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
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, 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.

Proof of
Let $\tuple {x, \lambda} \in A_+$.

Let $\mu \in \GF$.

Then, we have:

So we have proved.

Proof of
Let $\struct {x, \lambda}, \struct {y, \mu} \in A_+$.

Then, we have:

So we have proved.

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, 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:

Proof of $\norm { {\mathbf 1}_{A_+} }_{A_+} = 1$
We have:

So $\norm {\, \cdot \,}_{A_+}$ is an algebra norm.

So $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a unital normed algebra.