For Normed Alg. with Unit, the Unit Can be Assumed to Have Norm 1

For Normed Algebras with Unit, the Unit Can be Assumed to Have Norm 1

Recall that if $\mathfrak{A}$ is an algebra then $\mathfrak{A}$ is said to have a unit (or be unital) if there exists an element $e \in \mathfrak{A}$ such that $ea = ae = e$ for every $a \in \mathfrak{A}$.

If $\mathfrak{A}$ is a normed algebra with unit, then since $e = ee$, we have that:

(1)

\begin{align} \quad \| e \| = \| ee \| \leq \| e \| \| e \| \end{align}

Since $e \neq 0$, the above inequality implies that $\| e \| \geq 1$. For practical purposes it would be nice if $\| e \| = 1$. However, this is not true in general. Fortunately, $\mathfrak{A}$ can always be equipped with a new equivalent norm for which $\| e \| = 1$.

Proposition 1: Let $(\mathfrak{A}, \| \cdot \|)$ be a normed algebra with unit $e$. Then there exists an algebra norm $\| \cdot \|_*$ on $\mathfrak{A}$ that is equivalent to $\| \cdot \|$ and such that $\| e \|_* = 1$.

• Proof: For each $a \in \mathfrak{A}$ let $T_a : \mathfrak{A} \to \mathfrak{A}$ be defined for all $b \in \mathfrak{A}$ by:

(2)

\begin{align} \quad T_a(b) = ab \end{align}

• Note that for each $a \in \mathfrak{A}$, $T_a$ is linear since for all $b_1, b_2 \in \mathfrak{A}$ and for all $\alpha \in \mathbf{F}$ we have that:

(3)

\begin{align} \quad T_a(b_1 + b_2) = a(b_1 + b_2) = ab_1 + ab_2 = T_a(b_1) + T_a(b_2) \end{align}

(4)

\begin{align} \quad T_a(\alpha b_1) = a(\alpha b_1) = \alpha (ab_1) = \alpha T_a(b_1) \end{align}

• Furthermore, for each $a \in \mathfrak{A}$ we have that $T_a$ is bounded with $\| T_a \| \leq \| a \|$ since:

(5)

\begin{align} \quad \| T_a(b) \| = \| ab \| \leq \| a \| \| b \|, \quad \forall b \in \mathfrak{A} \end{align}

• Now note that the map from $\mathfrak{A}$ to $\mathrm{BL}(\mathfrak{A}, \mathfrak{A})$ defined by $a \to T_a$ is injective. Indeed, if $a_1, a_2 \in \mathfrak{A}$ are such that $T_{a_1}(b) = T_{a_2}(b)$ for all $b \in \mathfrak{A} $ ]] then [[$ a_1b = a_2b$ for all $b \in \mathfrak{A}$. So $a_1b - a_2b = (a_1 - a_2)b = 0$ for all $b \in \mathfrak{A}$. In particular, since $\mathfrak{A}$ is an algebra with unit, we have that $(a_1 - a_2)e = 0$. So $a_1 - a_2 = 0$, i.e., $a_1 = a_2$.

• For each $a \in \mathfrak{A}$ let:

(6)

\begin{align} \quad \| a \|_* := \| T_a \| \end{align}

• Note that $\| \cdot \|_*$ is well-defined by the injectivity of the map $a \to T_a$. Moreover, $\| a \|_* \leq \| a \|$. For the reverse inequality, observe that:

(7)

\begin{align} \quad \| a \|_* = \| T_a \| = \sup_{\| b \| \leq 1} \{ \| T_a(b) \| \} = \sup_{\| b \| \leq 1} \{ \| ab \| \} \geq \left \| a \frac{e}{\| e \|} \right \| = \frac{\| a \|}{\| e \|} \end{align}

• Let $c = \frac{1}{\| e \|}$ and let $C = 1$. Then for all $a \in \mathfrak{A}$ we have that:

(8)

\begin{align} \quad c \| a \| \leq \| a \|_* \leq C \| a \| \end{align}

• So $\| \cdot \|_*$ is norm equivalent to $\| \cdot \|$. Furthermore, we have that:

(9)

\begin{align} \quad \| e \|_* = \| T_e \| = \sup_{\| b \| \leq 1} \{ \| T_e(b) \| \} = \sup_{\| b \| \leq 1} \{ \| eb \| \} = \sup_{\| b \| \leq 1} \{ \| b \| \} = 1 \end{align}

• Lastly, we see that $\| \cdot \|_*$ is an algebra norm since for all $a, b \in \mathfrak{A}$ we have that:

(10)

\begin{align} \quad \| ab \|_* = \| T_{ab} \| = \| T_a \circ T_b \| \leq \| T_a \| \| T_b \| = \| a \|_* \| b \|_* \end{align}

• So $\| \cdot \|_*$ is an algebra norm that is equivalent to $\| \cdot \|$ and such that $\| e \|_* = 1$. $\blacksquare$