Bounded approximate identity of L1(G)

Remarks

Functions $e_U$ exist: Urysohn's lemma gives a nonzero continuous $g \ge 0$ vanishing outside $U$, and $e_U$ is a multiple of $g(x) + g(x^{-1})$. The idea of the proof: $e_U * f = \int_G e_U(y) \, L_y f \, d\mu(y)$ is an average of left translates $L_y f$ with $y \in U$. Hence $\norm{e_U * f - f}_1 \le \sup_{y \in U} \norm{L_y f - f}_1$, which tends to $0$ by the continuity of translation in $L^1(G)$. The right-hand case is similar, using right translations and the symmetry of $e_U$.

Example: on $G = \R$, the functions $e_n = \frac{n}{2} \mathbf{1}_{[-1/n, 1/n]}$, or continuous bumps of width $2/n$, give $e_n * f \to f$ in $L^1(\R)$. This is the familiar smoothing of a function by averaging over shrinking intervals.

The group algebra has a unit exactly when $G$ is discrete, and then the unit is the point mass at the identity. So for groups like $\R$ the approximate identity is the best substitute for a unit.

Going deeper

For a student

Bounded approximate identities are the standard replacement for a unit in Banach algebra arguments. With Cohen's factorization theorem, the result gives $L^1(G) = L^1(G) * L^1(G)$. It is the key step in proving that every bounded left multiplier of $L^1(G)$ is convolution by a measure (Wendel's theorem). It also shows that closed left ideals of $L^1(G)$ are exactly the closed left-translation-invariant subspaces.