The Unitization of a Normed Algebra
 Definition: Let $X$ be a normed algebra over $\mathbf{F}$. The Unitization of $X$ denoted by $X + \mathbf{F}$ is the normed algebra with unit on the set $X \times \mathbf{F}$ with: 1.(( The operation of addition defined for all $(x, \alpha), (y, \beta) \in X + \mathbf{F}$ by $(x, \alpha) + (y, \beta) = (x + y, \alpha + \beta)$.2.** The operation of scalar multiplication defined for all $(x, \alpha) \in X + \mathbf{F}$ and all $k \in \mathbf{F}$ by $k(x, \alpha) = (kx, k\alpha)$. 3. The operation of multiplication defined for all $(x, \alpha), (y, \beta) \in X + \mathbf{F}$ by $(x, \alpha)(y, \beta) = (xy + \alpha y + \beta x, \alpha \beta)$. And with norm $\| (x, \alpha) \| = \| x \| + |\alpha|$. The unit of $X + \mathbf{F}$ is $(0, 1)$.
 Proposition 1: Let $X$ be a normed algebra over $\mathbf{F}$. Then the unitization of $X$ is also a normed algebra over $\mathbf{F}$.
• Proof: Certainly $X + \mathbf{F}$ with the operations of addition and scalar multiplication is a linear space. We first show that $X + \mathbf{F}$ is an algebra.
• 1. Showing that $(x, \alpha)[(y, \beta)(z,\gamma)] = [(x, \alpha)(y, \beta)](z, \gamma)$$: Let$(x, \alpha), (y, \beta), (z, \gamma) \in X + \mathbf{F}. Then: (1) \begin{align} \quad (x, \alpha)[(y, \beta)(z, \gamma)] &= (x, \alpha)[(yz + \beta z + \gamma y, \beta \gamma) \\ &= (x[yz + \beta z + \gamma y] + \alpha [yz + \beta z + \gamma y] + \beta \gamma x, \alpha [\beta \gamma]) \end{align} • And also: (2) \begin{align} \quad [(x, \alpha)(y, \beta)](z, \gamma) &= (xy + \alpha y + \beta x, \alpha \beta)(z, \gamma) \\ &= ([xy + \alpha y + \beta x]z + \alpha \beta z + \gamma [xy + \alpha y + \beta x], [\alpha \beta]\gamma) \end{align} • Comparing the two equations above and we see that they are equal. Thus(x, \alpha)[(y, \beta)(z,\gamma)] = [(x, \alpha)(y, \beta)](z, \gamma)$. • 2. Showing that$(x, \alpha)[(y, \beta) + (z, \gamma)] = (x, \alpha)(y, \beta) + (x, \alpha)(z, \gamma)$: Let$(x, \alpha), (y, \beta), (z, \gamma) \in X + \mathbf{F}. Then: (3) \begin{align} \quad (x, \alpha)[(y, \beta) + (z, \gamma)] &= (x, \alpha)(y + z, \beta + \gamma) \\ &=(x[y + z] + \alpha[y + z] + [\beta + \gamma]x, \alpha[\beta + \gamma]) \\ \end{align} • And: (4) \begin{align} \quad (x, \alpha)(y, \beta) + (x, \alpha)(z, \gamma) = (xy + \alpha y + \beta x, \alpha \beta) + (xz + \alpha z + \gamma x, \alpha \gamma) \\ &= (x[y + z] + \alpha[y + z] + [\beta + \gamma]x, \alpha[\beta + \gamma]) \end{align} • Thus(x, \alpha)[(y, \beta) + (z, \gamma)] = (x, \alpha)(y, \beta) + (x, \alpha)(z, \gamma)$. • 3. Showing that$[k(x, \alpha)](y, \beta) = k[(x, \alpha)(y, \beta)] = (x, \alpha)[k(y, \beta)]$: