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.


  • $\begingroup$ @Just a user$:$ In this case we are done because then the first two properties fail to satisfy unless $a = 0.$ $\endgroup$
    – RKC
    Nov 15, 2021 at 8:02
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 15, 2021 at 8:05

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – 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 $2$ ($A^2 = \{0\}$).


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