R \K. From Lemma 1 we infer the next outcome. To this
R \K. From Lemma 1 we infer the subsequent result. To this aim, we observe that 0 is dominated by nonnegative continuous compactly supported functions : R R , (t) = 0 (t) = (t) for all t K. Then 1 applies Lemma 1. Lemma 2 (see [28], Lemma two). If K R is often a compact subset, and : K R a continuous function, then there exists a sequence ( pl )l N of polynomial functions, such that pl 0 on R , pl |K , l , uniformly on K. From Lemma two and Theorem five (exactly where X1 stands for C (K ) and X0 stands for the Betamethasone disodium Epigenetics subspace P of all polynomial functions), the following corollary follows simply. We recall a wellknown significant example of an order full Banach lattice Y of self-adjoint operators acting on a complicated or true Hilbert space H. Let A = A( H ) be the ordered vector space of all the self-adjoint operators acting on H, and let V A. The organic order relation on A is U W if, and only if: Uh, h Wh, h for all h in H.tSymmetry 2021, 13,7 ofOne can prove that A with this ordering just isn’t a lattice. Therefore, it truly is interesting to fix A A and define the following:Y1 (A) = U A; UA = AU, Y = Y(A) = W Y1 (A); WU = UW, U Y1 (A).(five)Then, Y ( A) is definitely an order full Banach lattice (in addition to a commutative real algebra), as discussed in [5]. If U A, we denote by (U ) the spectrum of U, and by dEU the spectral measure attached to U. As usual, one particular denotes j (t) = t j , j N, t R . We recall that any symmetric (Z)-Semaxanib Inhibitor linear operator in the whole space H to H is continuous, therefore is self-adjoint. This follows quite effortlessly from the closed graph theorem. Considering the fact that all symmetric operators appearing in what follows are defined on the complete Hilbert space H, they’re self-adjoint operators. Conversely, by definition, any self-adjoint operator acting on H is symmetric and continuous. Thus, within this paper, there might be no distinction among these two notions. As is well-known, the spectrum of such an operator is usually a compact subset of your genuine line. If H = Rn , n N, n 2, then A is isomorphic to the space Sym(n, R) of all n n symmetric matrices with genuine entries. See [29], p. 11 for an example associated to this space. Corollary 1 (see [28], Corollary 1). Together with the above notations, assume that A is often a positive selfadjoint operator acting on H, Y = Y ( A) will be the space defined by Equation (5), and Bj jN is usually a sequence of operators in Y ( A). The following statements are equivalent: (a) There exists a exclusive positive linear operator T : C ( ( A)) Y , such that T j = Bj , j N, T (b) (t)dE A , (C ( ( A))) , || T ||( A)For any polynomial m 0 j j 0 on ( A), the result is m 0 j Bj 0; if J0 N is an j= j= arbitrary finite subset, and j ; j J0 R, then the following inequalities hold:i,j Ji j Bi jli,j Ji j Ai jl , l 0, 1.Proof. (b) ( a) . We define T0 : P Y ( A) by T0 ( j j ) := j Bj , exactly where J0 Nj J0 j Jis an arbitrary finite subset, j R, j J0 . Then, T0 is linear and, in line with the very first condition (b), T0 ( p) 0 for all polynomials p with p(t) 0 t ( A). Alternatively, for each g C ( ( A)), there exists a continual function c, such that g(t) c for all t ( A). According to Theorem five, T0 features a linear constructive extension T : C ( ( A)) Y . Next, we prove that T is continuous (and its norm is usually determined). This could be shown for any constructive linear operator, within a more basic framework. Namely, any good linear operator acting involving two ordered Banach spaces is continuous (see [6] and/or [21]). Right here, we are interested only in our problem, when the norm of your involved good linear operator can be d.