Let \(M \subseteq L^1(\mathbb{R})\) be a closed, translation-invariant subspace. (1) Prove that \(M\) is a closed ideal in the convolution Banach algebra \((L^1(\mathbb{R}), *)\). (2) Explain the connection with Wiener’s Tauberian Theorem and the Fourier transform. (3) Discuss its relation to \(L^2(\mathbb{R})\) as a convolution module / algebra and the Plancherel transform.
Solution
1. Let \(\tau_y f(x)=f(x-y)\). If \(f\in M\) and \(g\in L^1(\mathbb R)\), then\[ g*f=\int_{\mathbb R} g(y)\,\tau_y f\,dy \]as a Bochner integral in \(L^1(\mathbb R)\).
The map \(y\mapsto \tau_y f\) is continuous from \(\mathbb R\) to \(L^1(\mathbb R)\), hence strongly measurable. Moreover \[ \|g(y)\tau_yf\|_1=|g(y)|\,\|f\|_1, \] which is integrable in \(y\). Thus the Bochner integral exists. Its value is the usual convolution by Fubini–Tonelli.
2. Since \(M\) is translation invariant and closed, \(g*f\in M\). Hence \(M\) is aclosed ideal of the commutative Banach algebra \((L^1(\mathbb R),*)\).
For every \(y\), \(\tau_yf\in M\), so the integrand in 1 is \(M\)-valued. A Bochner-integrable map with values in a closed linear subspace has its integral in that subspace: approximate it in \(L^1(\mathbb R;L^1(\mathbb R))\) by simple \(M\)-valued functions. Thus \(g*f\in M\). Closedness is part of the hypothesis, and convolution on \(L^1(\mathbb R)\) is commutative.
3. Fourier transform converts the ideal structure into multiplication:\[ \widehat{g*f}(\xi)=\widehat g(\xi)\widehat f(\xi). \]For a single \(f\in L^1(\mathbb R)\), Wiener’s Tauberian theorem says that the closedlinear span of its translates is all of \(L^1(\mathbb R)\) if and only if \(\widehat f\)is nowhere zero.
The convolution identity is the standard Fubini calculation for \(L^1\) functions. Wiener’s theorem is precisely the cyclicity criterion for the translation representation on \(L^1(\mathbb R)\).
4. Consequently, if \(M\) contains an \(f\) with nowhere-vanishing Fourier transform,then \(M=L^1(\mathbb R)\). More generally, if\[ Z(M)=\{\xi\in\mathbb R:\widehat f(\xi)=0\text{ for every }f\in M\}, \]then \(M\) is proper only if \(Z(M)\neq\varnothing\).
The first statement follows from 3 because \(M\) is closed and contains every translate of \(f\). For the second, a proper closed ideal in a commutative Banach algebra is contained in a maximal ideal. The characters of \(L^1(\mathbb R)\) are the Fourier evaluations \(f\mapsto\widehat f(\xi)\), so containment in a maximal ideal gives a common zero \(\xi\).
5. The zero set \(Z(M)\) is a spectral invariant, but it does notclassify closed ideals of \(L^1(\mathbb R)\).
Complete recovery of closed ideals from their hulls is the spectral-synthesis problem. For non-discrete locally compact abelian groups, spectral synthesis fails in general, so distinct closed ideals can have the same hull.
6. If \(a\in L^1(\mathbb R)\) and \(h\in L^2(\mathbb R)\), then\[ \|a*h\|_2\le \|a\|_1\|h\|_2, \]so \(L^2(\mathbb R)\) is a Banach module over \(L^1(\mathbb R)\).
This is Young’s convolution inequality in the \(L^1*L^2\to L^2\) case.
7. \(L^2(\mathbb R)\) is not closed under convolution.
Let \(u(x)=(1+|x|)^{-2/3}\). Then \(u\in L^2(\mathbb R)\). For \(x\ge 3\), restricting the convolution integral to \(y\in[x/3,2x/3]\) gives \[ (u*u)(x)\ge Cx\cdot x^{-2/3}x^{-2/3}=Cx^{-1/3}. \] Hence \(u*u\notin L^2(\mathbb R)\). Thus convolution does not make all of \(L^2\) into an algebra.
8. Under the unitary Fourier–Plancherel transform\(\mathcal F:L^2(\mathbb R)\to L^2(\mathbb R)\), the \(L^1\)-module action becomesmultiplication:\[ \mathcal F(a*h)=\widehat a\,\mathcal Fh. \]
First prove the identity for \(h\in L^1\cap L^2\). Since \(\widehat a\in L^\infty\) and convolution by \(a\) is bounded on \(L^2\) by 6, density of \(L^1\cap L^2\) in \(L^2\) extends the identity to every \(h\in L^2\).
9. Closed translation-invariant subspaces of \(L^2(\mathbb R)\) are exactly theinverse Fourier transforms of spaces \(L^2(E)\) for measurable sets\(E\subseteq\mathbb R\).
Translations become multiplication by the characters \(e^{-iy\xi}\). If \(V\subseteq L^2\) is closed and invariant under every translation, it is also invariant under every inverse translation, so its orthogonal projection commutes with all character multipliers. The von Neumann algebra generated by those multipliers is the algebra of multiplication operators by \(L^\infty\)-functions. Hence the projection is multiplication by an idempotent \(L^\infty\)-function, necessarily \(1_E\) for a measurable set \(E\). Therefore \(\mathcal F(V)=L^2(E)\).
10. Q.E.D.
Part (1) is step 2; the Fourier transform and Wiener’s theorem in part (2) are steps <1>3–<1>5; the \(L^2\) statements in part (3) are steps <1>6–<1>9.