Integral Equations

✷ Let ξ ∈ h p and Aξ = η. Then,∣ ∣∞∑∞∑ ∣∣∣∣∣ ∞ ∣∣∣∣∣n p+2 |η n | 2 = n p+2 ∑a nj ξ jn=1n=1 j=1≤ ‖ξ‖ 2 p∞∑∞∑n=1 j=1n p+2 |a nj| 2j p ≤ C 2 ‖ξ‖ 2 p,hence, η ∈ h p+2 . Now we will prove that A is completely continuous. In fact, operator A canbe represented as a limit with respect to the operator norm of a sequence of finite-dimensionaloperators A N , which are produced by the cut matrix A and defined by the matrices A N ={a nj } N n,j=1:∣ ∣N−1‖(A − A N )ξ‖ 2 ∑ ∣∣∣∣∣ ∞p+2 = n p+2 ∑ ∣∣∣∣∣ 2 ∣ ∣∞∑ ∣∣∣∣∣ ∞a nj ξ j + n p+2 ∑ ∣∣∣∣∣ 2a nj ξ j≤ ‖ξ‖ 2 p≤ ‖ξ‖ 2 p∞∑∞∑n=N j=1[ ∑ ∞ ( ∑ ∞n=Nj=1n=1n p+2 |a nj| 2j pj=Nn p+2 |a nj| 2j p+ ‖ξ‖ 2 pN−1 ∑n=N∞∑n=1 j=N) ∑ ∞ ( ∑ ∞+j=Nn=1|a nj | 2j=1j p n p+2n p+2 |a nj| 2j p )],and, consequently, the following estimates are valid with respect to the operator norm:‖A − A N ‖ ≤∞∑ ( ∑ Nn=Nj=1n p+2 |a nj| 2j p) ∑ ∞ ( ∑ ∞+j=Nn=1n p+2 |a nj| 2j p )→ 0, N → ∞. ✷Corollary 1 The sequence of operators A N strongly converges to operator A: ‖A − A N ‖ → 0,N → ∞, and ‖A‖ < C, C = const. In order to provide the complete continuity of A, it issufficient to require the fulfilment of the estimate∞∑ ∞∑n p+2 |a nj | 2 < C, C = const. (365)n=1 j=1Definition 9 Assume that condition (365) holds for the matrix elements of summation operatorA. Then, the summation operator F = L + A : h p → h p+2 is called the L-operator.Theorem 46 L-operator F = L + A : h p → h p+2 is a Fredholm operator; ind (L + A) = 0.20 Matrix Representation of Logarithmic Integral OperatorsConsider a class Φ of functions ϕ such that they admit in the interval (−1, 1) the representationin the form of the Fourier–Chebyshev seriesϕ(x) = 1 ∞∑2 T 0(x)ξ 0 + ξ n T n (x); −1 ≤ x ≤ 1, (366)n=174