Compact convolution operators commute with right translations and have conjugate-kernel adjoints

Fix $g\in K$ and $h\in L^2(K)$; for every $x$ the function $y\mapsto \phi (x{y}^{-1})h(yg)$ is integrable by Cauchy–Schwarz and Haar invariance, and the substitution $y^{\mathrm{\prime }}=yg$ together with right invariance of $\mu$ and the identity $x(y^{\mathrm{\prime }}{g}^{-1}{)}^{-1}=xg{y}^{\mathrm{\prime }-1}$ gives $(C_{\phi }\rho (g)h)(x)={\int }_K\phi (x{y}^{-1})h(yg)d\mu (y)={\int }_K\phi (xg{y}^{\mathrm{\prime }-1})h(y^{\mathrm{\prime }})d\mu (y^{\mathrm{\prime }})=(\rho (g)C_{\phi }h)(x)$ for almost every $x\in K$, whence $C_{\phi }\rho (g)=\rho (g)C_{\phi }$ in $L^2(K)$; this is (1).

First suppose $\phi \in C(K)$. Its kernel $(x,y)\mapsto \phi (x{y}^{-1})$ is product-measurable by the finite-rectangle product-measurability argument in L² convolution on a compact group is Hilbert–Schmidt. For $h_1,h_2\in L^2(K)$ the function $(x,y)\mapsto \phi (x{y}^{-1})h_1(y)\overline{h_2(x)}$ lies in $L^1(K\times K)$ because its absolute integral is at most $\mathrm{\parallel }\phi {\mathrm{\parallel }}_2\mathrm{\parallel }{h}_1{\mathrm{\parallel }}_2\mathrm{\parallel }{h}_2{\mathrm{\parallel }}_2$ and $\mu$ is a probability, so Fubini and the definition of $C_{\phi }$ give $⟨C_{\phi }{h}_1,h_2⟩={\int }_K{\int }_K\phi (x{y}^{-1})h_1(y)\overline{h_2(x)}d\mu (y)d\mu (x)$.

By definition of ${\phi }^{\ast }$ one has $\overline{{\phi }^{\ast }(y{x}^{-1})}=\phi (x{y}^{-1})$ for all $x,y\in K$, so expanding the definition of $C_{{\phi }^{\ast }}$ in the second slot and applying Fubini to the same $L^1$ function as in step 1.2 gives $⟨h_1,C_{{\phi }^{\ast }}{h}_2⟩={\int }_K{\int }_K{h}_1(y)\overline{{\phi }^{\ast }(y{x}^{-1})}\overline{h_2(x)}d\mu (x)d\mu (y)={\int }_K{\int }_K\phi (x{y}^{-1})h_1(y)\overline{h_2(x)}d\mu (y)d\mu (x)=⟨C_{\phi }{h}_1,h_2⟩$ for all $h_1,h_2\in L^2(K)$.

For continuous $\phi$, the identity of step 2.1 exhibits $C_{{\phi }^{\ast }}$ as an adjoint of the bounded operator $C_{\phi }$, so $C_{\phi }^{\ast }=C_{{\phi }^{\ast }}$ by uniqueness of the Hilbert adjoint. For general $\phi \in L^2(K)$ choose ${\phi }_n\in C(K)$ converging in $L^2$ to $\phi$, as in the convolution supplier [F4]. Haar inversion gives $\mathrm{\parallel }{\phi }_n^{\ast }-{\phi }^{\ast }{\mathrm{\parallel }}_2=\mathrm{\parallel }{\phi }_n-\phi {\mathrm{\parallel }}_2$, and Cauchy–Schwarz gives $\mathrm{\parallel }{C}_a\mathrm{\parallel }\le \mathrm{\parallel }a{\mathrm{\parallel }}_2$. Hence $C_{{\phi }_n}\to C_{\phi }$ and $C_{{\phi }_n^{\ast }}\to C_{{\phi }^{\ast }}$ in operator norm; the adjoint norm identity in [F8] passes $C_{{\phi }_n}^{\ast }=C_{{\phi }_n^{\ast }}$ to the limit. Thus $C_{\phi }^{\ast }=C_{{\phi }^{\ast }}$ for every $L^2$ kernel. If ${\phi }^{\ast }=\phi$ almost everywhere, independence of the representative gives $C_{{\phi }^{\ast }}=C_{\phi }$, and $C_{\phi }$ is self-adjoint.

Let $z\in \ker (C_{\phi }-\lambda I)$ for some $\lambda \in \mathbb{C}$ and let $g\in K$; by (1) the operators $C_{\phi }$ and $\rho (g)$ commute, hence $C_{\phi }(\rho (g)z)=\rho (g)C_{\phi }z=\lambda \rho (g)z$, so $\rho (g)z$ lies in the same eigenspace, and applying this to $g^{-1}$ gives $\rho (g)\ker (C_{\phi }-\lambda I)=\ker (C_{\phi }-\lambda I)$; the case $\lambda =0$ is the kernel $\ker C_{\phi }$. The Axiom of Choice is consumed through the normalized Haar probability and the cited Hilbert-space and adjoint suppliers; the computations above are choice-free apart from those inputs.