Proof. Fix \(\psi\in\mathscr S(\mathbb{R}^d)\). By definition of \(p_{N,0}\), \[|\psi(\omega)| \le p_{N,0}(\psi)(1+\|\omega\|_2)^{-N}, \qquad \omega\in\mathbb{R}^d .\] Therefore \[|Q(\omega)\psi(\omega)| \le p_{N,0}(\psi) \frac{|Q(\omega)|}{(1+\|\omega\|_2)^N}.\] The right-hand side is integrable by assumption, so \(\mathcal{I}_Q[\psi]\) is well-defined. Moreover, \[|\mathcal{I}_Q[\psi]| \le A_N p_{N,0}(\psi).\] This seminorm bound proves that \(\mathcal{I}_Q\) is continuous on \(\mathscr S(\mathbb{R}^d)\). Hence \(\mathcal{I}_Q\in\mathscr S'(\mathbb{R}^d)\). ◻

Proof. By independence, \[P_Y = P_X * P_V .\] Let \(A\subseteq\mathbb{R}^d\) be Borel. Since \(P_V(dv)=f_V(v)\,dv\), Tonelli’s theorem gives \[P_Y(A) = \iint \mathbf{1}_A(x+v) f_V(v)\,dv\,dP_X(x).\] For each fixed \(x\), the change of variables \(y=x+v\) gives \[\int \mathbf{1}_A(x+v) f_V(v)\,dv = \int_A f_V(y-x)\,dy .\] Hence \[P_Y(A) = \int_A \left\{ \int_{\mathbb{R}^d} f_V(y-x)\,dP_X(x) \right\} dy .\] Therefore \(P_Y\) is absolutely continuous with density \[f_Y(y) = \int_{\mathbb{R}^d} f_V(y-x)\,dP_X(x).\] ◻

Proof. Surjectivity follows from the definition of \(\mathcal{A}_V(\mathbb{R}^d)\): \[\mathcal{A}_V(\mathbb{R}^d) = \mathcal{K}_V\bigl(\mathcal{M}(\mathbb{R}^d)\bigr).\]

It remains to prove injectivity. Since \(\mathcal{K}_V\) is linear, it is enough to show that its kernel is trivial. Let \(\mu\in\mathcal{M}(\mathbb{R}^d)\) satisfy \[\mathcal{K}_V[\mu] = f_V * \mu = 0 \qquad \text{in } L^1(\mathbb{R}^d;\mathbb{C}).\] Let \[\varphi_\mu(\omega) \mathrel{\mathop{:}}= \int_{\mathbb{R}^d} e^{i\omega^\top x}\,d\mu(x), \qquad \omega\in\mathbb{R}^d,\] be the Fourier–Stieltjes transform of \(\mu\). Since \(\mu\) is finite, \(\varphi_\mu\) is bounded and continuous, with \[|\varphi_\mu(\omega)| \le \|\mu\|_{TV}, \qquad \omega\in\mathbb{R}^d .\]

Taking Fourier transforms yields \[\widetilde{\mathcal{K}_V[\mu]}(\omega) = \widetilde{(f_V * \mu)}(\omega) = \widetilde{f}_V(\omega)\varphi_\mu(\omega) = \varphi_V(\omega)\varphi_\mu(\omega), \qquad \omega\in\mathbb{R}^d .\] Here \(\widetilde{f}_V=\varphi_V\) because \(f_V\) is the density of \(P_V\). Since \(\mathcal{K}_V[\mu]=0\) in \(L^1(\mathbb{R}^d;\mathbb{C})\), its Fourier transform vanishes identically. Thus \[\varphi_V(\omega)\varphi_\mu(\omega) = 0, \qquad \omega\in\mathbb{R}^d .\]

Define \[D \mathrel{\mathop{:}}= \{\omega\in\mathbb{R}^d:\varphi_V(\omega)\neq 0\}.\] By Assumption 2, \(D^c\) has Lebesgue measure zero. Hence \(D\) is dense in \(\mathbb{R}^d\), because no nonempty open ball can be contained in \(D^c\). On \(D\), \[\varphi_\mu(\omega)=0.\] The continuity of \(\varphi_\mu\) extends this equality to all of \(\mathbb{R}^d\): \[\varphi_\mu(\omega)=0, \qquad \omega\in\mathbb{R}^d .\] By uniqueness of Fourier–Stieltjes transforms for finite complex measures, \(\mu=0\). Therefore \(\mathcal{K}_V\) is injective. ◻

