Integral Equations

A, being completely continuous in the Banach space E, is completely continuous in E 1 . Accordingto the corollary to Theorem (9), a completely continuous operator cannot have a boundedinverse in infinite-dimensional space. The contradiction thus obtained proves the theorem.Corollary. Every nonzero characteristic value of a completely continuous operator A in theBanach space E has only finite multiplicity and these characteristic values form a bounded setwhich cannot have a single limit point distinct from the origin of coordinates.We have thus obtained a characterization of the point spectrum of a completely continuousoperator A in the Banach space E.One can also show that a completely continuous operator A in the Banach space E cannothave a continuous spectrum.8 Linear operator equations: the Fredholm theory8.1 The Fredholm theory in Banach spaceTheorem 13 Let A be a completely continuous operator which maps a Banach space E intoitself. If the equation y = x − Ax is solvable for arbitrary y, then the equation x − Ax = 0 hasno solutions distinct from zero solution.Proof. Assume the contrary: there exists an element x 1 ≠ 0 such that x 1 − Ax 1 = T x 1 = 0Those elements x for which T x = 0 form a linear subspace E 1 of the banach space E. Denoteby E n the subspace of E consisting of elements x for which the powers T n x = 0.It is clear that subspaces E n form a nondecreasing subsequenceE 1 ⊆ E 2 ⊆ . . . E n ⊆ . . . .We shall show that the equality sign cannot hold at any point of this chain of inclusions. Infact, since x 1 ≠ 0 and equation y = T x is solvable for arbitrary y, we can find a sequence ofelements x 1 , x 2 , . . . , x n , . . . distinct from zero such thatT x 2 = x 1 ,T x 3 = x 2 ,. . . ,T x n = x n−1 ,. . . .The element x n belongs to subspace E n for each n but does not belong to subspace E n−1 . Infact,butT n x n = T n−1 x n−1 = . . . = T x 1 = 0,T n−1 x n = T n−2 x n−1 = . . . = T x 2 = x 1 ≠ 0.23