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!
Here is a proof without using the hint.
It suffices to prove the following statement: If $\pi:A \to B$ is a surjective *-homomorphism between $C^*$-algebras and $y \in B$, then there exists $x \in A$ such that $\pi(x)=y$ and $\|x\|=\|y\|$. Also, we just need to prove the statement when $\|y\|=1$.
Special Case: $\underline{y=y^* \quad and \quad \|y\|=1}\\$
Since $\pi$ is surjective, there exists $a \in A$ such that $\pi(a)=y$. Replacing $a$ by $\frac{a+a^*}{2}$, we may assume in addition that $a=a^*$.
Let $f: \mathbb{R} \to \mathbb{R}$ be the continuous function defined by \begin{equation*} f(t)= \begin{cases} t &\text{ , if } \quad |t| \leq 1 \\ \frac{t}{|t|} &\text{ , if} \quad |t| >1 \end{cases} \end{equation*}
Let $x=f(a)$. Then $\|x\| \leq \|f\|_\infty \leq 1$ and $\pi(x)=\pi(f(a))=f(\pi(a))=f(y)=y$. The last equation holds because $\sigma(y) \subseteq [-1,1]$ and $f(t)=t $ for all $t \in [-1,1]$.
General Case:$\underline{\|y\|=1} \\$
Let \begin{gather} Y= \begin{pmatrix} 0 & y\\ y^* & 0 \\ \end{pmatrix}. \end{gather} Then $\|Y\|=\max\{\|y\|,\|y^*\| \}=1$ and $Y=Y^*$. Apply the special case to the surjective *-homomorphism $\pi_2:M_2( A) \to M_2(B)$ given by \begin{gather} \pi_2 \begin{pmatrix} x_{11} & x_{12}\\ x_{21} & x_{22} \\ \end{pmatrix} = \begin{pmatrix} \pi(x_{11}) & \pi(x_{12})\\ \pi(x_{21}) & \pi(x_{22}) \\ \end{pmatrix}. \end{gather} and $Y$, there exists $X \in M_2(A)$ such that $\|X\| \leq 1$ and $\pi_2(X)=Y$. Let $x \in A$ be the $(1,2)$-entry of X. Then $\|x\| \leq \|X\| \leq 1$ and $\pi(x)=y$.
Since $\pi$ is norm-decreasing, $1=\|y\| =\|\pi(x)|\leq \|x\|$. This completes the proof.