 Solution

This covariance is also zero, but symmetry is essential to the proof. Write ${\epsilon }_t={\eta }_t|{\epsilon }_t|$, 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 $\mathbb{E}[X_{t}f(X_{t+h})]=0$, and henceThe 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.