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!



Proving a certain $ C^{*} $-algebraic inequality
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).