Proof. Define the linear functional \[I_\lambda:\mathcal{M}(\mathbb{R}^d)\to\mathbb{C}, \qquad I_\lambda[\mu] \mathrel{\mathop{:}}= \int_{\mathbb{R}^d}\lambda(x)\,d\mu(x).\] Since \(\lambda\in C_0(\mathbb{R}^d;\mathbb{C})\), it is bounded. Hence \[|I_\lambda[\mu]| \le \|\lambda\|_\infty \|\mu\|_{TV}, \qquad \mu\in\mathcal{M}(\mathbb{R}^d),\] so \(I_\lambda\) is continuous with respect to the total variation norm. By Proposition 2, the map \[\mathcal{K}_V: \bigl(\mathcal{M}(\mathbb{R}^d),\|\cdot\|_{TV}\bigr) \to \bigl(\mathcal{A}_V(\mathbb{R}^d),\|\cdot\|_{\mathcal{A}_V}\bigr)\] is an isometric isomorphism. We may therefore define \[\mathcal{T}_{\lambda,V} \mathrel{\mathop{:}}= I_\lambda\circ \mathcal{K}_V^{-1}.\] Equivalently, if \(f\in\mathcal{A}_V(\mathbb{R}^d)\) and \(\mu_f=\mathcal{K}_V^{-1}[f]\), then \[\mathcal{T}_{\lambda,V}[f] = \int_{\mathbb{R}^d}\lambda(x)\,d\mu_f(x).\] This definition is unambiguous because the representing measure \(\mu_f\) is unique.

The same bound proves continuity. Indeed, for \(f\in\mathcal{A}_V(\mathbb{R}^d)\), let \(\mu_f=\mathcal{K}_V^{-1}[f]\). Then \[|\mathcal{T}_{\lambda,V}[f]| = |I_\lambda[\mu_f]| \le \|\lambda\|_\infty \|\mu_f\|_{TV} = \|\lambda\|_\infty \|f\|_{\mathcal{A}_V}.\] Thus \[\mathcal{T}_{\lambda,V}\in\mathcal{A}'_V(\mathbb{R}^d).\]

Now let \(\mu\in\mathcal{M}(\mathbb{R}^d)\). Since \(\mathcal{K}_V^{-1}[\mathcal{K}_V[\mu]]=\mu\), \[\mathcal{T}_{\lambda,V}\!\left[\mathcal{K}_V[\mu]\right] = I_\lambda[\mu] = \int_{\mathbb{R}^d}\lambda(x)\,d\mu(x).\] This proves the representation formula. Under Assumptions 1 and 3, Proposition 1 gives \[f_Y = f_V * P_X = \mathcal{K}_V[P_X].\] Applying the representation formula with \(\mu=P_X\) yields \[\mathcal{T}_{\lambda,V}[f_Y] = \mathcal{T}_{\lambda,V}\!\left[\mathcal{K}_V[P_X]\right] = \int_{\mathbb{R}^d}\lambda(x)\,dP_X(x) = \Lambda_\lambda(P_X).\]

It remains to prove uniqueness. Let \(S\in\mathcal{A}'_V(\mathbb{R}^d)\) satisfy \[S\!\left[\mathcal{K}_V[\mu]\right] = \int_{\mathbb{R}^d}\lambda(x)\,d\mu(x), \qquad \mu\in\mathcal{M}(\mathbb{R}^d).\] Since \(\mathcal{K}_V\) is onto \(\mathcal{A}_V(\mathbb{R}^d)\), every \(f\in\mathcal{A}_V(\mathbb{R}^d)\) can be written as \(f=\mathcal{K}_V[\mu]\) for some \(\mu\in\mathcal{M}(\mathbb{R}^d)\). Therefore \[S[f] = S\!\left[\mathcal{K}_V[\mu]\right] = \int_{\mathbb{R}^d}\lambda(x)\,d\mu(x) = \mathcal{T}_{\lambda,V}\!\left[\mathcal{K}_V[\mu]\right] = \mathcal{T}_{\lambda,V}[f].\] Hence \(S=\mathcal{T}_{\lambda,V}\). ◻

Proof. We prove the three assertions separately.

