R of T0 , that is continuous on C0 ( F ) with respectR of

R of T0 , that is continuous on C0 ( F ) with respect
R of T0 , which is continuous on C0 ( F ) with respect towards the sup-norm. Thus, T includes a representation by implies of a positive Borel standard measure on F, such that T (x) = x (t)d x C0 ( F ).FSymmetry 2021, 13,11 ofLet p P be a non-negative polynomial function. There is a nondecreasing sequence p ( xm )m of continuous non-negative function with compact help, such that xm pointwise on F. Positivity of T and Lebesgue’s dominated convergence theorem for yield p(t)d = T ( p) supT ( xm ) = sup xm (t)d= p(t)d p P .FFFThanks to Haviland’s theorem, there exists a constructive Borel standard measure on F, such that ( p ) = ( p ) – ( p ) ( p ) = ( p ) ( p ), p P . Considering the fact that is assumed to become M-determinate, it follows that ( B) = ( B) B) for any Borel subset B of F. From this final assertion, approximating every single x L1 ( F , by a nondecreasing sequence of non-negative simple functions, as well as working with Lebesgue’s convergence theorem, a single obtains firstly for positive functions, then for arbitrary -integrable functions, :Fd =Fd Fd L1 ( F ).In distinct, we should have xd xd= T ( x ) = T0 ( x ) = Q1 ( x ). (7)FFThen, Equations (6) and (7) conclude the proof. Remark two. We recall that the preceding Lemma 3 is no additional valid when we replace L1 ( F ) using the Hilbert space L2 ( F ), F = Rn , n 2 (see [13], Theorem four.four, exactly where the authors construct such a measure ). The following theorem follows from Lemma three. It facilitates proving inequalities around the entire L1 ( F ) space verifying the involved inequalities only on polynomial functions (right here is really a moment determinate measure on F). Theorem 8 (see [27], Theorem 2). Let F be a closed unbounded subset of Rn , Y an order total Banach lattice, y j jNn a offered sequence in Y, and an M-determinate measure on F. Let T2 B L1 ( F ), Y be a Safranin Chemical linear constructive (bounded) operator from L1 ( F ) to Y. The following statements are equivalent: (a) (b) There exists a distinctive linear operator T B L1 ( F ), Y , such that T j = y j , j Nn , 0 T T2 on the good cone of L1 ( F ), and T T2 ; For any finite subset J0 Nn , and any a j j J R, we havej J0 a j jFj J0 a j y j j J0 a j Tj .Proof. Observe that the assertion (b) says that 0 T0 ( p) T2 ( p), p P , (eight)exactly where T0 : P Y is the special linear operator that verifies the interpolation situations of Equation (1). Thus, ( a) (b) is apparent. To prove the converse, contemplate the vector subspace X1 L1 ( F ) of all functions L1 ( F ), verifying| (t)| p(t) t FSymmetry 2021, 13,12 offor some polynomial p. Clearly, X1 contains the subspace of polynomials at the same time because the subspace of continuous compactly supported real-valued functions. On the other hand, the subspace of polynomials is usually a majorizing subspace in X1 , and according to the initial inequality of Equation (eight), T0 is constructive as a linear operator on P . Application of Theorem 5 yields the Pinacidil Membrane Transporter/Ion Channel existence of a optimistic linear extension T : X1 Y of T0 . Let x be a non-negative continuous compactly supported function on F, and ( pm )m a sequence of polynomials with all the properties specified in Lemma three. As outlined by the second inequality of Equation (8), we’ve got T ( pm ) = T0 ( pm ) T2 ( pm ) for all m N. Our next purpose is usually to prove that T ( x ) T2 ( x ). (9)Assuming the contrary, we ought to have T2 ( x ) – T ( x ) Y . Considering the fact that Y is closed, a / Hahn anach separation theorem results in the existence of a good linear kind y inside the dual Y of Y, verifying y ( T2 ( x ) – T ( x )) 0 (10) The good linear kind y T has a representing constructive standard Borel measure for wh.