Solutions to some exercises in Murphy’s C*-algebra and Operator Theory

Proof. (a)

(b)Let , then is a closed ideal in . By , . For all , so , hence and thus . Therefore, .

(c) is a unitary is equal to is a isomorphism in the sense is a surjection which preserves inner product. Denote by , then ;

thus is a unitary.

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Proof. For all positive linear functional on , is a positive linear functional on and thus bounded, i.e. for some positive constant .

Every bounded linear functional on is sum of 4 positive linear functionals on by Jordan Decomposition, thus for all for some positive constant . Hence for some positive constant by the Principle of Uniform Boundedness, i.e. is bounded. ◻

Proof. (a) If is invertible, then , thus is strictly positive.

Conversely, if is strictly positive, then . Since the set of all invertible elements is open, . Hence there exists some such that is invertible, thus is invertible.

(b)If is strictly positive in , that is , then there exists a sequence in such that Hence Therefore, and .

Conversely, if , then for all then there exists a sequence in such that , hence Therefore, every rank-one projection belongs to . Since all the rank-one projections is the total subset of , .

(c)Suppose is a positive functional such that , then so Hence for all , Therefore, for . ◻

Proof. Suppose is a basis for , such taht for all . and a bounded net in converges strongly to an operator . For every , there exists a finite subset of such that and there exists such that Therefore, for some constant . Hence . ◻

Proof. (1) Suppose and , then and , so .Hence that is,

(2)If and have a same dimension less than , choose orthonormal bases and

for and , respectively( may be infty). Let and let then (because they coincide on and ) and .

Conversely, if , then and for some , so

(3)Since , by (1). Because , , i.e.

(4)Since is an abelian -algebra with the identity , there is a -isomorphism , where is the compact set of all non-zero algebra homomorphism from to .

Suppose , and , then thus so .

(5)If is finite-demensional, then , so is finite, i.e., is finite.

If is infinite-demensional, suppose is an orthonormal basis for , and set Then Thus Let , then and , but , that is is infinite, so is the Hilbert space. ◻

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