Jtdylktuy

The following statement is Exercise I.26 in K. Davidson's book C-Algebras by Example*:

Let $mathfrak{A}$ be a C*-algebra, $mathfrak{J}$ be an (two-sided
and closed) ideal of $mathfrak{A}$ and $A in mathfrak{A}$. Then
there exists $J in mathfrak{J}$ such that $|A - J| = |A + mathfrak{J}|$.

The book gives a hint of applying the Jordan decomposition (Corollary I.4.2 in the book) of the element $|A|-|A+mathfrak{J}|I$. But I don't see how to proceed from here. Any help would be appreciated. Thank you!

I'll be happy to be proven wrong and see a proof using the Jordan decomposition. But I'm guessing that there is a typo in the book and that the intent was to say "Corollary I.5.6".

Assume first that $A=C(T)$ for some compact Hausdorff $T$. Then $J={f: f|_E=0}$ for some compact $Esubset T$. Fix $fin C(T)$. As $T$ is compact, there exists $t_0in T$ with $|f|=|f(t_0)|$.

• If $t_0in E$, then $|f-g|geq|f(t_0)|=|f|$ for any $gin J$; then $|f+J|=|f|=|f-0|$.




• If $t_0notin E$, we may use Urysohn's Lemma to construct $gin C(T)$ with $g|_E=0$, $g(t_0)=|f|-|f+J|$, and $|g|leq g(t_0)$. Then $$|f-g|geq|f(t_0)-g(t_0)|=|f+J|,$$
  so $|f+J|=|f-g|$.

Now consider the general case. By Corollary I.5.6,
$$tag1
frac{C^*(a)}{C^*(a)cap J}simeq frac{C^*(a)+J}{J}.
$$
So, as the isomorphism is canonical and, of course, isometric,
$$
|a+J|=|a+C^*(a)cap J|,
$$
where the second quotient norm happens in an abelian C$^*$-algebra from the left in $(1)$. From the abelian case above, there exists $bin C^*(a)cap J$ with $|a-b|=|a+C^*(a)cap J|geq|a+J|$. Thus
$$
|a-b|=|a+J|.
$$