July 16, 2017

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Extra resources for Appendix to Frigyes Riesz and Bela Sz. -Nagy Functional Analysis...

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6) and u = I „ I is satisfied if / = I ' I. If f$ = I ' _ j , then the corresponding solution us = I „'„_ J. One can see that a small perturbation of / produces a large perturbation of the solution. The Hubert matrix for all n > 0 is positive-definite, because it is a Gramian of a system of linearly independent functions: hij = / Jo xxJridx. BASIC DEFINITIONS. 2 Stable summation of the Fourier series and integrals with randomly perturbed coefficients. Suppose that oo / = 5>M*)> where (/i,,/ij) = δ^, where δ^; = < t (2-7) ' .

77). 76) is verified. 75), one uses the spectral theorem again and gets Jl = \\TâlTy - yf = ^IIT-VII = Γ " ^f^f \a "+" s) Jo One has limiß2(a) = \\Pjsyf o—»0 := ß2(a). 82) because y J. Λ/" by the assumption, and JV=(EQ- Ε-ο)Η. 82), one gets \\ua,8-y\\ 0. 74): The convergence can be as slow as one wishes for some y. The usual assumption which would guarantee some rate of decay of ß and, therefore, of ||u,5 — y\\ is the following one: y = Tbz, 0 < b < 1.

The answer, given in [116]-[118], is: One can recover the discontinuity curves of f and the sizes of the jumps of f across the discontinuity curves. 15 Inverse spectral problem. Consider the problem lu := -u" + q(x)u + Au, u(0) = u(l) = 0. 46) BASIC DEFINITIONS. , 1ϊπιλ η = οο. 48) η—» A natural question is: Does the set of eigenvalues {^j}j=i,2,... determine q{x) uniquely? The answer is: It does not, in general. , [137]): If q(x) is unknown on half of the interval 0 < x < \, then the knowledge of all {\J}VJ determines q(x) on the remaining half of the interval ^ < x < 1 uniquely.

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