Given an ideal $I\subset A$ and an element $a\in A$, is it true that $\sup\{\|ab\|:b\in I,\ \|b\|\leq 1\}=\|a\|$? Let $I$ be a 2-sided closed ideal in a C*-algebra $A$. Given $a\in A$, is it true that $$\sup_{\substack{b\in I\\ \|b\|\leq 1}}\|ab\|=\|a\|?$$ Or do we need assumptions on $I$? Note that the left-hand side of this equation is precisely the operator norm of the operator $L_{a}\colon I\to I$ given by $L_{a}b:=ab$.
Using that $K(H)$ contains the finite-rank operators I managed to prove this claim for $A=B(H)$ and $I=K(H)$.
 A: You have to require that $I\ne 0$, otherwise this trivially fails. But even then, the claim is not true.
For instance, let $A$ be any (non-zero) $C^*$-algebra. Then $I= A \oplus 0$ is an ideal in $A \oplus A$. However, if $a \in A$ is a non-zero element, then
$$\sup_{(b,0)}\|(0,a)(b,0)\|=0 \ne \|a\|.$$
You can then ask what goes wrong? Well, there is degeneracy in the above counterexample, in the sense that
$$(0,a)I = 0$$
yet $(0,a) \ne 0$. In other words, the ideal $I$ is NOT essential in $A\oplus A$. More generally, any non-essential ideal in $A$ will yield a counterexample.
On the other hand, if the ideal $I$ is actually essential in $A$, then we are in luck:
Theorem: Let $I$ be a closed two-sided ideal in $A$ with the property that $xI = 0$ implies $x =0$ for every $x \in A$ (such ideals are called essential). Then
$$\sup_{b \in I, \|b\|\le 1}\|ab\| =\|a\|.$$
Proof: By the universal property of the multiplier algebra, we may assume that $I\subseteq A \subseteq M(I)$. Hence, the claim reduces to proving the following lemma. $\quad \square$
Lemma: If $a \in M(B)$ where $B$ is a $C^*$-algebra, then
$$\|a\|= \sup_{\|b\|\le 1, b \in B}\|ab\|.$$
Proof: This is a well-known fact about multiplier algebras, and is for example immediately obvious when we identify $M(B) = \mathcal{L}_B(B)$, the adjointable operators $B\to B$ where we view $B$ as a Hilbert module over itself. $\quad \square$
In summary, your claim holds if and only if $I$ is an essential ideal of $A$.
