# 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.

Definition Let ${B}$ be a banach space, ${\mathcal{L}(B)}$ be the space of bounded linear operator from ${B \rightarrow B}$. If ${T \in \mathcal{L}(B)}$, then say a complex number ${\lambda}$ is in the resolvent set ${\rho(T)}$ of ${T}$ if ${\lambda I - T}$ has a bounded inverse. Then we define the resolvent of ${T}$ at ${\lambda}$ to be ${R_\lambda(T) = (\lambda I -T)^{-1}}$. If ${\lambda \not\in \rho(T)}$, then ${\lambda}$ is said to be in the spectrum ${\sigma(T)}$ of T.