Let $(\mathcal A,||\cdot||)$ be a normed algebra (with or without a unit). The **unitization** of $\mathcal A$ is the space $\mathcal A_+:=\mathcal A\oplus \Bbb C$ where the multiplication operation $\cdot$ and norm $|||\cdot |||$ are defined by
$$\begin{align}
(a,\lambda)\cdot(b,\mu) &:=(ab+\mu a+\lambda b,\lambda\mu) \\
|||(a,\lambda)||| &:= ||a||+|\lambda|
\end{align}$$
for any $a,b\in\mathcal A$ and $\mu,\lambda\in \Bbb C$. It can be easily verified that $(0,1)$ is a unit in $\mathcal A_+$ with norm $1$, and that $\mathcal A$ isometrically embeds into $\mathcal A_+$. This construction seems to be a very fundamental tool used to augment a unit into a space without one
, it came up in my first class of the chapter on Banach Algebra.

Our professor tried to convince us that unitization is done in the same spirit as completion of metric spaces. However, this construction troubles me for 2 main reasons.

Firstly, nothing stop me from unitizing a unital normed algebra. This gives me an essentially **different** space than the one I started with. This is not the case for the completion $\hat X$ of a Banach space $X$, in which we have $\hat X \cong X$. To add salt to the wound, the original unit in $\mathcal A$ is no longer a unit in $\mathcal A_+$.

Secondly, unlike the process of completion, the adjoined unit seems quite artificial. In metric space completion we only fill in "holes" but it seems like in unitization we artificially add genuine new direction into our space.

I know that my professor's analogy shouldn't be taken literally but in what sense does unitization resemble metric completion? Conceptually, what is unitization?

It also occurred to be that I'll learn to appreciate unitization as I encounter more and more results in this field. Nevertheless, I would like to have an intuitive understanding of unitization.

For example, take $(\mathcal A,+,\cdot,||\cdot||)$ to be $(L^1(\Bbb R),+,*,||\cdot||_{1})$, where $*$ is the convolution. This is a non-unital algebra since the Dirac distribution does not belong to $L^1$. In this case what does $\mathcal A_+$ look like?

intentis, that is, what we want/need of the "unitization". Just "making things" is not necessarily purposeful or intelligible, indeed, as is (to my mind) implicit in this question. $\endgroup$ – paul garrett Nov 17 '16 at 23:10regularthen $\|M_a\| = \|a\|$ for all $a$. In this case you can define $\tilde{\mathcal{A}}$ to be the span of $\{M_a: a \in \mathcal{A}\}$ and the identity operator. If $\mathcal{A}$ has a unit then you haven't added anything. $\endgroup$ – Nik Weaver Nov 18 '16 at 15:341more comment