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  "body": "\n## Definition\n\nLet $A$ be a Banach algebra, not necessarily with a unit.\nA *left approximate identity* of $A$ is a [[net|4DVJQJV0DG]] $(e_\\lambda)_{\\lambda \\in \\Lambda}$ in $A$ such that $\\norm{e_\\lambda a - a} \\to 0$ for every $a \\in A$.\nA *right approximate identity* is defined the same way with $a e_\\lambda$ in place of $e_\\lambda a$.\nAn *approximate identity* (or two-sided approximate identity) is a net that is both a left and a right approximate identity.\nIt is *bounded* if $\\sup_\\lambda \\norm{e_\\lambda} \u003c \\infty$.\n\n## Remarks\n\nExample: let $A = C_0(\\R)$, the continuous functions on $\\R$ that tend to $0$ at infinity, with pointwise product and the sup norm.\nFor $n \\ge 1$ let $e_n \\in A$ be continuous with $0 \\le e_n \\le 1$, $e_n = 1$ on $[-n, n]$ and $e_n = 0$ outside $[-n-1, n+1]$.\nFor $f \\in A$, $\\norm{e_n f - f}_\\infty \\le \\sup_{\\abs{x} \u003e n} \\abs{f(x)} \\to 0$.\nSo $(e_n)$ is a bounded approximate identity, with $\\norm{e_n} = 1$, although $A$ has no unit.\n\nNon-example: in $L^1(\\R)$ with convolution, let $e_n = \\tfrac{1}{2n} \\mathbf{1}_{[-n, n]}$.\nFor $f \\in L^1(\\R)$, $\\abs{(e_n * f)(x)} \\le \\norm{f}_1 / (2n)$, so $e_n * f \\to 0$ uniformly.\nIf also $e_n * f \\to f$ in $L^1$, a subsequence would converge to $f$ almost everywhere, so $f = 0$.\nThus the convergence condition fails for every $f \\ne 0$.\nSpikes $u_n = \\tfrac{n}{2} \\mathbf{1}_{[-1/n, 1/n]}$ do give a bounded approximate identity of $L^1(\\R)$.\n\nA unit $1$ of $A$ gives the constant approximate identity $e_\\lambda = 1$.\nEvery C*-algebra has an approximate identity of positive elements of norm at most $1$.\n",
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