The Completion of a Normed Algebra

# The Completion of a Normed Algebra

Definition: Let $\mathfrak{A}$ be a normed algebra. A Banach algebra $\mathfrak{B}$ is said to be a Completion of $\mathfrak{A}$ if there exists an isometric isomorphism $T$ from $\mathfrak{A}$ onto a dense subalgebra of $\mathfrak{B}$. |

The following theorem tells us that every normed algebra has a completion.

Theorem 1: If $\mathfrak{A}$ is a normed algebra, then $\mathfrak{A}$ has a completion. |

**Proof:**We know that as a normed space, $\mathfrak{A}$ has a completion $\mathfrak{B}$, i.e., a Banach space such that there exists an isometric isomorphism $T$ from $\mathfrak{A}$ onto a dense subspace of $\mathfrak{B}$. All that remains to show is that with the additional structure of product on $\mathfrak{A}$, that $\mathfrak{B}$ also has an analogous product.

- Let $b, b’ \in \mathfrak{B}$. Since $T(\mathfrak{A})$ is assumed to be dense in $\mathfrak{B}$, there exists sequences $(a_n), (a_n)’ \subset \mathfrak{A}$ such that:

\begin{align} b = \lim_{n \to \infty} T(a_n) \quad \mathrm{and} \quad b’ = \lim_{n \to \infty} T(a_n’) \end{align}

- Since $(T(a_n))$ converges to $b$, $(T(a_n))$ is a Cauchy sequence in $\mathfrak{B}$. That is, for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $\| T(a_m) - T(a_n) \| < \epsilon$. But $T$ is an isometric isomorphism. So for all $m, n \in \mathbb{N}$, $\| T(a_m) - T(a_n) \| = \| T(a_m - a_n) \| = \| a_m - a_n \|$. Thus, for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $\| a_m - a_n \| < \epsilon$, and so $(a_n)$ is a Cauchy sequence in $\mathfrak{A}$. A similar argument shows that since $(T(a_n’))$ converges to $b’$, we have that $(a_n’)$ is a Cauchy sequence in $\mathfrak{A}$ as well.

- Also, since $T(a_n)$ converges to $b$, we have that $(T(a_n))$ is a bounded sequence, i.e., there exists an $M > 0$ such that $\| T(a_n) \| \leq M$ for all $n \in \mathbb{N}$. Again, since $T$ is an isometry, this implies that $(a_n)$ is bounded with $\| a_n \| \leq M$ for all $n \in \mathbb{N}$. A similar argument shows that there exists an $M’ > 0$ such that $\| a_n’ \| \leq M’$ for all $n \in \mathbb{N}$.

- Consider the sequence $(a_na_n’) \subset \mathfrak{A}$. For each $m, n \in \mathbb{N}$, we have that:

\begin{align} \| a_ma_m’ - a_na_n’ \| &= \| a_ma_m’ - a_ma_n’ + a_ma_n’ - a_na_n’ \| \\ &= \| a_m(a_m’ - a_n’) \| + \| (a_m - a_n)a_n’ \| \\ &= \| a_m \| \| a_m’ - a_n’ \| + \| a_m - a_n \| \| a_n’ \| \\ & \leq M \| a_m’ - a_n’ \| + M’ \| a_m - a_n \| \end{align}

- Given $\epsilon > 0$, since $(a_n)$, $(a_n’)$ are both Cauchy, there exists $N_1, N_2 \in \mathbb{N}$ such that if $m, n \geq N_1$ then $\| a_m - a_n \| < \frac{\epsilon}{M’}$ and if $m, n \geq N_2$ then $\| a_m’ - a_n’ \| < \frac{\epsilon}{M}$. Then $N := \max \{ N_1, N_2 \}$ is such that if $m, n \geq N$ then $\| a_ma_m’ - a_na_n’ \| < \epsilon$ from the above inequality. So $(a_na_n’)$ is a Cauchy sequence.

- For a third time, since $T$ is an isomorphism, this implies that $(T(a_na_n’))$ is a Cauchy sequence in $\mathfrak{B}$. Since $\mathfrak{B}$ is complete, this sequence converges to some $c \in \mathfrak{B}$. So define the multiplication on $\mathfrak{B}$ by:

\begin{equation} bb’ := c \end{equation}

- Then the normed space isomorphism $T$ extends to an algebra isomorphism onto a dense subalgebra of $\mathfrak{B}$, since for each $a, a' \in \mathfrak{A}$, let $(a_n), (a_n') \subset \mathfrak{A}$ be the constant sequences with $a_n = a$ and $a_n' = a'$ for all $n \in \mathbb{N}$. Then:

\begin{align} \quad T(aa') = \lim_{n \to \infty} T(a_na_n') = c = bb' = \lim_{n \to \infty} T(a_n) \lim_{n \to \infty} T(a_n') = T(a)T(a') \end{align}

- It is clear that this multiplication on $\mathfrak{B}$ satisfies the multiplication axioms, and the norm on $\mathfrak{B}$ becomes an algebra norm too. So the normed algebra $\mathfrak{A}$ has a completion to a Banach algebra $\mathfrak{B}$. $\blacksquare$