Quotient norm of self-adjoint element can be achieved

Theorem Let {I} be a closed ideal of a C* algebra {\mathcal{A}}, then for any self-adjoint element {a \in \mathcal{A}}, there exists an {i \in I} such that

\displaystyle \|a-i\| = \inf \{ \|a -y\|: y \in I \}.

Proof: Since {I} is a closed ideal, {\mathcal{A}/I} is also a C* algebra and one can define a natural projection homomorphism {\pi : \mathcal{A} \rightarrow \mathcal{A}/I}, with {\| \pi(a)\| = \inf \{ \|a -y\|: y \in I \}}. Fix an element {a \in \mathcal{A}}, define a continuous function {f} as follow:

\displaystyle f(x) = \begin{cases} \|\pi(a)\|, &x > \|\pi(a)\| \\ x, & -\|\pi(a)\| \le x \le \|\pi(a)\|\\ -\|\pi(a)\|, &x < -\|\pi(a)\| \end{cases}

Let {g(x) := x - f(x)}, then {g(x)} is identical {0} on {[-\|\pi(a)\|, \|\pi(a)\|]}. By continuous functional calculus, {g(a) \in \mathcal{A}}. And since {g} can be approximated by polynomials, we know {\pi(g(a)) = g(\pi(a))}. The spectral radius {\nu(\pi(a)) \le \|\pi(a)\|}, so we know {g} is identical {0} on the spectrum of {\pi(a)}, which implies {\pi(g(a)) = g(\pi(a)) = 0}. Therefore {g(a)} is in the kernel of {\pi}, and hence in the ideal {I}. Since {\|a - g(a)\| = \|f(a)\|} and {\|f(a)\| \le \|\pi(a)\|} by the Gelfand Naimark isomorphism theorem, we know {\|a - g(a)\| \le \| \pi(a)\|} and hence {\|a - g(a)\| = \| \pi(a)\|}. So {g(a)} is the {i} we are looking for. \Box

Note: this is a homework exercise given by Dr. Carlen.

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