Let$(x, \alpha), (y, \beta) \in X + \mathbf{F}$and let$k \in \mathbf{F}. Then: (5) \begin{align} \quad [k(x, \alpha)](y, \beta) = (kx, k\alpha)(y, \beta) = (kxy + k\alpha y + k \beta x, k \alpha \beta) = k(xy + \alpha y + \beta x, \alpha \beta) = k[(x, \alpha)(y, \beta)] \end{align} • And also: (6) \begin{align} \quad (x, \alpha)[k(y, \beta)] = (x, \alpha)(ky, k\beta) = (kxy + k\beta x + k \alpha y, k \alpha \beta) = k (xy + \alpha y + \beta x, \alpha \beta) = k[(x, \alpha)(y, \beta)] \end{align} • Thus[k(x, \alpha)](y, \beta) = k[(x, \alpha)(y, \beta)] = (x, \alpha)[k(y, \beta)]$. • So$X + \mathbf{F}$is an algebra. • We will now show that the above defined norm is indeed a norm on$X + \mathbf{F}$. • 1. Showing that$\| (x, \alpha) \| = 0$if and only if$(x, \alpha) = (0, 0)$: Suppose that$\| (x, \alpha) \| = 0$. Then$\| x \| + |\alpha| = 0$which implies that$\| x \| = 0$and$|\alpha| = 0$, i.e.,$x = 0$and$\alpha = 0$. So$(x, \alpha) = (0, 0)$. On the other hand, if$(x, \alpha) = (0, 0)$then$\| (x, \alpha) \| = \| 0 \| + |0| = 0$. • 2. Showing that$\| k(x, \alpha) \| = |k| \| (x, \alpha) \|$: Let$(x, \alpha) \in X + \mathbf{F}$and let$k \in \mathbf{F}. Then: (7) \begin{align} \quad \| k(x, \alpha) \| = \| (kx, k\alpha) \| = \| kx \| + |k \alpha| = |k| \| x \| + |k||\alpha| = |k|[\| x \| + |\alpha|] = |k| \| (x, \alpha) \| \end{align} • 3. Showing that\| (x, \alpha) + (y, \beta) \| \leq \| (x, \alpha) \| + \| (y, \beta) \|$: Let$(x, \alpha), (y, \beta) \in X + \mathbf{F}. Then: (8) \begin{align} \quad \| (x, \alpha) + (y, \beta) \| = \| (x + y, \alpha + \beta) \| = \| x + y \| + |\alpha + \beta| \leq \| x \| + \| y \| + |\alpha| + |\beta| = [\| x \| + |\alpha|] + [\| y \| + |\beta|] = \| (x, \alpha) \| + \| (y, \beta) \| \end{align} • 4. Showing that\|(x, \alpha)(y, \beta) \| \leq \| (x, \alpha) \| \| (y, \beta) \|$: Let$(x, \alpha), (y, \beta) \in X + \mathbf{F}. Then: (9) \begin{align} \quad \| (x, \alpha)(y, \beta) \| &= \| (xy + \alpha y + \beta x, \alpha \beta) \| \\ &= \| xy + \alpha y + \beta x \| + |\alpha \beta| \\ & \leq \| x \| \| y \| + |\alpha| \| y \| + |\beta| \| x \| + |\alpha||\beta| = [\| x \| + |\alpha|][\| y \| + |\beta|] = \| (x, \alpha) \| \| (y, \beta) \| \end{align} • So indeedX + \mathbf{F}$is an normed algebra.$\blacksquare$ Proposition 2: Let$X$be a normed algebra over$\mathbf{F}$and let$X + \mathbf{F}$be the unitization of$\mathbf{F}$. Then the subset$\{ (x, 0) : x \in X \}$of$X + \mathbf{F}$is a subalgebra of$X + \mathbf{F}$, and this subalgebra is isometrically isomorphic to$X$by the isomorphism$f : X \to X + \mathbf{F}$defined by$f(x) = (x, 0)\$.