There is the following result on the completion of a normed algebra:

Suppose $X$ is a normed algebra. Then there is a Banach algebra $Y$ and a map $T$ from $X$ onto a dense subspace $M \subset Y$ such that the map $T$ is an isometric isomorphism for which $T(x_1 \cdot x_2) = T(x_1) \cdot T(x_2)$ whenever $x_1, x_2 \in X$.

The following should prove the above mentioned result:

Let $X$ be a normed algebra. Then, in particular, $X$ is a normed space, and hence there is an isometric isomorphism $T$ from $X$ into a Banach space $Y$ such that $T(X)$ is dense. Since $T(Y) \subset Y$ is dense, there are, for any $x, y \in Y$, sequences $(x_n)_{n \in \mathbb{N}}, (y_n)_{n \in \mathbb{N}} \subset X$ such that $x = \lim_{n \in \mathbb{N}} Tx_n$ and $y = \lim_{n \in \mathbb{N}} Ty_n$.

Define now $x \cdot y = \lim_{n \in \mathbb{N}} T(x_n \cdot y_n)$, which is well-defined since $(T(x_n \cdot y_n))_{n \in \mathbb{N}}$ is Cauchy in $Y$.

The above stated proof is rather straightforward. However, I am wondering if there are other ways to complete a normed algebra.

Thanks in advance; any comments are welcome.

  • $\begingroup$ You did not pose a question but you gave an answer. $\endgroup$ – Jochen Aug 18 '16 at 10:22
  • $\begingroup$ @Jochen. There is the tag proof-verification. I am curious if this is the correct way to do the completion. $\endgroup$ – user342207 Aug 18 '16 at 10:23

In your proof, you still have to show that the choice of sequences $(x_n)$ and $(y_n)$ converging to $x$ and $y$ do not interfere in the definition of $xy$. You also need to verify that this multiplication satisfies all the necessary properties (associativity, scalars moving inside products, etc...), although this is easy.

As a comment, there are other ways of constructing $Y$. You can take $\ell^\infty(\mathbb{N},X)$ to be the algebra of bounded sequences in $X$ with supremum norm, and $c_0(\mathbb{N},X)$ the ideal of those sequences converging to $0$. $\ell^\infty(\mathbb{N},X)$ is not necessarily complete, but the quotient $Y=\ell^\infty(\mathbb{N},X)/c_0(\mathbb{N},X)$ is (the proof is not hard, but a little cumbersome), and it is a Banach algebra.

The map $X\to Y$, sending $x\in X$ to the class of the constant sequence $(x,x,\ldots)$ is an isometric homomorphism, so the closure of its image has the desired properties.

  • $\begingroup$ Thanks. I have verified that $Y$ satisfies the axioms of a Banach algebra, which follows in fact readily from the fact that $X$ is a normed algebra. Regarding your first point: what do you mean with interference of the sequences? Are you referring to the uniqueness such that the definition of $x \cdot y$ is unambiguous? $\endgroup$ – user342207 Aug 18 '16 at 20:23
  • $\begingroup$ @jvnv Precisely. To define $x\cdot y$, you started by choosing sequences $(x_n)$ and $(y_n)$ with $(Tx_n)$ and $(Ty_n)$ converging to $x$ and $y$, respectively. What would happen if we chose other sequences $(x_n')$ and $(y_n')$? You need to show that $\lim T(x_ny_n)=\lim T(x_n'y_n')$ (the fact that both these limits exist is no problem). In fact, it is necessary to do this before verifying the axioms for a Banach algebra. $\endgroup$ – Luiz Cordeiro Aug 19 '16 at 2:35
  • $\begingroup$ I thought about this, but did not include it, which I should have done. Just to be sure it is correct: Given the sequences $(x_n), (y_n) \subset X$, let $(x'_n), (y'_n)$ such that $\lim_n (x_n - x'_n) = 0$ and $\lim_n (y_n - y'_n) = 0$. Then $$ \| x'_n y'_n - x_n y_n \| \leq \|x'_n \| \|y'_n - y_n \| + \|y_n \| \|x'_n - x_n \| \to 0$$ as $n \to \infty$. As $T$ is isometric, it follows that $\lim_n T(x_ny_n) = \lim_n T(x'_n y'_n)$. $\endgroup$ – user342207 Aug 19 '16 at 8:55
  • $\begingroup$ @jvnv Yes, that is ok (but I would mention that $(x_n')$ and $(y_n)$ are bounded, to make sure that I understand why the second term goes to zero). $\endgroup$ – Luiz Cordeiro Aug 19 '16 at 11:17

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