The Unitization of a Normed Algebra

# The Unitization of a Normed Algebra

 Definition: Let $\mathfrak{A}$ be a normed algebra over $\mathbf{F}$. The Unitization of $\mathfrak{A}$** denoted by $\mathfrak{A} + \mathbf{F}$ is the normed algebra with unit on the set $\mathfrak{A} \times \mathbf{F}$ with the following operations: 1. The operation of addition defined for all $(a, \alpha), (b, \beta) \in \mathfrak{A} + \mathbf{F}$ by $(a, \alpha) + (b, \beta) = (a + b, \alpha + \beta)$. 2. The operation of scalar multiplication defined for all $(a, \alpha) \in \mathfrak{A} + \mathbf{F}$ and all $k \in \mathbf{F}$ by $k(a, \alpha) = (ka, k\alpha)$. 3. The operation of multiplication defined for all $(a, \alpha), (b, \beta) \in \mathfrak{A} + \mathbf{F}$ by $(a, \alpha)(b, \beta) = (ab + \alpha b + \beta a, \alpha \beta)$. And with norm $\| (a, \alpha) \| = \| a \| + |\alpha|$. The unit of $\mathfrak{A} + \mathbf{F}$ is the point $(0, 1)$.

The multiplication in $\mathfrak{A} + \mathbf{F}$ can be remembered as follows. For $(a, \alpha), (b, \beta) \in \mathfrak{A} + \mathbf{F}$, multiply out $(a + \alpha)(b + \beta)$ to get $ab + \beta a + \alpha b + \alpha \beta$. The first coordinate of the multiplication of $(a, \alpha)(b, \beta)$ with be $ab + \beta a + \alpha b$ and the second coordinate will be $\alpha \beta$.

The following proposition verifies that the unitization of any normed algebra is indeed a normed algebra with unit.

