Theorem: Completion Theorem (Normed Algebra)

Let $\GF \in \set {\R, \C}$. 
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$.
Then there exists a Banach algebra $\struct {\widetilde A, \widetilde {\norm {\, \cdot \,} } }$ and a isometric algebra homomorphism $\phi : A \to \widetilde A$ such that $\phi \sqbrk A$ is dense in $\widetilde A$.
This Banach algebra $\struct {\widetilde A, \widetilde {\norm {\, \cdot \,} } }$ is unique up to isometric algebra isomorphism.
Further, if $\struct {A, \norm {\, \cdot \,} }$ is unital, then so is $\struct {\widetilde A, \widetilde {\norm {\, \cdot \,} } }$.

Proof:
=== Proof of Existence ===
Let $\struct {A^{\ast \ast}, \norm {\, \cdot \,}_{A^{\ast \ast} } }$ be the second normed dual of $\struct {A, \norm {\, \cdot \,} }$.
From Normed Dual Space is Banach Space, $\struct {A^{\ast \ast}, \norm {\, \cdot \,}_{A^{\ast \ast} } }$ is a Banach space. 
Let $\phi : A \to A^{\ast \ast}$ be the evaluation transformation.
Let $\widetilde A = \map \cl {\phi \sqbrk A}$, where $\cl$ denotes the closure in $\struct {A^{\ast \ast}, \norm {\, \cdot \,}_{A^{\ast \ast} } }$.
From Completion Theorem (Normed Vector Space), $\phi$ is a linear isometry and $\phi \sqbrk A$ is dense in $\widetilde A$.
It remains to assign a multiplicative structure to $\widetilde A$ and show that with this structure both makes $\widetilde A$ a Banach algebra and $\phi$ an algebra homomorphism.
Let $a, b \in \map \cl {\phi \sqbrk A}$. 
From Point in Closure of Subset of Metric Space iff Limit of Sequence, there exists sequences $\sequence {a_n}_{n \in \N}$ and $\sequence {b_n}_{n \in \N}$ in $A$ such that:
:$\map \phi {a_n} \to a$
and:
:$\map \phi {b_n} \to b$ 
We define:
:$\ds a \circ b = \lim_{n \mathop \to \infty} \map \phi {a_n b_n}$
We first verify that $\circ$ is well-defined.
Since $\phi$ is a linear isometry, we have:
:$\widetilde {\norm {\map \phi {a_n} - \map \phi {a_m} } } = \widetilde {\norm {\map \phi {a_n - a_m} } } = \norm {a_n - a_m}$
and similarly:
$\widetilde {\norm {\map \phi {b_n} - \map \phi {b_m} } } = \norm {b_n - b_m}$
From Convergent Sequence is Cauchy Sequence, both $\sequence {\map \phi {a_n} }_{n \mathop \in \N}$ and $\sequence {\map \phi {b_n} }_{n \mathop \in \N}$ are Cauchy in $\struct {\widetilde A, \widetilde {\norm {\, \cdot \,} } }$.
Hence $\sequence {a_n}_{n \mathop \in \N}$ and $\sequence {b_n}_{n \mathop \in \N}$ are Cauchy in $\struct {\widetilde A, \widetilde {\norm {\, \cdot \,} } }$.
From Product of Cauchy Sequences in Normed Algebra is Cauchy, we have: 
:$\sequence {a_n b_n}_{n \mathop \in \N}$ is Cauchy.
Since $\phi$ is a linear isometry again, we have:
:$\sequence {\map \phi {a_n b_n} }_{n \mathop \in \N}$ is Cauchy.
Since $\struct {\widetilde A, \widetilde {\norm {\, \cdot \,} } }$ is a Banach space as in Completion Theorem (Normed Vector Space), we have that:
:$\sequence {\map \phi {a_n b_n} }_{n \mathop \in \N}$ is convergent.
Hence:
:$\ds \lim_{n \mathop \in \N} \map \phi {a_n b_n}$ is well-defined. 
We now show that the value of:
:$\ds \lim_{n \mathop \in \N} \map \phi {a_n b_n}$
is independent of the choice of sequences $\sequence {a_n}_{n \mathop \in \N}$ and $\sequence {b_n}_{n \mathop \in \N}$.
