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Let $ A $ be a non-unital $ C^{*} $-algebra. Is there an ‘elementary’ way to prove, for all $ (a,\lambda) \in A \times \mathbb{C} $, the inequality $$ |\lambda| \leq \sup_{b \in A, ~ \| b \| \leq 1} \| a b + \lambda \cdot b \|_{A}? $$ I have a proof of this, but it is simply overkill.

Proof

Firstly, define a linear functional $ \phi $ on the unitization $ A^{\sim} $ of $ A $ by $$ \forall (a,\lambda) \in A \times \mathbb{C}: \quad \phi(a,\lambda) \stackrel{\text{df}}{=} \lambda. $$ As all positive elements of $ A^{\sim} $ must have a non-negative scalar component, it follows that $ \phi $ is a positive linear functional, in which case, by a well-known result about positive linear functionals on $ C^{*} $-algebras, $ \phi $ is automatically continuous and $ \| \phi \| = \phi(0,1) = 1 $. Hence, for all $ (a,\lambda) \in A \times \mathbb{C} $, we have $$ |\lambda| = |\phi(a,\lambda)| \leq \| (a,\lambda) \|_{A^{\sim}} \stackrel{\text{df}}{=} \sup_{b \in A, ~ \| b \| \leq 1} \| a b + \lambda \cdot b \|_{A}. \quad \blacksquare $$


I believe that one can avoid theorems about positive linear functionals on $ C^{*} $-algebras and invent a proof that is mostly Banach $ * $-algebraic in nature, with the finishing blow provided by the $ C^{*} $-identity. However, I do not see the light.

I hope that my request is not too vague. Thanks!

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The following argument seems easier, but there might be a still more fundamental one.

Notice that $ \phi: A^{\sim} \to \mathbb{C} $ above is also a $ C^{*} $-algebraic homomorphism. As $ C^{*} $-algebraic homomorphisms are automatically contractive (which is a consequence of a not-too-difficult spectrum argument), we have $$ \forall (a,\lambda) \in A \times \mathbb{C}: \quad |\lambda| = |\phi(a,\lambda)| \leq \| (a,\lambda) \|_{A^{\sim}} = \sup_{b \in A, ~ \| b \|_{A} \leq 1} \| a b + \lambda \cdot b \|_{A}. $$ No hard facts about positive linear functionals on $ C^{*} $-algebras were used.


Additional Information:

The inequality is valid if we only assume that $ A $ is a Banach algebra that satisfies the following conditions:

  1. $ A $ has no two-sided identity.
  2. $ A $ has a right approximate identity (r.a.i.) norm-bounded by $ 1 $, which we denote by the net $ (e_{i})_{i \in I} $.

Let $ A $ be a Banach algebra satisfying Conditions (1) and (2).

Let $ \mathbb{L}(A) $ denote the Banach algebra of all bounded homomorphisms from $ A $ to itself, where multiplication is defined by composition and the norm is simply the operator norm.

The only leap (not too great, I hope!) of imagination required is to notice that for $ (a,\lambda) \in A \times \mathbb{C} $, $$ \sup_{b \in A, ~ \| b \|_{A} \leq 1} \| a b + \lambda \cdot b \|_{A} $$ represents the operator norm of $ L_{a} + \lambda \cdot \text{id}_{A} \in \mathbb{L}(A) $, where $ L_{a} $ denotes left-multiplication by $ a $.

Define

  • $ A^{\sim} \stackrel{\text{df}}{=} \{ L_{a} + \lambda \cdot \text{Id}_{A} \mid (a,\lambda) \in A \times \mathbb{C} \} $,
  • $ L_{A} \stackrel{\text{df}}{=} \{ L_{a} \mid a \in A \} $.

Then clearly $ A^{\sim} $ is a sub-algebra of $ \mathbb{L}(A) $.

Claim 1: $ L_{A} $ is complete w.r.t. the operator norm on $ \mathbb{L}(A) $.

