L1 convolution smooths bounded functions into UCB

Choose Borel representatives of $f$ and $\phi$, and let $M:=\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}$. For each $x\in G$, the preimage of a Borel null set $N$ under $y\mapsto y^{-1}x$ is $x{N}^{-1}$, which is null by inversion and left invariance [F2]; thus changing either representative changes the integrand only on a null set. For every $\eta >0$, the set where $\mathrm{\mid }\phi \mathrm{\mid }>M+\eta$ is null by [F1], so its pullback is null and $y\mapsto f(y)\phi (y^{-1}x)$ is integrable. The integral is therefore defined for every $x$, independent of representatives, and [F4] gives $\mathrm{\mid }(f\ast \phi )(x)\mathrm{\mid }\le (M+\eta )\mathrm{\parallel }f{\mathrm{\parallel }}_1$; letting $\eta \downarrow 0$ yields $\mathrm{\parallel }f\ast \phi {\mathrm{\parallel }}_{\sup }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{1}M$.

First let $f,b\in C_c(G)$, set $K:=\mathrm{supp}f$, $L:=\mathrm{supp}b$, and $C:=KL$, and fix $x\in G$. These are compact sets of finite Haar measure, and $K,C$ are Borel by [F2, F7, F8]. On $K\times C$ the function $\mathrm{\Psi }(y,z):=b(y^{-1}z)$ is continuous. Compactness gives, for each $\delta >0$, a finite Borel partition $E_1,\dots ,E_m$ of $K$ and points $y_j\in K$ such that $\mathrm{\mid }\mathrm{\Psi }(y,z)-\mathrm{\Psi }(y_j,z)\mathrm{\mid }<\delta$ for $y\in E_j$ and $z\in C$. Replacing $\mathrm{\Psi }$ by ${\sum }_j{\mathbf{1}}_{E_j}(y)b(y_j^{-1}z)$ makes the kernel a finite sum of product-measurable terms, so Fubini on the finite restricted measures applies [F3, F7, F8]. The error in either iterated integral is at most $\delta \mathrm{\parallel }f{\mathrm{\parallel }}_1\mu (C)M$, and hence tends to zero with $\delta$. Thus ${\int }_C{\int }_{K}f(y)b(y^{-1}z)\phi (z^{-1}x)d\mu (y)d\mu (z)={\int }_K{\int }_{C}f(y)b(y^{-1}z)\phi (z^{-1}x)d\mu (z)d\mu (y)$. For each fixed $y$, substituting $z=yw$ in the inner integral on the right and using left invariance gives ${\int }_{L}b(w)\phi (w^{-1}{y}^{-1}x)d\mu (w)=(b\ast \phi )(y^{-1}x)$. The left iterated integral is $((f\ast b)\ast \phi )(x)$ by the compactly supported convolution formula, and the right one is $f\ast (b\ast \phi )(x)$. This proves associativity for $f,b\in C_c(G)$.

Linearity of the integral gives bilinearity in $f$ and $\phi$. By substituting $y=gz$ and using left invariance, $(L_{g}f)\ast \phi (x)={\int }_{G}f(z)\phi (z^{-1}{g}^{-1}x)d\mu (z)=(f\ast \phi )(g^{-1}x)=L_g(f\ast \phi )(x)$. Applying the same formula to the difference gives $\mathrm{\mid }{L}_g(f\ast \phi )(x)-(f\ast \phi )(x)\mathrm{\mid }\le \mathrm{\parallel }{L}_{g}f-f{\mathrm{\parallel }}_{1}M$ for every $x$; taking the supremum proves the stated defect estimate.

By [A1], the AC hypothesis of [F5] is met, so $\mathrm{\parallel }{L}_{g}f-f{\mathrm{\parallel }}_1\to 0$ as $g\to e$. Step 2.2 therefore gives $\mathrm{\parallel }{L}_g(f\ast \phi )-f\ast \phi {\mathrm{\parallel }}_{\sup }\to 0$, which is exactly membership in $\mathrm{U}\mathrm{C}\mathrm{B}(G)$ by [F8]. That definition also gives continuity, and step 1.1 gives boundedness.

For general $f,b\in L^1(G)$, [A1] meets the AC hypothesis of [F6], so choose sequences $f_n,b_n\in C_c(G)$ converging in $L^1$. The convolution bound in [F6] and the smoothing bound of step 1.1 imply that both $(f_n\ast b_n)\ast \phi$ and $f_n\ast (b_n\ast \phi )$ converge uniformly, respectively, to $(f\ast b)\ast \phi$ and $f\ast (b\ast \phi )$: each difference is bounded by $(\mathrm{\parallel }{f}_n-f{\mathrm{\parallel }}_1\mathrm{\parallel }{b}_n{\mathrm{\parallel }}_1+\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }{b}_n-b{\mathrm{\parallel }}_1)M$. Since the two expressions agree for every $n$ by step 2.1, their limits agree, proving associativity for arbitrary $L^1$ data. Together with steps 1.1–3.1 this proves the remaining assertions.