Part (a): completeness of \(\Xi^k\). Let \((\psi_n)_{n\ge1}\) be a Cauchy sequence in \(\Xi^k\). Since \[\|\psi\|_\infty\le \|\psi\|_{\Xi^k}, \qquad \|\psi\|_1\le \|\psi\|_{\Xi^k},\] the sequence \((\psi_n)\) is Cauchy both in \((C_0(\mathbb{R}^d;\mathbb{C}),\|\cdot\|_\infty)\) and in \((L^1(\mathbb{R}^d;\mathbb{C}),\|\cdot\|_1)\). These spaces are Banach, so there exist \[g_0\in C_0(\mathbb{R}^d;\mathbb{C}), \qquad h\in L^1(\mathbb{R}^d;\mathbb{C}),\] such that \[\|\psi_n-g_0\|_\infty\to0, \qquad \|\psi_n-h\|_1\to0.\] Uniform convergence gives \(\psi_n(x)\to g_0(x)\) for every \(x\). By \(L^1\)-convergence, a subsequence converges to \(h\) almost everywhere [72], Theorem 2.30. Hence \(g_0=h\) almost everywhere. We therefore identify \(h\) with the continuous representative \(g_0\), and obtain \[g_0\in L^1(\mathbb{R}^d;\mathbb{C}), \qquad \|\psi_n-g_0\|_1\to0.\]

For each multi-index \(\alpha\) with \(|\alpha|\le k\), the sequence \((\partial^\alpha\psi_n)\) is Cauchy in \(C_0(\mathbb{R}^d;\mathbb{C})\), because \[\|\partial^\alpha\psi_n-\partial^\alpha\psi_m\|_\infty \le \|\psi_n-\psi_m\|_{\Xi^k}.\] Let \(g_\alpha\in C_0(\mathbb{R}^d;\mathbb{C})\) denote its uniform limit: \[\|\partial^\alpha\psi_n-g_\alpha\|_\infty\to0.\]

It remains to identify \(g_\alpha\) with \(\partial^\alpha g_0\). If \(k=0\), there is nothing to prove. Suppose \(k\ge1\). Fix a multi-index \(\alpha\) with \(|\alpha|\le k-1\), a coordinate \(j\in\{1,\ldots,d\}\), a point \(x\in\mathbb{R}^d\), and \(t\in\mathbb{R}\). Here \(e_j\) denotes the \(j\)-th coordinate vector. For every \(n\), the fundamental theorem of calculus gives \[\partial^\alpha\psi_n(x+t e_j)-\partial^\alpha\psi_n(x) = \int_0^t \partial^{\alpha+e_j}\psi_n(x+s e_j)\,ds .\] Letting \(n\to\infty\) and using uniform convergence of the derivatives yields \[g_\alpha(x+t e_j)-g_\alpha(x) = \int_0^t g_{\alpha+e_j}(x+s e_j)\,ds .\] Since \(g_{\alpha+e_j}\) is continuous, this identity implies \[\partial_j g_\alpha(x)=g_{\alpha+e_j}(x).\] Applying this argument for all \(|\alpha|\le k-1\) and all \(j\), induction on \(|\alpha|\) gives \[g_0\in C^k(\mathbb{R}^d;\mathbb{C}), \qquad \partial^\alpha g_0=g_\alpha \quad\text{for all }|\alpha|\le k.\] Since \(g_\alpha\in C_0(\mathbb{R}^d;\mathbb{C})\) for every \(|\alpha|\le k\) and \(g_0\in L^1(\mathbb{R}^d;\mathbb{C})\), we have \(g_0\in\Xi^k\). Finally, \[\|\psi_n-g_0\|_{\Xi^k} = \max_{|\alpha|\le k} \|\partial^\alpha\psi_n-g_\alpha\|_\infty + \|\psi_n-g_0\|_1 \to0.\] Thus \(\Xi^k\) is complete.

Part (b): continuous inclusion of \(\mathcal{A}_V\) into \(\Xi^k\). Let \(f\in\mathcal{A}_V(\mathbb{R}^d)\). By definition of \(\mathcal{A}_V(\mathbb{R}^d)\), and using the unique representing measure, write \[f=f_V*\mu, \qquad \mu\in\mathcal{M}(\mathbb{R}^d).\]

First, \(f\in L^1(\mathbb{R}^d;\mathbb{C})\). Indeed, by Tonelli’s theorem applied to the total variation measure \(|\mu|\), \[\begin{align} \|f\|_1 &= \int_{\mathbb{R}^d} \left| \int_{\mathbb{R}^d} f_V(x-z)\,d\mu(z) \right|dx \\ &\le \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |f_V(x-z)|\,d|\mu|(z)\,dx \\ &= \int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d}|f_V(x-z)|\,dx \right)d|\mu|(z) \\ &= \|f_V\|_1\|\mu\|_{TV}. \end{align}\]

