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Let $A$ be a non-unital C*-algebra, and let $\pi: A\to B(A)$ defined by $\pi(a)(x)=ax$ be the left representation of A. Is the following inequality correct?

$$\lVert I+ \pi(a) \rVert\ge 1$$ for all $a \in A$. ($I$ is the identity operator and $\lVert\cdot\rVert$ is the operator norm.)

If the inequality is correct, is it a known inequality?

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$\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}$This is true. Let $a \in A$. Then $|a|$ also belongs to $A$, and since $A$ is nonunital a little bit of functional calculus shows that there exists $x \in A$ with $\norm x = 1$ and $\norm{\abs a x}$ arbitrarily small. (I guess you have to consider the cases where $0$ is or is not isolated in the spectrum of $\abs a$ separately, unless there's an easier argument I'm missing.) Supposing $A$ is sitting inside $B(H)$, we can write $a = u\abs a$ where $u \in B(H)$ is a partial isometry, and then calculate $$\norm{(I + a)x} = \norm{(I + u\abs a)x} \geq \norm x - \norm{u\abs a x} = 1 - \norm{\abs a x}.$$ Since $\norm{\abs a x}$ can be made arbitrarily small, this shows that the norm of $I + a$ in $B(A)$ is at least $1$.

I'm sure this is something many people know, but I don't have any reference.

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  • $\begingroup$ @LSpice I see, thank you. Feel free to edit any of my answers for style! $\endgroup$
    – Nik Weaver
    Commented Apr 22, 2022 at 0:21
  • $\begingroup$ OK, I have made the edit. Thanks! $\endgroup$
    – LSpice
    Commented Apr 22, 2022 at 0:43
  • $\begingroup$ Alternatively, $\tilde A := $span$\{I\} \cup \pi(A)$ is a $C^*$-algebra (the unitisation of $A$), see for instance the proof of Prop. 1.1.3 in G. K. Pedersen's book on $C^*$-algebras. The map $\tilde A \to \mathbb C$ which annihilates $\pi(A)$ is a $\ast$-homomorphism, and such are always contractive (a fundamental consequence of the $C^*$-identity). $\endgroup$
    – Jamie Gabe
    Commented Apr 22, 2022 at 7:58
  • $\begingroup$ @JamieGabe ah, I thought it would be a C*-algebra but I didn't remember why. I guess it's been a while since I read Pedersen ... $\endgroup$
    – Nik Weaver
    Commented Apr 22, 2022 at 11:23
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    $\begingroup$ An alternative argument, very much along @Jamie's comment, is that the operator norm is equal to the norm on the multiplier algebra $M(A)$ and, since $A$ is a proper ideal in $M(A)$, and hence contains no invertible elements, the distance from 1 to $A$ must be 1. $\endgroup$
    – Ruy
    Commented Apr 22, 2022 at 13:52

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