= Solution
Write $J_{n,h}$ for the squared integral norm in question. The <scaling property of the Fourier transform> gives $\mathcal F K_h(u)=\mathcal F K(hu)$, which vanishes outside $[-1/h,1/h]$. Also $|\mathcal F K(v)|\leq\|K\|_1$. The condition $\int K=1$ does not imply $\|K\|_1=1$, because $K$ need not be nonnegative.
Part (b), now for observations of $Y$, gives $\mathbb E|\varphi_n^Y(u)-\varphi^Y(u)|^2\leq1/n$. Apply the <Tonelli theorem> to this nonnegative integrand, use the bound on $1/\varphi_\epsilon$, and change variables $v=hu$:
$$
\mathbb E J_{n,h}
\leq\frac{\widetilde M_h^2}{n}\int_{-1/h}^{1/h}|\mathcal F K(hu)|^2\,du
=\frac{\widetilde M_h^2}{nh}\int_{-1}^1|\mathcal F K(v)|^2\,dv
\leq\frac{2\|K\|_1^2\widetilde M_h^2}{nh}.
$$
The <characteristic function> $\varphi_\epsilon$ is continuous and nonzero on the compact interval of integration, so $\widetilde M_h<\infty$ for every $h>0$. The <Markov inequality> gives, for every $z>0$,
$$
\mathbb P\left(J_{n,h}^{1/2}>z\frac{\widetilde M_h}{\sqrt{nh}}\right)
\leq\frac{2\|K\|_1^2}{z^2}.
$$
Consequently \b[$J_{n,h}^{1/2}=O_{\mathbb P}(\widetilde M_h/\sqrt{nh})$], including for any deterministic bandwidth sequence $h=h_n>0$. The <stochastic order> follows directly from this uniform probability bound. Independence of $X$ and $\epsilon$ gives $\varphi^Y=\varphi^X\varphi_\epsilon$, explaining why division by $\varphi_\epsilon$ is the <Fourier deconvolution> operation.
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