Introduction to Modern Analysis by Shmuel Kantorovitz

By Shmuel Kantorovitz

This article relies on lectures given via the writer on the complicated undergaraduate and graduate degrees in degree thought, practical research, Bannach Algebras, Spectral conception (of bounded and unbounded operators), Semigroups of Operators, likelihood and Mathematical records, and Partial Differential Equations. the 1st 10 chapters speak about theoretical tools in degree conception and useful research and include over one hundred twenty finish of bankruptcy workouts. the ultimate chapters practice idea to functions in likelihood idea and Partial Differential Equations.

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Letting z := x − y, the existence part of the theorem will follow if we show that (z, u) = 0 for all u ∈ Y . Since Y is a subspace, and Y = {0} without loss of generality, every u ∈ Y is a scalar multiple of a unit vector in Y , so it suffices to prove that (z, u) = 0 for unit vectors u ∈ Y . 34), z − λu The left-hand side is 2 = z 2 ¯ u)] + |λ|2 . − 2 [λ(z, x − (y + λu) 2 ≥ d2 , since y + λu ∈ Y . Since z = d, we obtain ¯ u)] + |λ|2 . 0 ≤ −2 [λ(z, Choose λ = (z, u). Then 0 ≤ −|(z, u)|2 , so that (z, u) = 0 as claimed.

This proves the existence part of the theorem. Suppose now that y, y ∈ X are such that φ(x) = (x, y) = (x, y ) for all x ∈ X. Then (x, y − y ) = 0 for all x, hence in particular (y − y , y − y ) = 0, which implies that y = y . 8 The Lebesgue–Radon–Nikodym theorem We shall apply the Riesz representation theorem to prove the Lebesgue decomposition theorem and the Radon–Nikodym theorem for (positive) measures. We start with a measure-theoretic lemma. The positive measure space (X, A, µ) is σ-finite if there exists a sequence of mutually disjoint measurable sets Xj with union X, such that µ(Xj ) < ∞ for all j.

Since Aj ⊂ Ej , j = 1, 2, . . and Ej are mutually disjoint, so are the sets Aj . Hence ν(E) := µ(A) = ν(Ej ), µ(Aj ) := j j and we conclude that (X, M, ν) is a measure space extending (X, A, µ). It is complete, because if E ∈ M is ν-null and A, B are as in (6), then for any F ⊂ E, we have ∅ ⊂ F ⊂ B, and since µ(B − A) = 0, µ(B − ∅) = µ(B) = µ(A) := ν(E) = 0, so that F ∈ M. Finally, suppose (X, N , σ) is any complete extension of (X, A, µ), let E ∈ M, and let A, B be as in (6). Write E = A∪(E −A).

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