The Unitization of a Normed Algebra

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 $(0, 1)$.

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

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 indeed $X + \mathbf{F}$ 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 $f(x) = (x, 0)$.
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