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

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!

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.

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).

• 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 $B(H)$ Feb 10 '15 at 21:08
• @YemonChoi: I agree. Anyway, I’m trying hard not to rely too much on knowledge about $C^{*}$-unitizations. Feb 12 '15 at 4:51