Next fix \(\alpha\) with \(|\alpha|\le k\), and define \[g_\alpha(x) \mathrel{\mathop{:}}= \int_{\mathbb{R}^d} \partial^\alpha f_V(x-z)\,d\mu(z).\] This integral is well-defined because \(\partial^\alpha f_V\in C_0(\mathbb{R}^d;\mathbb{C})\subset L^\infty(\mathbb{R}^d;\mathbb{C})\), and \[|g_\alpha(x)| \le \|\partial^\alpha f_V\|_\infty\|\mu\|_{TV}.\]

We first show that \(g_\alpha\in C_0(\mathbb{R}^d;\mathbb{C})\). If \(x_n\to x\), then \[\partial^\alpha f_V(x_n-z)\to \partial^\alpha f_V(x-z) \qquad \text{for every }z\in\mathbb{R}^d,\] and \[|\partial^\alpha f_V(x_n-z)| \le \|\partial^\alpha f_V\|_\infty .\] The bound is \(|\mu|\)-integrable because \(|\mu|(\mathbb{R}^d)<\infty\). Dominated convergence gives \(g_\alpha(x_n)\to g_\alpha(x)\), so \(g_\alpha\) is continuous.

To prove that \(g_\alpha\) vanishes at infinity, set \[M_\alpha\mathrel{\mathop{:}}=\|\partial^\alpha f_V\|_\infty .\] Fix \(\varepsilon>0\). Since \(|\mu|\) is finite, choose a compact set \(K\subset\mathbb{R}^d\) such that \[|\mu|(K^c) < \frac{\varepsilon}{2(1+M_\alpha)}.\] Since \(\partial^\alpha f_V\in C_0(\mathbb{R}^d;\mathbb{C})\), choose \(R>0\) such that \[\|y\|_2>R \quad\Longrightarrow\quad |\partial^\alpha f_V(y)| < \frac{\varepsilon}{2(1+|\mu|(K))}.\] Because \(K\) is compact, there exists \(M>0\) such that \[\|x\|_2>M,\;z\in K \quad\Longrightarrow\quad \|x-z\|_2>R.\] For \(\|x\|_2>M\), \[\begin{align} |g_\alpha(x)| &\le \int_K |\partial^\alpha f_V(x-z)|\,d|\mu|(z) + \int_{K^c}|\partial^\alpha f_V(x-z)|\,d|\mu|(z) \\ &\le \frac{\varepsilon}{2(1+|\mu|(K))}\,|\mu|(K) + M_\alpha|\mu|(K^c) \\ &< \varepsilon . \end{align}\] Thus \(g_\alpha\in C_0(\mathbb{R}^d;\mathbb{C})\).

It remains to identify \(g_\alpha\) as \(\partial^\alpha f\). We argue by induction on \(|\alpha|\). For \(\alpha=0\), the identity \(g_0=f\) is the definition of \(f\). Suppose that, for some \(\alpha\) with \(|\alpha|\le k-1\), \[\partial^\alpha f(x) = \int_{\mathbb{R}^d} \partial^\alpha f_V(x-z)\,d\mu(z).\] Fix \(j\in\{1,\ldots,d\}\). For \(t\neq0\), \[\frac{\partial^\alpha f(x+t e_j)-\partial^\alpha f(x)}{t} = \int_{\mathbb{R}^d} \frac{ \partial^\alpha f_V(x+t e_j-z)-\partial^\alpha f_V(x-z) }{t} \,d\mu(z).\] For each \(z\), the integrand converges to \(\partial^{\alpha+e_j}f_V(x-z)\) as \(t\to0\). Moreover, the fundamental theorem of calculus gives \[\frac{ \partial^\alpha f_V(x+t e_j-z)-\partial^\alpha f_V(x-z) }{t} = \int_0^1 \partial^{\alpha+e_j}f_V(x+s t e_j-z)\,ds,\] and therefore \[\left| \frac{ \partial^\alpha f_V(x+t e_j-z)-\partial^\alpha f_V(x-z) }{t} \right| \le \|\partial^{\alpha+e_j}f_V\|_\infty .\] The right-hand side is \(|\mu|\)-integrable. Dominated convergence yields \[\partial^{\alpha+e_j}f(x) = \int_{\mathbb{R}^d} \partial^{\alpha+e_j}f_V(x-z)\,d\mu(z) = g_{\alpha+e_j}(x).\] By induction, \[\partial^\alpha f=g_\alpha\in C_0(\mathbb{R}^d;\mathbb{C}) \qquad \text{for all }|\alpha|\le k.\] Together with the \(L^1\) bound above, this proves \[f\in\Xi^k(\mathbb{R}^d;\mathbb{C}).\]