 Proposition 1: Let $\mathfrak{A}$ be a normed algebra over $\mathbf{F}$. Then the unitization of $\mathfrak{A}$ is also a normed algebra over $\mathbf{F}$ with unit $(0, 1)$.
• Proof: Certainly $\mathfrak{A} + \mathbf{F}$ with the operations of addition and scalar multiplication is a linear space. All that remains to show is that (A) $\mathfrak{A} + \mathbf{F}$ with the operation of multiplication defined above is an algebra and that (B) The norm defined above is an algebra norm on $\mathfrak{A}+ \mathbf{F}$.
• (A) Showing that $\mathfrak{A} + \mathbf{F}$ is an algebra:
• 1. Showing that $(a, \alpha)[(b, \beta)(c,\gamma)] = [(a, \alpha)(b, \beta)](c, \gamma)$$: Let$(a, \alpha), (b, \beta), (c, \gamma) \in \mathfrak{A} + \mathbf{F}. Then: (1) \begin{align} \quad (a, \alpha)[(b, \beta)(c, \gamma)] &= (a, \alpha)[(bc + \beta c + \gamma b, \beta \gamma) \\ &= (a[bc + \beta c + \gamma b] + \alpha [bc + \beta c + \gamma b] + \beta \gamma a, \alpha [\beta \gamma]) \end{align} • And also: (2) \begin{align} \quad [(a, \alpha)(b, \beta)](c, \gamma) &= (ab + \alpha b + \beta a, \alpha \beta)(c, \gamma) \\ &= ([ab + \alpha b + \beta a]c + \alpha \beta c + \gamma [ab + \alpha b + \beta a], [\alpha \beta]\gamma) \end{align} • Comparing the two equations above and we see that they are equal. Thus(a, \alpha)[(b, \beta)(c,\gamma)] = [(a, \alpha)(b, \beta)](c, \gamma)$. • 2. Showing that$(a, \alpha)[(b, \beta) + (c, \gamma)] = (a, \alpha)(b, \beta) + (a, \alpha)(c, \gamma)$: Let$(a, \alpha), (b, \beta), (c, \gamma) \in \mathfrak{A} + \mathbf{F}. Then: (3) \begin{align} \quad (a, \alpha)[(b, \beta) + (c, \gamma)] &= (a, \alpha)(b + c, \beta + \gamma) \\ &=(a[b + c] + \alpha[b + c] + [\beta + \gamma]a, \alpha[\beta + \gamma]) \\ \end{align} • And: (4) \begin{align} \quad (a, \alpha)(b, \beta) + (a, \alpha)(c, \gamma) = (ab + \alpha b + \beta a, \alpha \beta) + (ac + \alpha c + \gamma a, \alpha \gamma) \\ &= (a[b + c] + \alpha[b + c] + [\beta + \gamma]a, \alpha[\beta + \gamma]) \end{align} • Thus(a, \alpha)[(b, \beta) + (c, \gamma)] = (a, \alpha)(b, \beta) + (a, \alpha)(c, \gamma)$. • 3. Showing that$[k(a, \alpha)](b, \beta) = k[(a, \alpha)(b, \beta)] = (a, \alpha)[k(b, \beta)]$: Let$(a, \alpha), (b, \beta) \in \mathfrak{A} + \mathbf{F}$and let$k \in \mathbf{F}. Then: (5) \begin{align} \quad [k(a, \alpha)](b, \beta) = (ka, k\alpha)(b, \beta) = (kab + k\alpha b + k \beta a, k \alpha \beta) = k(ab + \alpha b + \beta a, \alpha \beta) = k[(a, \alpha)(b, \beta)] \end{align} • And also: (6) \begin{align} \quad (a, \alpha)[k(b, \beta)] = (a, \alpha)(kb, k\beta) = (kab + k\beta a + k \alpha b, k \alpha \beta) = k (ab + \alpha b + \beta a, \alpha \beta) = k[(a, \alpha)(b, \beta)] \end{align} • Thus[k(a, \alpha)](b, \beta) = k[(a, \alpha)(b, \beta)] = (a, \alpha)[k(b, \beta)]$. • So$\mathfrak{A} + \mathbf{F}$is an algebra. • (B) Showing that the norm defined above is an algebra norm on$\mathfrak{A} + \mathbf{F}$: • 1. Showing that$\| (a, \alpha) \| = 0$if and only if$(a, \alpha) = (0, 0)$: Suppose that$\| (a, \alpha) \| = 0$. Then$\| a \| + |\alpha| = 0$which implies that$\| a \| = 0$and$|\alpha| = 0$, i.e.,$a = 0$and$\alpha = 0$. So$(a, \alpha) = (0, 0)$. On the other hand, if$(a, \alpha) = (0, 0)$then$\| (a, \alpha) \| = \| 0 \| + |0| = 0$. • 2. Showing that$\| k(a, \alpha) \| = |k| \| (a, \alpha) \|$: Let$(a, \alpha) \in \mathfrak{A} + \mathbf{F}$and let$k \in \mathbf{F}. Then: (7) \begin{align} \quad \| k(a, \alpha) \| = \| (ka, k\alpha) \| = \| ka \| + |k \alpha| = |k| \| a \| + |k||\alpha| = |k|[\| a \| + |\alpha|] = |k| \| (a, \alpha) \| \end{align} • 3. Showing that\| (a, \alpha) + (b, \beta) \| \leq \| (a, \alpha) \| + \| (b, \beta) \|$: Let$(a, \alpha), (b, \beta) \in \mathfrak{A} + \mathbf{F}. Then: (8) \begin{align} \quad \| (a, \alpha) + (b, \beta) \| = \| (a + b \alpha + \beta) \| = \| a + b \| + |\alpha + \beta| \leq \| a \| + \| b \| + |\alpha| + |\beta| = [\| a \| + |\alpha|] + [\| b \| + |\beta|] = \| (a, \alpha) \| + \| (b, \beta) \| \end{align} • 4. Showing that\|(a, \alpha)(b, \beta) \| \leq \| (a, \alpha) \| \| (b, \beta) \|$: Let$(a, \alpha), (b, \beta) \in \mathfrak{A} + \mathbf{F}. Then: (9) \begin{align} \quad \| (a, \alpha)(b, \beta) \| &= \| (ab + \alpha b + \beta a, \alpha \beta) \| \\ &= \| ab + \alpha b + \beta a \| + |\alpha \beta| \\ & \leq \| a \| \| b \| + |\alpha| \| b \| + |\beta| \| a \| + |\alpha||\beta| = [\| a \| + |\alpha|][\| b \| + |\beta|] = \| (a, \alpha) \| \| (b, \beta) \| \end{align} • So indeed\mathfrak{A} + \mathbf{F}$is an normed algebra with unit$(0, 1)\blacksquare$ Proposition 2: Let$\mathfrak{A}$be a normed algebra over$\mathbf{F}$and let$\mathfrak{A} + \mathbf{F}$be the unitization of$\mathbf{F}$. Then the subset$\{ (a, 0) : a \in \mathfrak{A} \}$of$\mathfrak{A} + \mathbf{F}$is a subalgebra of$\mathfrak{A} + \mathbf{F}$, and this subalgebra is isometrically isomorphic to$\mathfrak{A}$by the isomorphism$f : \mathfrak{A} \to \mathfrak{A} + \mathbf{F}$defined by$f(a) = (a, 0)\$.