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**Sample text**

If is a linear then under a convolution space and if L: x~x is an endomorphism of of L it is natural to understand every bilinear, commutative and associative operation *: X ~ such that the relation L(x * y)=(Lx) * y is fulfilled for all x, y EX. This definition proposed by the author [30] in 1966 is only an inverting of the definition of a multiplier of commutative and associative algebra. In a sense the purpose of convolutional calculus is converse to the purpose of the multiplier theory. The basic problem of the latter theory is to characterize the multipliers of a given convolutional algebra.

A convolution * in x is said to be separately contintlOUS, iff xn -+ x implies xn * Y -+ x *Y for earh Y E[t. 11. If * is a separately continuous and annihilators-free convolution of an endomorphism L: -+ of a Frechet space with a cyclic element, then tke multiplier ring of the convolutional algebra (,£, *) coincides with tke commutant of tke operator L in [t. Under the commutant of L in x it is understood the set of all continuous endomorphisms M: [t -+ x which commute with L. Proof. Let k E[t be a cyclic element of the operator L.

0 L e m m a 2. Por an annihilators-free c07zmutative and associative algebra mwith (9) forallxEm. mUltiplication * the relation (1) is equivalent to relation L(x*x)=(Lx)*x 22 CHAPTERl Proof. It is obvious that (1) implies (9). Now, let us prove the converse. If x, y Em are arbitrary, then from (9) it follows L[(x+ y)*2] =(x+ y) *L(x+ y). After some transformations, this equality takes the form 2L(x*y)=(Lx)*y+x*(Ly). (10) By means of (10) we transform the expression 4L(x * y * z) in two ways, using the ~representations L (x *y * z) = L[x * (y * z)] = L [(x * y) *z].