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The scaling property of the Fourier transform extends by duality toConsequently a homogeneous distribution of degree $\sigma$ transforms into one of degree $-n-\sigma$.

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 $|\mathcal{F}K(v)|\le \Vert K{\Vert }_1$. The condition $\int K=1$ does not imply $\Vert K{\Vert }_1=1$, because $K$ need not be nonnegative.

Part (b), now for observations of $Y$, gives $\mathbb{E}|{\phi }_n^Y(u)-{\phi }^Y(u){|}^2\le 1/n$. Apply the Tonelli theorem to this nonnegative integrand, use the bound on $1/{\phi }_{ϵ}$, and change variables $v=hu$:The characteristic function ${\phi }_{ϵ}$ is continuous and nonzero on the compact interval of integration, so ${\tilde{M}}_h<\infty$ for every $h>0$. The Markov inequality gives, for every $z>0$,Consequently $J_{n,h}^{1/2}=O_{\mathbb{P}}({\tilde{M}}_h/\sqrt{nh})$, including for any deterministic bandwidth sequence $h=h_n>0$. The stochastic order follows directly from this uniform probability bound. Independence of $X$ and $ϵ$ gives ${\phi }^Y={\phi }^X{\phi }_{ϵ}$, explaining why division by ${\phi }_{ϵ}$ is the Fourier deconvolution operation.

The scaling property of the Fourier transform gives, first for Schwartz functions and then by duality,If the homogeneous distribution $u$ has degree $\sigma$, thenPutting $s=1/t$ yields ${\delta }_s\hat{u}=s^{-n-\sigma }\hat{u}$. Thus the Fourier transform of a homogeneous distribution has degree $\boxed{-n-\sigma }$.

Choose a nonzero $f\in L^p({\mathbb{R}}^n)\cap L^1({\mathbb{R}}^n)$ whose transform belongs to $L^q$, for example a Gaussian by the Fourier transform of a Gaussian. For $\lambda >0$, set $f_{\lambda }(x)=f(\lambda x)$. The scaling property of the Fourier transform givesThe assumed estimate, applied for every $\lambda >0$, becomesSince both norms are nonzero, this can hold as $\lambda$ tends both to zero and to infinity only if the powers agree. Thus the scaling necessity for an Lp to Lq Fourier bound yieldsHence $q$ is uniquely the conjugate exponent of $p$.