Unitization of Normed Algebra is Banach Algebra iff Original Algebra is Banach Algebra

Theorem

Let $\GF \in \set {\R, \C}$.

Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$ that is not unital as an algebra.

Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the normed unitization of $\struct {A, \norm {\, \cdot \,} }$.

Then $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a Banach algebra if and only if $\struct {A, \norm {\, \cdot \,} }$ is a Banach algebra.

Proof

Necessary Condition

Suppose that $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a Banach algebra.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {A, \norm {\, \cdot \,} }$.

We then have, for each $n, m \in \N$:

$\norm {\tuple {x_n, 0} - \tuple {x_m, 0} }_{A_+} = \norm {\tuple {x_n - x_m, 0} }_{A_+} = \norm {x_n - x_m}$

Let $\epsilon > 0$.

Since $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence, there exists $N \in \N$ such that:

$\norm {x_n - x_m} < \epsilon$ for $n, m \ge N$:

so that:

$\norm {\tuple {x_n, 0} - \tuple {x_m, 0} }_{A_+} < \epsilon$ for $n, m \ge N$.

Since $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a Banach algebra, there exists $\tuple {x, \lambda} \in A_+$ such that $\tuple {x_n, 0} \to \tuple {x, \lambda}$.

From Norm on Unitization of Normed Algebra is Equivalent to Direct Product Norm, $\norm {\, \cdot \,}_{A_+}$ is equivalent to the direct product norm $\norm {\, \cdot \,}_{A \times \GF}$.

From Convergence in Direct Product Norm and Convergent Sequences in Vector Spaces with Equivalent Norms Coincide, we have $x_n \to x$ and $\lambda = 0$.

Then, we have:

$\norm {\tuple {x_n, 0} - \tuple {x, 0} }_{A_+} \to 0$

while:

$\norm {\tuple {x_n, 0} - \tuple {x, 0} }_{A_+} = \norm {\tuple {x_n - x, 0} }_{A_+} = \norm {x_n - x}$

so that:

$\norm {x_n - x} \to 0$

From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we have $x_n \to x$ in $A$.

So every Cauchy sequence in $\struct {A, \norm {\, \cdot \,} }$ converges.

$\Box$

Sufficient Condition

Suppose that $\struct {A, \norm {\, \cdot \,} }$ is a Banach algebra.

In particular, $\struct {A, \norm {\, \cdot \,} }$ is a Banach space.

From:

Real Number Line is Banach Space if $\GF = \R$

Complex Plane is Banach Space if $\GF = \C$

we have that $\GF = \C$ is also a Banach space.

From Direct Product of Banach Spaces is Banach Space, the direct product $A \times \GF = A_+$ equipped with the direct product norm $\norm {\, \cdot \,}_{A \times \GF}$ is a Banach space.

From Norm on Unitization of Normed Algebra is Equivalent to Direct Product Norm, $\norm {\, \cdot \,}_{A_+}$ and $\norm {\, \cdot \,}_{A \times \GF}$ are equivalent.

From Norm Equivalence Preserves Completeness, $\struct {A, \norm {\, \cdot \,}_{A_+} }$ is therefore a Banach space, hence a Banach algebra.

$\blacksquare$