Does \(L^1\) have a natural multiplication making it an algebra? Distinguish pointwise multiplication on a general measure space from convolution on a locally compact group.
Solution
There is no single answer for a bare symbol \(L^1\): the multiplication depends on the underlying measure-theoretic structure.
1. Pointwise multiplication does not make a general \(L^1(X,\mu)\) into an algebra.If \(f,g\in L^1\), the product \(fg\) need not lie in \(L^1\). For example, on \((0,1)\) with Lebesgue measure,\[ f(x)=x^{-2/3} \]lies in \(L^1(0,1)\), but\[ f(x)^2=x^{-4/3}\notin L^1(0,1). \]Thus \(L^1(X,\mu)\) is not generally closed under pointwise multiplication.
A standard pointwise algebra is instead \[ L^1\cap L^\infty, \] since \[ \|fg\|_1\le \|f\|_1\|g\|_\infty. \]
2. Convolution gives the canonical \(L^1\) algebra on a locally compact group.Let \(G\) be a locally compact group with a left Haar measure. For \(f,g\in L^1(G)\) define\[ (f*g)(x)=\int_G f(y)g(y^{-1}x)\,dy. \]Then \(f*g\in L^1(G)\) and Young’s inequality gives\[ \|f*g\|_1\le \|f\|_1\|g\|_1. \]Convolution is associative, so\[ L^1(G) \]is a Banach algebra under convolution.
If \(G\) is discrete, this specializes to the usual convolution algebra \(\ell^1(G)\). If \(G\) is abelian, the convolution algebra is commutative.
Thus a general \(L^1\) space has no canonical multiplication from the notation alone, while group structure plus Haar measure supplies the natural convolution product.