Existence of sequential approximate identity of non-unital Banach algebras through left regular representation

Question

Suppose $(A, \|\cdot\|, *)$ is a non-unital Banach algebra.

Consider the left regular representation $\iota: A \hookrightarrow \mathcal{B}(A)$, defined by $\iota: a \mapsto (x\mapsto a*x)$. Let's also denote $\hat{a}:=\iota(a): x\mapsto a*x$, for all $x\in A$.

Then, by Banach algebra inequality, we have $\|\hat{a}\|_{\mathcal{B}(A)}=\sup \frac{\|a*x\|}{\|x\|}\leqslant \|a\|.$ So the representation $\iota$ is bounded, since we always have $\|\iota\|_{\text{OP}}=\sup \frac{\|\hat{a}\|_{\mathcal{B}(A)}}{\|a\|}\leqslant 1$.

Recall the definition of left approximate identity: a net $(e_\lambda)\subset A$ such that $\|e_\lambda * a -a\|\to 0$ for all $a\in A$. Then, by left regular representation defined above, we can also write $$\|e_\lambda * a -a\|=\|\hat{e_\lambda}(a)-Id(a)\|=\|(\hat{e_\lambda}-Id)(a)\|\to0.$$ But this can be equivalently stated as

There is a net of bounded operators $(\hat{e_\lambda})\subset \mathcal{B}(A)$ converging to the identity map $Id$ under the strong operator topology (SOT).

So the following proposition is trivially true.

Proposition. Nonunital Banach algebra $A$ admits a left approximate identity $(e_\lambda)$ iff the identity map lies in the SOT-closure of the image of left regular representation: $Id\in cl_{SOT}(\iota(A))$.

Assume the condition above is satisfied, it appears the following additional assumption would suffice as a sufficient condition for the existence of a sequential left approximate identity.

On $\mathcal{B}(A)$, the SOT is first countable. (By the Birkhoff-Kakutani theorem, this is equivalent as the SOT being metrizable)

But, according to this answer, the above condition only occurs when $\mathcal{B}(A)$ is finite dimensional. But this metrizability condition can work if we only restrict the SOT on a norm-bounded set: a norm-closed ball containing the identity map.

Alternatively, since the SOT is weaker than the norm topology, we can simply reduce it to norm-closure.

Theorem If the identity map lies in the norm closure of the image of left regular representation: $Id\in cl_{\|\cdot\|_{\mathcal{B}(A)}}(\iota(A))$, then $A$ admits a sequential left approximate identity.

A similar discussion for the right regular representation and right approximate identity also stands.

Question 1: is the above reasoning correct? Did I miss or misunderstand anything??

Question 2: if it does make sense, are there established/known sufficient conditions for the identity to be in the norm closure of the image???

Answer 1

If $a$ is an element of $A$, then the map $i(a):x\in A\mapsto ax\in A$ is clearly a map of right $A$-modules. A strong limit of maps of right $A$-modules is a map of right $A$-modules, so if the image of $i$ is dense, then all maps in $B(A)$ are right $A$-linear.

Now if $b$ is an element of $A$, then map $j(b):x\in A\mapsto xb\in A$ is bounded, so we see that it is right $A$-linear: this implies that $b$ is central in $A$, and this that the algebra $A$ is commutative.

Since the map $a\in A\mapsto i(a)\in B(A)$ is a map of algebras with dense image, this implies that $B(A)$ is commutative, and therefore $A$ has dimension $1$.

Answer 2

The answer is based on

Lemma For a Banach space, if $T\in B(X)$ is not a multiple of $I,$ there exists $S\in B(X),$ such that $TS\neq ST.$

Let $\dim(A)\ge 2.$ Assume that $l(A)$ is dense in $B(A).$ Then the operators of the form $r(a)x=xa$ commute with all operators in $B(A),$ as they commute with $l(b)$ for any $b\in A.$ Fix $a\in A$ such that $r(a)$ is not a multiple of the identity operator (if such $a$ does not exists the algebra is one-dimensional). Then, applying the lemma there is $S\in B(A)$ such that $Sr(a)\neq r(a)S,$ a contradiction.

Answer 3

Let the Banach Algebra $A$ be finite ($n$) dimensional. In this case the question is does $A$ admit a representation in the form a $n \times n$ matrix over the base field. If $\{v_1,....,v_n\}$ are the basis then we can write left multiplication by an element $a$ as a matrix $[a.v_1,a.v_2,...,a.v_n]$. Hence each element $a \in A$ has a matrix representation: $[a.v_1,a.v_2,...,a.v_n]$. The question is does this matrix uniquely represent an element $a \in A$. If not then $[(a_1-a_2).v_1,(a_1-a_2).v_2,...,(a_1-a_2).v_n] = 0$ implying that $(a_1-a_2).x = 0$, $\forall x \in A$. Since $A$ is non-unital, its not clear that $a_1=a_2$. May be you want to start with this. If $A$ doesnt have zero divisors then $a_1=a_2$ and hence its enough to look at matrix algebras for your case when $A$ doesnt have zero divisors. check under what conditions $a.x = 0$, $\forall x \in A$ implies $a=0$. This is a weaker condition than requiring no zero divisors. Let $a=\sum_i g_i v_i$. Now $a.x = 0$ implies, $a.v_j=\sum_i g_i v_i.v_j = 0$. Hence the matrix $M_k=[v_{ij}]$ with $v_{ij} = v_k.v_j$ is a singular matrix. Under these conditions the matrix representation need not be unique and when $M_k$ for some $k \in [n]$ is non-singular then matrix representation is unique. If matrix representation is possible then your question boils down to analyzing matrix subalgebras over fields. Interesting question.