Basic Theorems Regarding the Group of Inv. Elements in an Algebra

# Basic Theorems Regarding the Group of Invertible Elements in an Algebra

 Lemma 1: Let $\mathfrak{A}$ be a normed algebra with unit. If $a, b \in \mathrm{Inv}(\mathfrak{A})$ and $\displaystyle{\| b - a \| \leq \frac{1}{2\| a^{-1} \|}}$ then $\| b^{-1} - a^{-1} \| \leq 2 \| a^{-1} \|^2 \| a - b \|$.
• Proof: By the reverse triangle inequality we have that $| \| b^{-1} \| - \| a^{-1} \| \leq \| b^{-1} - a^{-1} \|$. Using this and given that $\displaystyle{\| b - a \| \leq \frac{1}{2\| a^{-1} \|}}$, we have that:
(1)
\begin{align} \quad \| b^{-1} \| - \| a^{-1} \| \leq | \|b^{-1} \| - \| a^{-1} \|| \leq \| b^{-1} - a^{-1} \| &= \| b^{-1}aa^{-1} - b^{-1}ba^{-1} \| \\ &= \| b^{-1}(a - b)a^{-1} \| \\ & \leq \| b^{-1} \| \| a - b \| \| a^{-1} \| \\ & \leq \| b^{-1} \| \frac{1}{2 \| a^{-1} \|} \| a^{-1} \| \\ & \leq \frac{\| b^{-1} \|}{2} \end{align}
• Rearranging the above inequality gives us:
(2)
\begin{align} \frac{\| b^{-1} \|}{2} &\leq \| a^{-1} \| \\ \| b^{-1} \| &\leq 2 \| a^{-1} \| \quad (*) \end{align}
• Therefore, using $(*)$ and and we get:
(3)
\begin{align} \quad \| b^{-1} - a^{-1} \| &=\| b^{-1}aa^{-1} - b^{-1}ba^{-1} \| \\ &= \| b^{-1}(a - b)a^{-1} \| \\ & \leq \| b^{-1} \| \| a - b \| \| a^{-1} \| \\ & \leq 2 \| a^{-1} \| \| a - b \| \| a^{-1} \| \\ & \leq 2 \| a^{-1} \|^2 \| a - b \| \quad \blacksquare \end{align}
 Corollary 2: Let $\mathfrak{A}$ be a normed algebra with unit. Equip $\mathfrak{A}$ with the topology induced by its norm. Then the function $f : \mathrm{Inv}(\mathfrak{A}) \to \mathrm{Inv}(\mathfrak{A})$ defined for each $a \in \mathfrak{A}$ by $f(a) = a^{-1}$ is a homeomorphism of $\mathrm{Inv}(\mathfrak{A})$ onto $\mathrm{Inv}(\mathfrak{A})$.

Recall that if $X$ is a topological space then a homeomorphism from $X$ onto $X$ is a function $f : X \to X$ that is bijective and such that $f$, $f^{-1}$ are continuous.

• Proof: We need to show that $f$ is bijective, $f$ is continuous, and $f^{-1}$ is continuous.
• 1. Showing that $f$ is bijective: It it clear that $f$ is bijective. To quickly verify this, let $a, b \in \mathrm{Inv}(\mathfrak{A})$ and suppose that $f(a) = f(b)$. Then $a^{-1} = b^{-1}$. So $aa^{-1} = ab^{-1}$ translates to $1 = ab^{-1}$. So $b^{-1}$ is the inverse of $a$. By the uniqueness of inverses, we have that $a^{-1} = b^{-1}$ and thus $f$ ]] is injective. On the other hand, let [[$a \in X$. Then $a^{-1} \in \mathrm{Inv}(\mathfrak{A})$ is such that $f(a^{-1}) = (a^{-1})^{-1} = a$. So $f$ is surjective.
• 2. Showing that $f$ is continuous: We will now show that $f$ is continuous on $\mathrm{Inv}(\mathfrak{A})$ by showing that $f$ is continuous at every $a \in \mathrm{Inv}(\mathfrak{A})$. Let $\epsilon > 0$ be given and $a \in \mathrm{Inv}(\mathfrak{A})$. Let:
(4)
\begin{align} \quad \delta = \min \left \{ \frac{1}{2\| a^{-1} \|}, \frac{\epsilon}{2 \| a^{-1} \|^2} \right \} \end{align}
• Then if $\| b - a \| \leq \delta$ we have that both $\displaystyle{\| b - a \| \leq \frac{1}{2 \| a^{-1} \|}}$ and $\displaystyle{\| b - a \| \leq \frac{\epsilon}{2 \| a^{-1} \|^2}}$. With this first inequality, Proposition 1 tells us that
(5)
\begin{align} \quad \| f(b) - f(a) \| = \| b^{-1} - a^{-1} \| \leq 2 \| a^{-1} \|^2 \| a - b \| \end{align}
• And the second inequality gives us that:
(6)
\begin{align} \quad \| f(b) - f(a) \| \leq 2 \| a^{-1} \|^2 \| a - b \| \leq 2 \| a^{-1} \|^2 \frac{\epsilon}{2 \| a^{-1} \|^2} = \epsilon \end{align}
• Therefore $f$ is continuous at $a \in \mathrm{Inv}(\mathfrak{A})$ and hence continuous on $\mathrm{Inv}(\mathfrak{A})$.
• 3. Showing that $f^{-1}$ is continuous: Observe that $f(f(a)) = a$ for every $a \in \mathrm{Inv}(\mathfrak{A})$, that is, $f$ is an involution, i.e., $f = f^{-1}$. So $f$ is continuous by (2).
• So we conclude that $f$ is a homeomorphism of $\mathrm{Inv}(\mathfrak{A})$ onto $\mathrm{Inv}(\mathfrak{A})$. $\blacksquare$