Proof of Claim 1

Let $ a \in A $. We already know that $ \| L_{a} \| \leq \| a \|_{A} $. However, $ \| e_{i} \|_{A} \leq 1 $ for all $ i \in I $ and $$ \lim_{i \in I} \| {L_{a}}(e_{i}) \|_{A} = \lim_{i \in I} \| a e_{i} \|_{A} = \| a \|_{A}, $$ so we actually have $ \| L_{a} \| = \| a \|_{A} $. The map $ x \mapsto L_{x} $ is therefore a bijective isometry from $ A $ to $ L_{A} $, so $ L_{A} $ is complete w.r.t. the operator norm on $ \mathbb{L}(A) $. $ \quad \blacksquare $

Define a (not a priori continuous) linear functional $ \phi $ on $ A^{\sim} $ by $$ \phi(L_{a} + \lambda \cdot \text{Id}_{A}) \stackrel{\text{df}}{=} \lambda. $$ To prove that $ \phi $ is well-defined, we must show that given $ (a,\lambda) \in A \times \mathbb{C} $, if $ L_{a} + \lambda \cdot \text{Id}_{A} = 0_{\mathbb{L}(A)} $, then it must follow that $ \lambda = 0 $.

Assume the contrary. If we define $ e \stackrel{\text{df}}{=} - \dfrac{1}{\lambda} \cdot a $, then $ L_{e} = \text{Id}_{A} $, which means that $ e $ is a left identity of $ A $. Now, for all $ x \in A $, we have \begin{align} x e - x & = \lim_{i \in I} ~ (x e - x) e_{i} \qquad (\text{As $ (e_{i})_{i \in I} $ is an r.a.i..}) \\ & = \lim_{i \in I} ~ (x e e_{i} - x e_{i}) \\ & = \lim_{i \in I} ~ (x e_{i} - x e_{i}) \qquad (\text{As $ e $ is a left identity.}) \\ & = \lim_{i \in I} ~ 0_{A} \\ & = 0_{A}. \end{align} Hence, $ e $ is a right identity of $ A $ as well. This contradicts our assumption that $ A $ has no two-sided identity, so we indeed have $ \lambda = 0 $.

Claim 2: $ A^{\sim} $ is complete w.r.t. the operator norm on $ \mathbb{L}(A) $.

Proof of Claim 2

As $ L_{A} $ is a complete linear subspace of $ A^{\sim} $, it is automatically closed. The quotient vector space $ A^{\sim} / L_{A} $ can thus be given a norm, which then has to be complete because $ \ker(\phi) = L_{A} $ and so $$ A^{\sim} / L_{A} = A^{\sim} / \ker(\phi) \cong \mathbb{C}. $$ Exploiting the result that a normed vector space is complete if its quotient by a complete linear subspace is complete, we conclude that $ A^{\sim} $ is complete w.r.t. the operator norm on $ \mathbb{L}(A) $. $ \quad \blacksquare $

We now see that $ A^{\sim} $ is a unital Banach algebra w.r.t. the operator norm on $ \mathbb{L}(A) $. As $ \phi $ is a multiplicative linear functional on $ A^{\sim} $, it follows that $ \phi $ must be bounded with norm $ \leq 1 $ (this is an easy fact whose proof can be found in Rudin’s Real and Complex Analysis).

The inequality is therefore established.


Notes:

If Condition (1) is violated, i.e., $ A $ has a two-sided identity $ e $, then the inequality is false because then $$ 1 \nleq \sup_{b \in A, ~ \| b \|_{A} \leq 1} \| e b - 1 \cdot b \|_{A} = 0. $$ I do not know what happens if Condition (2) is violated, however.

If $ A $ is a $ C^{*} $-algebra, then we only need Condition (1) because Condition (2) automatically holds. This can be misleading, however, because it is not at all obvious that non-unital $ C^{*} $-algebras should have an r.a.i. norm-bounded by $ 1 $. Fortunately, it turns out that in proving Claims 1 and 2, the $ C^{*} $-identity is all that is needed as it takes over the pivotal role played by Condition (2).

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  • $\begingroup$ I think the hard part of the problem is somehow bound up with the following: it is not obvious that the $C^*$-unitization exists, if you don't know of an embedding into $B(H)$ $\endgroup$ – Yemon Choi Feb 10 '15 at 21:08
  • $\begingroup$ @YemonChoi: I agree. Anyway, I’m trying hard not to rely too much on knowledge about $ C^{*} $-unitizations. $\endgroup$ – Transcendental Feb 12 '15 at 4:51

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