here, fer) is an arbitrary function from C(Q».

This implies that this sequence converges to f(t) almost everywhere with respect to a(E, r) for almost all r with respect to p (E). For such r, by virtue of the Lebesgue theorem on majorized convergence, we have ---7) n~oo ff(t)dta(E t, r). 30) exists. Then one can construct a sequence of step functions 0:::;, f I (t) :::;, fz (t) :::;, '" :::;, f( r) that converges to f(t) almost everywhere with respect to ",(E). But then this sequence converges to f(t) a(E, r)-almost everywhere for p(E)-almost every r.

Let us show that the operator Luis selfadjoint. In fact, L: = L u*' But L u*f= u* * f = u * f = L uf for an arbitrary function f ELI n LOQ' Therefore, the selfadjointness of L u follows from the denseness of LIn L in L2 and the fact that the operator L u is bounded. Now let u (r) belong, in addition, to the radical ~ of the algebra L 1 (Q, m). Then, clearly, the element 0 e + u belongs to the radical of the algebra 1; 1 (Q, m). We show that the operator L u = O. Indeed, let R z be the resolvent of the operator L u' It is easy OQ to see that R zf = (u - Z e) -I *f (f E L 2).