Let $\sequence {a_n}_{n \mathop \in \N}$, $\sequence {a_n'}_{n \mathop \in \N}$, $\sequence {b_n}_{n \mathop \in \N}$ and $\sequence {b_n'}_{n \mathop \in \N}$ be such that:
:$\ds \lim_{n \mathop \to \infty} \map \phi {a_n} = \lim_{n \mathop \to \infty} \map \phi {a_n'} = a$
and:
:$\ds \lim_{n \mathop \to \infty} \map \phi {b_n} = \lim_{n \mathop \to \infty} \map \phi {b_n'} = b$
Hence, we have from Sum Rule for Sequence in Normed Vector Space:
:$\widetilde {\norm {\map \phi {a_n} - \map \phi {a_n'} } } \to 0$
and:
:$\widetilde {\norm {\map \phi {b_n} - \map \phi {b_n'} } } \to 0$
Since $\phi$ is a linear isometry, we again obtain:
:$\norm {a_n - a_n'} \to 0$
and:
:$\norm {b_n - b_n'} \to 0$
Recall that $\sequence {a_n}_{n \mathop \in \N}$, $\sequence {b_n}_{n \mathop \in \N}$, $\sequence {a_n'}_{n \mathop \in \N}$ and $\sequence {b_n'}_{n \mathop \in \N}$ are also Cauchy.
Hence from Cauchy Sequence is Bounded: Normed Vector Space:
:there exists $M > 0$ such that $\max \set {\norm {a_n}, \norm {b_n}, \norm {a_n'}, \norm {b_n'} } \le M$ for each $n \in \N$. 
Let:
:$\ds L_1 = \lim_{n \mathop \in \N} \map \phi {a_n b_n}$
and:
:$\ds L_2 = \lim_{n \mathop \in \N} \map \phi {a_n' b_n'}$
We show that $L_1 = L_2$.
We have:
{{begin-eqn}}
{{eqn	| l = \norm {a_n b_n - a_n' b_n'}
		| r = \norm {a_n b_n - a_n' b_n + a_n' b_n - a_n' b_n'}
}}
{{eqn	| r = \norm {b_n \paren {a_n - a_n'} + a_n' \paren {b_n - b_n'} }
}}
{{eqn	| o = \le
		| r = \norm {b_n} \norm {a_n - a_n'} + \norm {a_n'} \norm {b_n - b_n'}
		| c = {{NormAxiomVector|3}}, {{Defof|Norm on Algebra}}
}}
{{eqn	| o = \le
		| r = M \paren {\norm {a_n - a_n'} + \norm {b_n - b_n'} }
}}
{{eqn	| o = \to
		| r = 0
}}
{{end-eqn}}
Hence, since $\phi$ is a linear isometry, we have:
:$\widetilde {\norm {\map \phi {a_n b_n} - \map \phi {a_n' b_n'} } } \to 0$
That is, from Norm is Continuous, we have:
:$\widetilde {\norm {L_1 - L_2} } = 0$
Hence, from {{NormAxiomVector|2}}:
:$L_1 = L_2$
We now show that $\circ$ is bilinear.
We show that:
:$a \circ \paren {b + \lambda c} = a \circ b + \lambda \paren {a \circ c}$ for $a, b, c \in \widetilde A$ and $\lambda \in \GF$.
The fact that:
:$\paren {a + \lambda b} \circ c = a \circ c + \lambda \paren {b \circ c}$
can be proven virtually identically.
Let $a, b, c \in \widetilde A$ and $\lambda \in \GF$.
Let $\sequence {a_n}_{n \mathop \in \N}$, $\sequence {b_n}_{n \mathop \in \N}$ and $\sequence {c_n}_{n \mathop \in \N}$ be sequences with:
:$\map \phi {a_n} \to a$, $\map \phi {b_n} \to b$ and $\map \phi {c_n} \to c$. 
