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.

  • 1
    $\begingroup$ Unless I misunderstood the question, there is no such example: $|\lambda|\leq\|(a,\lambda)\|$ holds because the map $A^+\to\mathbb{C}$ is a $*$-homomorphism, and every $*$-homomorphism is norm-nonincreasing. $\endgroup$ Apr 28, 2016 at 15:12
  • $\begingroup$ Doesn't this also follow from Hahn-Banach and the fact that in any unital algebra we assume the identity element has norm $1$? $\endgroup$
    – Yemon Choi
    Apr 28, 2016 at 16:13
  • $\begingroup$ @TobiasFritz Thank you, you did not misunderstand it: Your concise reply settles my question. Would (or could) one denote the *-hom $A^+\twoheadrightarrow\mathbb{C}$ as an augmentation homomorphism? As I found in the meantime, there's the similar but extensive post. $\endgroup$
    – Hanno
    Apr 29, 2016 at 10:36


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