The same estimates give the continuity of the inclusion. If \(\mu=\mathcal{K}_V^{-1}[f]\), then \[\|f\|_1 \le \|f_V\|_1\|\mu\|_{TV},\] and, for every \(|\alpha|\le k\), \[\|\partial^\alpha f\|_\infty \le \|\partial^\alpha f_V\|_\infty\|\mu\|_{TV}.\] Hence \[\begin{align} \|f\|_{\Xi^k} &= \max_{|\alpha|\le k}\|\partial^\alpha f\|_\infty + \|f\|_1 \\ &\le \left( \max_{|\alpha|\le k}\|\partial^\alpha f_V\|_\infty + \|f_V\|_1 \right)\|\mu\|_{TV}. \end{align}\] Since \(\|\mu\|_{TV}=\|f\|_{\mathcal{A}_V}\), we obtain \[\|f\|_{\Xi^k} \le C_{V,k}\|f\|_{\mathcal{A}_V}, \qquad C_{V,k} \mathrel{\mathop{:}}= \max_{|\alpha|\le k}\|\partial^\alpha f_V\|_\infty + \|f_V\|_1 .\] Thus the embedding \[\mathcal{A}_V(\mathbb{R}^d)\hookrightarrow \Xi^k(\mathbb{R}^d;\mathbb{C})\] is continuous.

Part (c): continuous and dense inclusion of the Schwartz space. We first prove continuity of the inclusion \[\mathscr S(\mathbb{R}^d)\hookrightarrow \Xi^k(\mathbb{R}^d;\mathbb{C}).\] For \(\varphi\in\mathscr S(\mathbb{R}^d)\), define the Schwartz seminorm \[Q_{d,k}(\varphi) \mathrel{\mathop{:}}= \max_{|\alpha|\le k} \sup_{x\in\mathbb{R}^d} (1+\|x\|_2)^{d+1}|\partial^\alpha\varphi(x)|.\] Then \[\max_{|\alpha|\le k}\|\partial^\alpha\varphi\|_\infty \le Q_{d,k}(\varphi),\] and \[\begin{align} \|\varphi\|_1 &= \int_{\mathbb{R}^d} (1+\|x\|_2)^{-d-1} (1+\|x\|_2)^{d+1} |\varphi(x)|\,dx \\ &\le \left( \int_{\mathbb{R}^d} (1+\|x\|_2)^{-d-1}\,dx \right) Q_{d,k}(\varphi). \end{align}\] The integral is finite. Hence \[\|\varphi\|_{\Xi^k} \le C_{d,k}Q_{d,k}(\varphi),\] which proves continuity of the inclusion.

It remains to prove density. Let \(\psi\in\Xi^k(\mathbb{R}^d;\mathbb{C})\). Choose \(\chi\in C_c^\infty(\mathbb{R}^d)\) such that \(0\le\chi\le1\), \(\chi\equiv1\) on \(B(0,1)\), and \(\chi\equiv0\) outside \(B(0,2)\). For \(R\ge1\), define \[\chi_R(x)\mathrel{\mathop{:}}=\chi(x/R), \qquad \psi_R(x)\mathrel{\mathop{:}}=\chi_R(x)\psi(x).\] Then \(\psi_R\in C_c^k(\mathbb{R}^d;\mathbb{C})\).