From Linear Combination of Convergent Sequences in Topological Vector Space is Convergent, we have:
:$\map \phi {b_n} + \lambda \map \phi {c_n} = \map \phi {b_n + \lambda c_n} \to b + \lambda c$ 
Then, we have:
{{begin-eqn}}
{{eqn	| l = a \circ \paren {b + \lambda c}
		| r = \lim_{n \mathop \to \infty} \map \phi {a_n \paren {b_n + \lambda c_n} }
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \map \phi {a_n b_n + \lambda b_n c_n}
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \paren {\map \phi {a_n b_n} + \lambda \map \phi {b_n c_n} }
		| c = $\phi$ is linear
}}
{{eqn	| r = a \circ b + \lambda \paren {b \circ c}
		| c = Linear Combination of Convergent Sequences in Topological Vector Space is Convergent
}}
{{end-eqn}}
We conclude that $\struct {\widetilde A, \circ}$ is an algebra.
We now show that $\phi$ is an algebra homomorphism.
Let $a, b \in A$ so that $\map \phi a, \map \phi b \in \phi \sqbrk A$.
From Constant Sequence in Topological Space Converges, we have:
:$\map \phi a \to \map \phi a$
and:
:$\map \phi b \to \map \phi b$
Hence from our definition of $\circ$ with $a_n = a$ and $b_n = b$ for all $n \in \N$, we obtain:
:$\ds \map \phi a \circ \map \phi b = \lim_{n \mathop \to \infty} \map \phi {a b} = \map \phi {a b}$
for each $a, b \in A$. 
That is, $\phi$ is an algebra homomorphism.
We now show that $\widetilde {\norm {\, \cdot \,} }$ is a algebra norm on $\widetilde A$.
Let $a, b \in \widetilde A$.
Then there exists sequences $\sequence {a_n}_{n \mathop \in \N}$ and $\sequence {b_n}_{n \mathop \in \N}$ in $A$ such that:
:$\map \phi {a_n} \to a$
and:
:$\map \phi {b_n} \to b$
Since $\phi$ is a linear isometry and $\norm {\, \cdot \,}$ is a norm on $A$, we have:
{{begin-eqn}}
{{eqn	| l = \widetilde {\norm {\map \phi {a_n b_n} } }
		| r = \norm {a_n b_n}
		| c = since $\phi$ is a linear isometry
}}
{{eqn	| o = \le
		| r = \norm {a_n} \norm {b_n}
		| c = {{Defof|Norm on Algebra}}
}}
{{eqn	| r = \widetilde {\norm {\map \phi {a_n} } } \widetilde {\norm {\map \phi {b_n} } }
		| c = since $\phi$ is a linear isometry
}}
{{end-eqn}}
Taking $n \to \infty$ and using Norm is Continuous, we have:
:$\widetilde {\norm {a \circ b} } \le \widetilde {\norm a} \widetilde {\norm b}$
Recall again that $\struct {\widetilde A, \widetilde {\norm {\, \cdot \,} } }$ is a Banach space. 
Hence $\struct {\widetilde A, \widetilde {\norm {\, \cdot \,} } }$ is a Banach algebra.
Finally, we assure ourselves that if $A$ is unital in the sense of a normed algebra, then so is $\struct {\widetilde A, \widetilde {\norm {\, \cdot \,} } }$.
Suppose that $A$ is unital in the sense of a normed algebra.
Then the identity element ${\mathbf 1}_A$ of $A$ satisfies:
:$\norm { {\mathbf 1}_A} = 1$
We show that $\map \phi { {\mathbf 1}_A}$ is an identity element for $\widetilde A$.
Since $\phi$ is an isometry, we have:
:$\widetilde {\norm {\map \phi { {\mathbf 1}_A} } } = 1$
Now, we have, for $a \in A$ and an appropriate sequence $\sequence {a_n}_{n \mathop \in \N}$:
{{begin-eqn}}
{{eqn	| l = \map \phi { {\mathbf 1}_A} \circ a
		| r = \lim_{n \mathop \to \infty} \map \phi { {\mathbf 1}_A a_n}
		| c = by definition of $\circ$
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \map \phi {a_n}
		| c = {{Defof|Identity Element}}
}}
{{eqn	| r = a
		| c = by choice of $\sequence {a_n}_{n \mathop \in \N}$
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \map \phi {a_n}
		| c = by choice of $\sequence {a_n}_{n \mathop \in \N}$
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \map \phi {a_n {\mathbf 1}_A}
		| c = {{Defof|Identity Element}}
}}
{{eqn	| r = a \circ \map \phi { {\mathbf 1}_A}
		| c = by definition of $\circ$
}}
{{end-eqn}}
The proof of existence is complete. 
{{qed|lemma}}