Examples of algebras that have a bounded approximate identity 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 $\{e_{r}\}_{r&gt;0}\subset L^{1}$ such that $\|e_{r}\|_{L^{1}}\leq C$ for all $r&gt;0$ 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?
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.)
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.