Urable actual function h L ( F ), satisfying the moment conditions tk
Urable genuine function h L ( F ), satisfying the moment circumstances tk h(t)d= mk , k i d i = 1, . . . , n,Fand 0 h b1, for some positive quantity b. 3.4. Extension of Linear Operators, Markov Moment Challenge, and Mazur rlicz PHA-543613 In stock theorem The objective of this section is usually to prove a result on an operator valued Markov moment challenge whose resolution is defined on a space of analytic functions, and an operator valued Mazur rlicz theorem involving the space of all real continuous functions defined around the spectrum of a self-adjoint operator as the domain space of the solution. Both these outcomes stick to as applications of a PK 11195 site general theorem on extension for linear operators, with two constraints. They do not use any result on polynomial approximation. Within the next theorem, X are going to be a real vector space, Y an order-complete vector lattice, A, B X convex subsets, Q : A Y a concave operator, P : B Y a convex operator, M X a vector subspace, and T0 : M Y a linear operator. All vector spaces and linear operators are regarded more than the genuine field. Theorem 13 (see [26]). Assume that T0 ( x ) Q( x ) x M A, T0 ( x ) P( x ) x M B. The following statements are equivalent: (a) There exists a linear extension T : X Y from the operator T0 such that T | A Q, T | B P; (b) There exists P1 : A Y, convex, and Q1 : B Y concave operator such that for all , t, , a1 , a , b1 , b , v [0, 1]2 (0, ) A2 B2 M, the following implication holds:(1 – t) a1 – tb1 = v ((1 – ) a – b ) (1 – t) P1 ( a1 ) – tQ1 (b1 ) T0 (v) ((1 – ) Q( a ) – P(b )).It is worth noticing that the extension T of Theorem 13 satisfies the following circumstances: it can be an extension of T0 , it can be dominated by P on B, and it dominates Q on A. Right here the convex subsets A, B are arbitrary, with no restriction around the existence of relative interior points or on their position with respect towards the subspace M. As a consequence of Theorem 13, 1 can prove the following result: Theorem 14 (see [26]). Let X be a locally convex space, Y an order total vector lattice with powerful order unit u0 , and S X a vector subspace. Let C X be a convex subset with all the following properties: (a) (b) There exists a neighborhood D with the origin such that (S D ) C = (that may be, by definition, C and S are distanced); C is bounded.Symmetry 2021, 13,20 ofThen for any equicontinuous loved ones of linear operators y Y \0, there exists an equicontinuous family members Tjj Jfjj JL(S, Y ) and for anyL( X, Y ) such thatTj (s) = f j (s), s S, Tj y, C, j J. Furthermore, if D is a convex balanced neighbourhood with the origin such that f j ( D S) [-u0 , u0 ], (S D ) C = and if 0 is such that PD ( a) a C and 1 0 is substantial enough such that y 1 u0 , then the following relations hold: Tj ( x ) (1 1 ) PD ( x ), x X, j J. We’ve got denoted by PD the gauge (Minkowski functions) attached to D. Let ( Bk )kNn be a sequence of operators in Y = Y ( A) defined by (five) and || B|| Y \0. On the other hand, let n = 0 be a all-natural quantity and X be the space of all definitely convergent power series h inside the unit closed polydisk D1 = 1, i 1, . . . , n, with genuine coefficients. The norm on X is defined by:||h|| = sup |h(z)| : z D1 .We denote hk (z) = z11 zkn , k = (k1 , . . . , k n ) Nn , z D1 ,|k | = k1 k n . n Theorem 15 (see [29], Theorem three.1, p. 178). Assume that A1 , . . . , An are elements of Y ( A) such that there exists a true quantity M 0, with all the propertyk| Bk | MA1 1 A2kn n , k Nn , k1 ! kn !2kp =nA2 I, pwhere I : H H is the identity operator. Let gk kN.