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7330hw7

# 7330hw7 - Math 7330 Functional Analysis Notes 7 The...

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Math 7330: Functional Analysis Fall 2002 Notes 7: The Spectral Theorem A. Sengupta Let B be a complex, commutative B * algebra, with Δ its Gelfand spectrum. Then, as we have seen in class, (i) the Gelfand transform B C (Δ) : x 7→ ˆ x satisfies ˆ x * = ˆ x for every x B ; (ii) the spectral radius ρ ( x ) equals the norm | x | for every x B . Fact (ii) was proven first for hermitian elements in any B * algebra and then, using the Gelfand transform, for all elements in a commutative B * algebra. If a B is hermitian then ρ ( a ) = lim n →∞ | a n | 1 /n while | a 2 | = | aa * | = | a | 2 which implies | a 2 k | = | a | 2 k , and so, letting n → ∞ through powers of 2 we get ρ ( a ) = | a | for every hermitian a in any B * algebra. For a commutative B * algebra B we have for a general x B , ρ ( xx * ) = | ˆ xx * ) | sup ≤ | ˆ x | sup | ˆ x * | sup = ρ ( x ) ρ ( x * ) ρ ( x ) | x * | Since xx * is hermitian, ρ ( xx * ) = | xx * | , which is equal to | x || x * | . So we have | x | ≤ ρ ( x ) But we already know the opposite inequality. So ρ ( x ) = | x | . By (i) and (ii) and other properties we have studied before, the Gelfand transform is a * –algebra homomorphism and is also an isometry. Its image ˆ B in C (Δ) is therefore a subalgebra of C (Δ) which is preserved under conjugation. Moreover, since the Gelfand transform is an isometry it follows that ˆ B is a closed subset of C (Δ): for if x n B are such that ˆ x n f for some f C (Δ) then (ˆ x n ) n is Cauchy in C (Δ) and so, by isometricity, ( x n ) n is Cauchy in B and so is convergent, say to x and then by continuity ofˆit follows that f = ˆ x , and so f is in the image of the Gelfand transform. Finally, ˆ B separates points of Δ because if h 1 and h 2 are distinct elements of Δ, then, by definition of Δ, there must be some x B for which h 1 ( x ) 6 = h 2 ( x ), i.e. ˆ x ( h 1 ) 6 = ˆ h ( x 2 ). The Stone-Weierstrass theorem now implies that ˆ B = C (Δ) This proves the Gelfand-Naimark theorem: Theorem . For a complex commutative B * –algebra B , the Gelfand transform is an isometric isomorphism of B onto C (Δ), where Δ is the Gelfand spectrum of B . 1

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1. Let H be a complex vector space and F : H × H C a mapping such that F ( x, y ) is linear in x and conjugate-linear in y . (i) Prove the polarization formula F ( x, y ) = 1 4 F ( x + y, x + y ) - 1 4 F ( x - y, x - y )+ i 4 F ( x + iy, x + iy ) - i 4 F ( x - iy, x - iy ) (1) (ii) Use this to prove that sup x,y H, | x | , | y |≤ 1 | F ( x, y ) | ≤ 4 sup v H, | v |≤ 1 | F ( v, v ) | (2) [Hint: In (1), the first term equals F ( a, a ) with a = ( x + y ) / 2 and | a | ≤ 1 if | x | , | y | ≤ 1. Similarly for the other terms.] 2
(iii) If y H then show that sup v H, | v |≤ 1 | ( y, v ) | = | y | (iv) If T : H H is a linear map for which sup v H, | v |≤ 1 | ( Tv, v ) | < , show that T is a bounded linear map and | T | ≤ 4 sup v H, | v |≤ 1 | ( Tv, v ) | (Recall that the norm of T is | T | = sup x H, | x |≤ 1 | Tx | .) 3

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2. Let H
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