I am studying Jen-Pierre Demailly's "Complex Analytic and Diferential Geometry" and in definition (5.39) the author implies that a locally convex topological algebra $A$ (i.e. a locally convex space that has a continuous commutative multiplication operation and a multiplicative identity) has a completion, i.e., there is a complete locally convex topological algebra $\tilde{A}$ that has $A$ as a dense subspace. However, I'm not being able to find any reference with a theorem that guarantees this. I know that for locally convex spaces, there is a nice theorem about the existence of completion, and I assume there is also one for locally convex topological algebras, since it seems to be used in the book.
Can anyone help me? Thank you in advance :)
First of all, the product in a (locally convex) topological algebra $\mathfrak{A}$ (we assume from now on that $\mathfrak{A}$ is Hausdorff) is usually only required to be separately continuous, i.e. if you fix either argument of the product you end up with a continuous linear map from $\mathfrak{A}$ to itself. If $\mathfrak{A}$ is barrelled and metrizable (e.g. if $\mathfrak{A}$ is Fréchet), then its product is (jointly) continuous, but there are examples of locally convex algebras where this fails - take e.g. the tensor algebra generated by the space of smooth real-valued functions with compact support on a non-void open subset of $\mathbb{R}^n$.
Joint continuity of the multiplication indeed is a sufficient condition for it to extend to the completion, and in this case the extension also happens to be jointly continuous, see e.g. Theorème 1, pp. III.50-III.51 of N. Bourbaki, Topologie Générale, Chapitres 1-4 (Springer-Verlag, 2007). If the multiplication is only separately continuous, it may not be possible to extend the multiplication to the completion.
A middle-ground result can be obtained after some preparation. Let $E,F,G$ be topological vector spaces, and $\mathfrak{M},\mathfrak{N}$ be families of bounded subsets of respectively $E,F$. We say that a bilinear map $T:E\times F\rightarrow G$ is $(\mathfrak{M},\mathfrak{N})$-hypocontinuous if for any $0$-neighborhood $V$ in $G$ there are $U_1\subset E$, $U_2\subset F$ $0$-neighborhoods and $K_1\in\mathfrak{M}$, $K_2\in\mathfrak{N}$ such that $T(U_1,K_2),T(K_1,U_2)\subset V$. If $\mathfrak{M}$ (resp $\mathfrak{N}$) covers $E$ (resp. $F$), then a $(\mathfrak{M},\mathfrak{N})$-hypocontinuous bilinear map $T:E\times F\rightarrow G$ is, of course, separately continuous - this is the case if e.g. $\mathfrak{M}$ (resp $\mathfrak{N}$) is the family of all finite or all bounded subsets of $E$ (resp. $F$) - and $(\mathfrak{M},\mathfrak{N})$-(bi)bounded, that is, if $K_1\in\mathfrak{M}$, $K_2\in\mathfrak{N}$, then $T(K_1,K_2)$ is a bounded subset of $G$. If $\mathfrak{M}$ (resp. $\mathfrak{N}$) is the family of all bounded subsets of $E$ (resp. $F$), we simply say that a $(\mathfrak{M},\mathfrak{N})$-hypocontinuous bilinear map $T$ is hypocontinuous. If $E,F$ are barrelled, then any separately continuous bilinear map $T:E\times F\rightarrow G$ is hypocontinuous.
Now we are in a position to state the result (see e.g. Theorem 4.2, pp. 30 of the book of A. Mallios, Topological Algebras, Selected Topics (North-Holland, 1986)):
Sometimes it happens that $\widetilde{\mathfrak{A}}_\sigma$ coincides with the completion of $\mathfrak{A}$ even if the latter's topology is not first countable. An important example is the space of smooth functions with compact support on a non-void open subset $U$ of $\mathbb{R}^n$ endowed with the strong topology when dually paired with itself. This topology is clearly not first countable, but the sequential completion of this space is the space of distributions on $U$ endowed with the strong topology, which is complete.