By S. Schwabik, M. Tvrdý, O. Vejvoda
Ebook through Schwabik, S., Tvrdý, M., Vejvoda, O.
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Extra resources for Differential and Integral Equations: Boundary Value Problems and Adjoints
E. |/(*)| < C for all t e [a, b]. If we set S = SxnS2 then S € /5^[a, b] and for any >1 e A(S) the inequalities (i), (ii) are satisfied. 4 set k(t) = \la g(x) dh(t), t e [a, b]. Then for A e A(S) we have by (i) and (ii) ЬÁЛ) - £ fgdh I /(T,) 1=1 -í ^dfc- lAtj) 1=1 fgáh dh - f(xj) g(xj) [h(aj) - % , _ . ) ] Jfltj- i '5 + Ij(^Л%,)-%,->)]- jffdh 1=1 < cJ=il \g(xj)[h(aj)-h(a . - . ) ] - f " fldЛ JOLj- 1 є Cє є "}--< +-=£. 5 the integral \a f dk = ja f(t) d(\a g dh) exists and equals \ba fg dh. Using the same technique the second implication can be also proved.
4*7y V> x = [xJxA*/y > ] * . Thus B = JxA*JY 1 G B(Ky, £ x ). 20 remain valid. 4* also for representations of the adjoint operator to A. )x and the corresponding norm x e K -• ||x|| x = (x, x)x'2. 24. Definition. Given AeL(X, Y) and yeY, solution to (3,1) if ||Au-y|| y < | | A x - y | | y ueX is said to be a least square for all xeK. 25. Proposition. 6) (Ax, Au0 - y)y = 0 for all x e X , then u0 is a least square solution to (3,1). Furthermore, xe X is a least square solution to (3,1) if and only if x — u0 e N(A).
T e [a, b] and x e Lp[a, b], 1 < p < oo. e. on [a, b] and belongs to LF\a, b\ Proof. Clearly, cp(s) is defined for any se\c,d\ c = s0 < s, < ... e. e. \f(t) h(t, G)\ < \f(t)\ x(t) t e [a, b] and any a e [c, d] . Applying the Lebesgue Dominated Convergence Theorem we obtain immediately (4,38). e. , k) (cf. e. , k). e. on [a, b] se[c,d] and hence rj e Lp\a, b] for any a e C\c, d] (or a e BV\c, d]). 38. Theorem (Cameron, Martin). 37. Then for any fel3\a,b\ where q -= p/(p — 1) z/p > 1 ana7 q = cc if p = I, and any g e C\c, d] (or a e BV[c, d]) the integrals м ds[h(t, s)] g(s) I dr anď d.