= Solution
\b[This <covariance> is also zero, but symmetry is essential to the proof.] Write $\varepsilon_t=\eta_t|\varepsilon_t|$, where $\eta_t$ is an independent fair sign. The <normal distribution> makes that sign independent of its magnitude and of every other driving variable. Conditional on the magnitudes and all noise except this sign, changing $\eta_t$ flips $X_t$ and leaves $X_t^2$ unchanged. Every future conditional scale uses only squared past observations, so $X_{t+h}$ is unchanged for $h>0$.
Thus $f(X_{t+h})$ is independent of the remaining fair sign in $X_t$. Averaging that sign gives $\mathbb E[X_tf(X_{t+h})]=0$, and hence
$$
\boxed{\operatorname{Cov}(X_t,f(X_{t+h}))=0.}
$$
The <Cauchy-Schwarz inequality> again guarantees integrability. For odd $h$, the <parity decomposition of a lag-two ARCH process> also gives <independence> directly. For even $h$, the <sign symmetry of an ARCH process> supplies the argument; the <martingale difference sequence> property alone would not justify this reversed <covariance>.
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