Let $A$ be a $C^*$-algebra without unit. Its unitisation $A^+$ carries the $C^*$-norm

$$\|(a,\lambda)\|_{A^+} = \sup \{\|ax+\lambda x\|_A \;\big|\; \|x\|_A = 1 \},$$

which is the operator norm of $L_a + \lambda\,\text{id}_A$ in $\mathscr{L}(A)$, with $L_a$ denoting left multiplication by $a$.

My question

Does a there exist an $A$ without unit together with concrete $a\in A$ and $\lambda\in\mathbb{C}$ satisfying $$\|(a,\lambda)\|_{A^+}\: < \:|\lambda |\; ?$$

If there's no such example, then let's put it the other way round: Can one prove that the lower bound $$ |\lambda| \le \|(a,\lambda)\|_{A^+}$$ holds true $\forall\,(a,\lambda)\in A^+\:$?

If that is the case the title I chose for my question ought to be rectified to

"Trivial lower bound for the $C^*$-unitisation norm" I guess ...

Possibly, this latter estimate may be deduced from the canonical short exact sequence
$$A\hookrightarrow A^+\twoheadrightarrow\mathbb{C}$$
of $C^*$-algebras $\,-\,$ *but that's not clear to me*.

augmentationhomomorphism? As I found in the meantime, there's the similar but extensive post. $\endgroup$