L² convolution on a compact group is Hilbert–Schmidt

Let $u\in C(K)$ and put $q_u(x,y):=u(x{y}^{-1})$. Inversion and multiplication are continuous, so $q_u$ is continuous on the compact space $K\times K$.

For each $ϵ>0$, continuity gives an open-rectangle cover of $K\times K$ on each member of which $q_u$ varies by less than $ϵ$; compactness gives a finite subcover. Disjointify its rectangles by successive differences and on each nonempty piece use a value of $q_u$ from its rectangle. These pieces are measurable in the product sigma-algebra, so this gives a measurable simple function within $ϵ$ of $q_u$ everywhere. Taking $ϵ=1\mathrm{/}n$ and applying pointwise-limit measurability to the real and imaginary parts shows that $q_u$ is product-measurable.

For fixed $x$, the map $y\mapsto x{y}^{-1}$ preserves $\mu$ by [F5]. Tonelli applied to the measurable function $\mathrm{\mid }{q}_u{\mathrm{\mid }}^2$ therefore gives $\mathrm{\parallel }{q}_u{\mathrm{\parallel }}_{L^2(\mu \times \mu )}^2={\int }_K{\int }_K\mathrm{\mid }u(x{y}^{-1}){\mathrm{\mid }}^{2}d\mu (y)d\mu (x)=\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$.

The map $u\mapsto q_u$ is an isometry from $C(K)$ into $L^2$ of the completed product measure by step 3.1. Since $C(K)$ is dense in $L^2(K)$ and the completed-product $L^2$ space is complete, it extends to an isometry $U:L^2(K)\to L^2(\overline{\mu \times \mu })$. Choose $u_n\in C(K)$ with $\mathrm{\parallel }f-u_n{\mathrm{\parallel }}_2\to 0$ and put $\kappa :=Uf={\lim }_n{q}_{u_n}$; then $\mathrm{\parallel }\kappa {\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_2$, and the class $\kappa$ is independent of the approximating sequence.

Choose measurable representatives of $f$ and $h\in L^2(K)$ and define $C_{f}h(x):={\int }_{K}f(x{y}^{-1})h(y)d\mu (y)$. For each $x$, the change of variables $y\mapsto x{y}^{-1}$ preserves $\mu$, so Cauchy–Schwarz makes this integral finite and gives $\mathrm{\mid }{C}_{f}h(x)-C_{u_n}h(x)\mathrm{\mid }\le \mathrm{\parallel }f-u_n{\mathrm{\parallel }}_2\mathrm{\parallel }h{\mathrm{\parallel }}_2$ for every continuous approximant $u_n$ chosen in step 4.1, uniformly in $x$. Each $C_{u_n}h$ is continuous: joint continuity of $q_{u_n}$ from step 1.1 and compactness of $K$ give ${\sup }_y\mathrm{\mid }{q}_{u_n}(x,y)-q_{u_n}(x_0,y)\mathrm{\mid }\to 0$ as $x\to x_0$, and $h\in L^1(K)$ since $\mu (K)=1$. Thus $C_{f}h$ is a measurable uniform limit. By [F2], $T_{q_{u_n}}=C_{u_n}$ and $\mathrm{\parallel }{T}_{q_{u_n}}-T_{\kappa }\mathrm{\parallel }\le \mathrm{\parallel }{q}_{u_n}-\kappa {\mathrm{\parallel }}_2\to 0$, so the uniform limit also gives $C_{f}h=T_{\kappa }h$ as an $L^2$ class.

The kernel theorem [F2] makes $T_{\kappa }$ Hilbert–Schmidt with $\mathrm{\parallel }{T}_{\kappa }{\mathrm{\parallel }}_{HS}=\mathrm{\parallel }\kappa {\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_2$. By step 5.1 this operator is exactly $C_f$, so $C_f$ is bounded and has the asserted Hilbert–Schmidt norm.

If $f$ is changed on a null set $N$, then for each $x$ the convolution integrand changes only for $y\in N^{-1}x$, since $x{y}^{-1}\in N$ iff $y\in N^{-1}x$; this set is null by inversion and right invariance of $\mu$. Changing the representative of the input $h$ also leaves every section integral unchanged. Thus $C_f$ is representative-independent; since Countable Choice holds, [F3] makes this Hilbert–Schmidt operator compact.