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We note that $L^{1}(\mathbb R)$ is an algebra with respect to convolution without identity element. However, regarded as a Banach algebra, $(L^{1}(\mathbb R), \ast) $ has a bounded approximate identity with respect to convolution, that is, there is a set $\{e_{r}\}_{r>0}\subset L^{1}$ such that $\|e_{r}\|_{L^{1}}\leq C$ for all $r>0$ and $C$ is some constant and $\|e_{r}\ast f- f\|_{L^{1}} \to 0$ as $r\to 0$ for $f\in L^{1}.$

My Questions are: (1) What are other examples of Banach algebras (preferably function spaces) that have a bounded approximate identity? (2) Is $L^{1}$ the only convolution algebra which has a bounded approximate identity?

Tomasz Kania
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Inquisitive
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    A non-function space: $A:= \mathcal{K}(\ell^2)$, the space of compact operators on $\ell^2$. For each $n\in \mathbb{N}$, let $P_n$ be the projection onto the first $n$ components, then ${P_n}$ forms an approximate identity for $A$. – Prahlad Vaidyanathan Aug 18 '15 at 16:19
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    Consider $\mathcal{A}=C_{0}(\mathbb{R})$ consisting of all continuous functions on $\mathbb{R}$ that vanish at $\pm\infty$. – Disintegrating By Parts Aug 18 '15 at 20:38

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Well, of course if $G$ is any locally compact abelian group then $L^1(G)$ is a Banach algebra under convolution. If $G$ is not discrete then $L^1(G)$ has no identity, while if $G$ is first-countable then there is a bounded approximate identity. (If $G$ is not first countable there's still a net that gives a bounded approximate identity, but perhaps not a sequence.)

Say $K$ is a locally compact Hausdorff space, $K$ is not compact, but $K$ is a countable union of compact sets. Let $A=C_0(K)$, the space of functions that vanish at infinity. Then $A$ is a Banach algebra (with pointwise multiplication) with no identity but with a bounded approximate identity. (Again, if you settle for a net instead of a sequence you don't need to assume that $K$ is a countable union of compact sets.)

David C. Ullrich
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  • Every (non-unital) C*-algebra has a bounded approximate identity consisting of self-adjoint elements.

  • Every amenable Banach algebra has a bounded approximate identity and this class of algebras is quite substantial.

  • If $X$ is a Banach space with the bounded approximation property, then the algebra $\mathscr{K}(X)$ of compact operators on $X$ has a bounded left approximate identity.

See

H.G. Dales, Banach Algebras and Automatic Continuity, Clarenton Press, 2001.

Tomasz Kania
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I am interested in the Sobolev algebra W^{1,2} of absolutely continuous functions f on the non negative half line such that both f and f' are square integrable. This is a Banach algebra without a unit under pointwise multiplication, and it does have an approximate identity.

However, there is no bounded approximate identity in W^{1,2}.

As a matter of fact, if A is a Banach algebra without a unit, that is reflexive regarded as a Banach space, and whose product is separately weakly continuous, then every approximate identity in A is unbounded. The reason is that for a bounded approximate identity (e_n) there would be a weakly convergent subsequence, still an approximate identity, that would converge to a unit, a contradiction.

Miguel Lacruz
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