= Solution
The implication is true in every dimension despite the failure of part (iii). The hypothesis gives $\mathcal Ff\in L^2$, since its weight is at least one. Let $g$ be its inverse Fourier-Plancherel transform; then $g\in H^s$ by construction.
Use the <Gaussian approximate identity> $G_\varepsilon(x)=\varepsilon^{-n}e^{-\pi|x|^2/\varepsilon^2}$. Its <Fourier transform> is $e^{-\pi\varepsilon^2|\xi|^2}$. <Fourier inversion> of this integrable multiplier times the bounded transform of $f$, justified by Fubini, gives the same function both as $G_\varepsilon*f$ and as $G_\varepsilon*g$. The first converges to $f$ in $L^1$, while the second converges to $g$ in $L^2$ by Plancherel and dominated convergence. Along a common sequence, select subsequences converging <almost everywhere> for both norms. Their limits agree, giving $f=g$ <almost everywhere>. Consequently
$$
\boxed{f\in H^s(\mathbb R^n),\qquad\widehat f=\mathcal Ff\text{ almost everywhere}.}
$$
This proves the requested conclusion without relying on the incorrect general-dimensional kernel assertion.
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