5
$\begingroup$

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?

$\endgroup$
0

1 Answer 1

4
$\begingroup$

$\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}$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.

$\endgroup$
5
  • $\begingroup$ @LSpice I see, thank you. Feel free to edit any of my answers for style! $\endgroup$
    – Nik Weaver
    Apr 22, 2022 at 0:21
  • $\begingroup$ OK, I have made the edit. Thanks! $\endgroup$
    – LSpice
    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
    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
    Apr 22, 2022 at 11:23
  • 1
    $\begingroup$ 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
    Apr 22, 2022 at 13:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.