Unitization of Normed Algebra is Unital Normed Algebra

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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.

$\blacksquare$

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