# Question on non-unital finite dimensional Banach algebras.

Let $$A$$ be a non-unital finite dimensional Banach algebra. Then show that $$A$$ cannot have any approximate identity.

What is required to show is that if a finite dimensional Banach algebra contains an approximate identity then it is necessarily unital. What is known is that an approximate identity in a Banach algebra $$A$$ is a net $$(e_{\alpha})$$ satisfying the following properties $$:$$

$$(1)$$ $$\lim\limits \|ae_{\alpha} - a\| = 0,$$

$$(2)$$ $$\lim\limits \|e_{\alpha} a - a\| = 0,$$

$$(3)$$ $$\|e_{\alpha}\| \leq 1$$ for all $$\alpha.$$

Also since we are talking about finite dimensional Banach algebras, those algebras are precisely $$\mathbb C^n$$ as a vector space. So in order to specify a Banach algebraic structure on $$\mathbb C^n$$ we need to investigate on the possible ways of defining product of two vectors which satisfies norm inequality. If the norm is the usual Euclidean norm on $$\mathbb C^n$$ then one such product would be dot product which clearly satisfies the requisite norm inequality due to Cauchy-Schwarz inequality. But the problem is that the underlying Banach algebra thus obtained would be unital, $$(1,1,\cdots,1)$$ being an identity. Another way which comes into my mind is to define convolution product of two vectors. But again it would give rise to a unital algebra which is not what I wished to have. So how to find a valid product on $$\mathbb C^n$$ which makes $$\mathbb C^n$$ a non-unital Banach algebra? I am searching for it but couldn't succeed. I am trying solve the problem as a whole in the general setting. Any help would be much appreciated.

Rabin.

• @Just a user$:$ In this case we are done because then the first two properties fail to satisfy unless $a = 0.$
– RKC
Nov 15, 2021 at 8:02
• I expand the comment to an answer. Of course there is no "better" example as it would disprove the true statement you want to prove. Nov 15, 2021 at 8:05

For an example of non-unital Banach algebra, define $$a\cdot b = 0$$ for any $$a,b\in\mathbb C^n$$.

To prove this statement, note that the closed unital ball of any finite dimensional Banach space is compact, therefore by the third property $$\|e_{\alpha}\|\le 1$$, there exists a convergent subnet $$e_{\alpha_\beta}\rightarrow e$$. And by the first two properties we can show that $$e$$ is an identity of the Banach algebra.

• The first two properties would imply that $\|ae - a\| = \|ea - a\| = 0.$ So we have $ae = ea = a,$ for all $a \in A.$ By the uniqueness of the identity in a unital ring we are through. Many many thanks for your kind help.
– RKC
Nov 15, 2021 at 8:07

Perhaps it is interesting to mention how Wedderburn classified finite dimensional Banach algebras.

First, every bilinear function $$A\times A\to A$$ is continuous on a finite dimensional Banach space $$A$$. Thus, any multiplication is continuous on $$A$$. Hence, the problem reduces to classifying all finite dimensional algebras over $$\mathbb{C}$$ (or $$\mathbb{R}$$ alike).

The following are included in Wedderburn's structure theorem.

1. Let $$J$$ be the Jacobson radical of $$A$$. $$J$$ is a maximal nilpotent ideal of $$A$$.
2. There exists a subalgebra $$B$$ isomorphic to the quotient algebra $$A/J$$. Furthermore, $$A = J\oplus B$$.
3. $$B$$ is semisimple. Furthermore, $$B$$ is a direct sum of full matrix algebras $$M_n(\mathbb{C})$$: $$B = M_{n_1}(\mathbb{C})\oplus M_{n_2}(\mathbb{C})\oplus\dots\oplus M_{n_r}(\mathbb{C})$$

Summarizing, $$A = J\oplus M_{n_1}(\mathbb{C})\oplus M_{n_2}(\mathbb{C})\oplus\dots\oplus M_{n_r}(\mathbb{C})$$. Consequently, $$A$$ is unital iff $$B\neq\{0\}$$. $$A$$ is not unital iff $$A$$ is nilpotent.

An example is given above where $$A$$ is nilpotent of order $$2$$ ($$A^2 = \{0\}$$).