Unitization of Normed Algebra is Unital Normed Algebra
Theorem
Let $\GF \in \set {\R, \C}$.
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra that is not unital as an algebra.
Let $A_+$ be the unitization of $A$.
Define $\norm {\, \cdot \,}_{A_+} : A_+ \to \hointr 0 \infty$ by:
- $\norm {\tuple {x, \lambda} }_{A_+} = \norm x + \cmod \lambda$
for each $\tuple {x, \lambda} \in A_+$.
Then $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a unital normed algebra.
Proof
From Unitization of Algebra over Field is Unital Algebra over Field, $A_+$ is a unital algebra.
We show that $\norm {\, \cdot \,}_{A_+}$ is an algebra norm and that $\norm { {\mathbf 1}_{A_+} } = 1$.
Proof of Norm Axiom $\text N 1$: Positive Definiteness
Suppose that $\tuple {x, \lambda} \in A_+$ is such that:
- $\norm {\tuple {x, \lambda} }_{A_+} = 0$
Then, we have:
- $\norm x + \cmod \lambda = 0$
So $\norm x = 0$ and $\cmod \lambda = 0$.
From Norm Axiom $\text N 1$: Positive Definiteness, we obtain that $x = {\mathbf 0}_X$.
Since $\cmod \lambda = 0$, we have $\lambda = 0$.
So we have $\tuple {x, \lambda} = {\mathbf 0}_{A_+}$.
Hence we have proved Norm Axiom $\text N 1$: Positive Definiteness.
$\Box$
Proof of Norm Axiom $\text N 2$: Positive Homogeneity
Let $\tuple {x, \lambda} \in A_+$.
Let $\mu \in \GF$.
Then, we have:
\(\ds \norm {\mu \tuple {x, \lambda} }_{A_+}\) | \(=\) | \(\ds \norm {\tuple {\mu x, \lambda \mu} }_{A_+}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\mu x} + \cmod {\lambda \mu}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cmod \mu \norm x + \cmod \mu \cmod \lambda\) | Norm Axiom $\text N 2$: Positive Homogeneity | |||||||||||
\(\ds \) | \(=\) | \(\ds \cmod \mu \norm {\tuple {x, \lambda} }_{A_+}\) |
So we have proved Norm Axiom $\text N 2$: Positive Homogeneity.
$\Box$
Proof of Norm Axiom $\text N 3$: Triangle Inequality
Let $\struct {x, \lambda}, \struct {y, \mu} \in A_+$.
Then, we have:
\(\ds \norm {\tuple {x, \lambda} + \tuple {y, \mu} }_{A_+}\) | \(=\) | \(\ds \norm {\tuple {x + y, \lambda + \mu} }_{A_+}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {x + y} + \cmod {\lambda + \mu}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm x + \norm y + \cmod \lambda + \cmod \mu\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\tuple {x, \lambda} }_{A_+} + \norm {\tuple {y, \mu} }_{A_+}\) |
So we have proved Norm Axiom $\text N 3$: Triangle Inequality.
$\Box$
Proof of Submultiplicativity
Let $\struct {x, \lambda}, \struct {y, \mu} \in A_+$.
Suppose that $\norm {\tuple {x, \lambda} }_{A_+} \norm {\tuple {y, \mu} }_{A_+} = 0$.
Then we have $\norm {\tuple {x, \lambda} }_{A_+} = 0$ or $\norm {\tuple {y, \mu} }_{A_+} = 0$.
From Norm Axiom $\text N 1$: Positive Definiteness, we have $\tuple {x, \lambda} = {\mathbf 0}_{A_+}$ or $\tuple {y, \mu} = {\mathbf 0}_{A_+}$.
In this case, we have $\tuple {x, \lambda} \tuple {y, \mu} = {\mathbf 0}_{A_+}$.
Now take $\tuple {x, \lambda} \ne {\mathbf 0}_{A_+}$ and $\tuple {y, \mu} \ne {\mathbf 0}_{A_+}$.
We have:
\(\ds \norm {\tuple {x, \lambda} \tuple {y, \mu} }_{A_+}\) | \(=\) | \(\ds \norm {\tuple {x y + \lambda y + \mu x, \lambda \mu} }_{A_+}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {x y + \lambda y + \mu x} + \cmod {\lambda \mu}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x y} + \norm {\lambda y} + \norm {\mu x} + \cmod {\lambda \mu}\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm x \norm y + \cmod \lambda \norm y + \cmod \mu \norm x + \cmod \lambda \cmod \mu\) | Definition of Norm on Algebra, Norm Axiom $\text N 1$: Positive Definiteness | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\norm x + \cmod \lambda} \paren {\norm y + \cmod \mu}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\tuple {x, \lambda} }_{A_+} \norm {\tuple {y, \mu} }_{A_+}\) |
$\Box$
Proof of $\norm { {\mathbf 1}_{A_+} }_{A_+} = 1$
We have:
\(\ds \norm { {\mathbf 1}_{A_+} }\) | \(=\) | \(\ds \norm {\tuple { {\mathbf 0}_X, 1} }_{A_+}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm { {\mathbf 0}_X} + \cmod 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
$\Box$
So $\norm {\, \cdot \,}_{A_+}$ is an algebra norm.
So $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a unital normed algebra.
$\blacksquare$