-known result (see, for example ([15], Theorem four.1.1, p. 106)). Theorem 7. Let (X , ) be
-known result (see, as an illustration ([15], Theorem 4.1.1, p. 106)). Theorem 7. Let (X , ) be a Banach space. Assume K : X X to become a bounded compact operator and Km : X X , m N, to become a sequence of offered bounded operators with lim K f – Km f = 0 for all f X . Take into account the operator equations in the following:m1 mr ( K f , t ) u tdt Cf Cu , msC = C(m, f ).( I – K) f ( I – Km ) f m= g, = g,m(32) (33)exactly where I could be the identity operator in X and g X . If lim(K – Km )Km = 0, i.e., the sequenceKm m is collectively compact, then , for all sufficiently significant m, ( I – Km )-1 exists and is Diversity Library Description uniformly bounded with the following. ( I – K m ) -1 + ( I – K ) -1 K m . 1 – ( I – K ) -1 ( K – K m ) K mMoreover, denoted by f X and f m X , the unique solutions of (32) and (33). respectively, the following outcomes. f – f m C ( K – Km ) f ,C = C(m, f , g).Proof of Theorem five. The invertibility of I – K follows below the assumption of Theorem 1, while the uniformly boundedness of K2m+1 ( f )m and also the following limit conditionmlim(K – K2m+1 ) fCu= 0,f Cu ,(34)hold beneath the assumptions of Theorem 3.Mathematics 2021, 9,16 ofIt remains to become proven that the sequence K2m+1 m is collectively compact. That is equiv alent (see ([15], p. 114) and ([19], p. 44)) to prove that lim sup sup EN (K2m+1 f )u = 0.N + m fCu =Hence, noting that the following may be the case: EN (K2m+1 f )u K2m+1 f – PCu(K – K2m+1 ) fCu+ Kf – PCuthe thesis is simply proved Goralatide Protocol utilizing (34) plus the uniform boundedness with the operator K. Proof of Theorem six. So as to prove (28), we initially define the following sequence: Fn ( x ) = f 2n ( x ) , f 2n +1 ( x ) , n = two, 4, . . . n = three, 5, . . . (35)obtained by composing two sub-sequences of these defined by the Nystr strategies ONM in (17) and ENM in (22) . As proved in Theorems four and five, beneath the assumptions of Theorems 1, four and 5, these sequences are each convergent to the unique option with the integral Equation (1). Therefore, each of the sub-sequences are convergent for the same limit function f , and also the speed of convergence is definitely the similar. Consequently, we are able to conclude that with m = 2n , the following may be the case:r , C = C(m, f ). mr Therefore, what remain to become carried out as a way to receive (28) is usually to estimate the following distance: [ f 2m+1 – f 2m+1 ]u .[ f – Fn ]uCfW (u)In the definition of the two polynomial sequences, we are able to create the following: f 2m+1 (y) – f 2m+1 (y) = (y)k =mm +1 b – b a – c k k k Ak (y) + k B (y) . u( xk ) u(yk ) k k =(36)We promptly recognize that, by the definition of a and c , using estimates (18) m m and (23), we obtain the following.| a – c | = | f 2m+1 ( xk )u( xk ) – f m ( xk )u( xk )| k k | f 2m+1 ( xk )u( xk ) – f ( xk )u( xk )| + | f ( xk )u( xk ) – f m ( xk )u( xk )| C f Wr (u) , k = 1, . . . , m. mrConsequently, the following may be the case: a – c m mC f mrWr (u) ,C = C(m, f ),(37)exactly where d = maxk |dk |, d = [d1 , d2 , . . . , dm ] T , denotes the infinity norm in Rm . Now, we remark that by (21) and (25) and below the assumption that D2,2 is invertible, the following identity holds accurate: D2,two (b +1 – b +1 ) = D2,1 (a – c ). m m m m Thus, we’ve got the following: b +1 – b +1 m m-1 D2,D2,a – c m m.If we denote by D2m+1 the matrix of coefficients in (21), we note that D2,1 is actually a submatrix of it. By using normal arguments (see as an illustration [15]), it can be feasible to show thatMathematics 2021, 9,17 ofD2m+1 I – K2m+1 Cu Cu as well as the operator norm in the right-hand side is uniformly bounded with respect.