We first show that \(\psi_R\to\psi\) in \(\Xi^k\). Since \(\chi_R(x)\to1\) for every \(x\) and \[|(\chi_R(x)-1)\psi(x)|\le |\psi(x)|,\] dominated convergence gives \[\|\psi_R-\psi\|_1\to0.\] Now fix \(\alpha\) with \(|\alpha|\le k\). By Leibniz’s rule, \[\partial^\alpha(\psi_R-\psi) = (\chi_R-1)\partial^\alpha\psi + \sum_{0<\beta\le\alpha} {\alpha\choose\beta} (\partial^\beta\chi_R)\partial^{\alpha-\beta}\psi .\] Because \(\chi_R=1\) on \(B(0,R)\) and \(\partial^\alpha\psi\in C_0(\mathbb{R}^d;\mathbb{C})\), \[\|(\chi_R-1)\partial^\alpha\psi\|_\infty\to0.\] For \(0<\beta\le\alpha\), \[\partial^\beta\chi_R(x) = R^{-|\beta|}(\partial^\beta\chi)(x/R),\] and \[\operatorname{supp}(\partial^\beta\chi_R) \subset \{x\in\mathbb{R}^d:R\le \|x\|_2\le 2R\}.\] Thus \[\|(\partial^\beta\chi_R)\partial^{\alpha-\beta}\psi\|_\infty \le R^{-|\beta|} \|\partial^\beta\chi\|_\infty \sup_{\|x\|_2\ge R} |\partial^{\alpha-\beta}\psi(x)|.\] Since \(\partial^{\alpha-\beta}\psi\in C_0(\mathbb{R}^d;\mathbb{C})\), the right-hand side tends to zero. Therefore \[\max_{|\alpha|\le k} \|\partial^\alpha(\psi_R-\psi)\|_\infty \to0,\] and hence \[\|\psi_R-\psi\|_{\Xi^k}\to0.\]

Now choose a standard compactly supported mollifier \(\rho\in C_c^\infty(\mathbb{R}^d)\) with \(\int\rho=1\), and set \[\rho_\varepsilon(x) \mathrel{\mathop{:}}= \varepsilon^{-d}\rho(x/\varepsilon), \qquad \psi_{R,\varepsilon} \mathrel{\mathop{:}}= \rho_\varepsilon*\psi_R .\] For fixed \(R\), we have \[\psi_{R,\varepsilon}\in C_c^\infty(\mathbb{R}^d)\subset\mathscr S(\mathbb{R}^d).\] Standard properties of mollifiers give, for every \(|\alpha|\le k\), \[\partial^\alpha\psi_{R,\varepsilon} = \rho_\varepsilon*\partial^\alpha\psi_R,\] and \[\|\psi_{R,\varepsilon}-\psi_R\|_1\to0, \qquad \|\partial^\alpha\psi_{R,\varepsilon} -\partial^\alpha\psi_R\|_\infty\to0 \quad (|\alpha|\le k),\] as \(\varepsilon\downarrow0\); see [72], Propositions 8.10 and 8.14. Hence \[\|\psi_{R,\varepsilon}-\psi_R\|_{\Xi^k}\to0 \qquad (\varepsilon\downarrow0).\]

For each \(n\), choose \(R_n\) such that \[\|\psi_{R_n}-\psi\|_{\Xi^k}<\frac{1}{2n},\] and then choose \(\varepsilon_n>0\) such that \[\|\psi_{R_n,\varepsilon_n}-\psi_{R_n}\|_{\Xi^k}<\frac{1}{2n}.\] The functions \[\varphi_n\mathrel{\mathop{:}}=\psi_{R_n,\varepsilon_n}\] belong to \(\mathscr S(\mathbb{R}^d)\), and \[\|\varphi_n-\psi\|_{\Xi^k}\le \frac{1}{n}.\] Thus \(\mathscr S(\mathbb{R}^d)\) is dense in \(\Xi^k(\mathbb{R}^d;\mathbb{C})\). ◻

Proof. We first prove part (a). Each \(\tau_x\) is linear. Moreover, for every multi-index \(\alpha\) with \(|\alpha|\le k\), \[\partial^\alpha(\tau_x\psi) = \tau_x(\partial^\alpha\psi).\] Thus translation preserves both the derivative sup norms and the \(L^1\)-norm: \[\|\partial^\alpha(\tau_x\psi)\|_\infty = \|\partial^\alpha\psi\|_\infty, \qquad \|\tau_x\psi\|_1 = \|\psi\|_1.\] Therefore \[\|\tau_x\psi\|_{\Xi^k} = \|\psi\|_{\Xi^k}, \qquad \psi\in\Xi^k(\mathbb{R}^d;\mathbb{C}).\] Each \(\tau_x\) is an isometry. Since \[\tau_0=\mathrm{Id}, \qquad \tau_x\tau_y=\tau_{x+y},\] the family \((\tau_x)_{x\in\mathbb{R}^d}\) is a group.

It remains to prove strong continuity. Fix \(f\in\Xi^k(\mathbb{R}^d;\mathbb{C})\). At the origin, \[\|\tau_x f-f\|_{\Xi^k} = \max_{|\alpha|\le k} \|\tau_x(\partial^\alpha f)-\partial^\alpha f\|_\infty + \|\tau_x f-f\|_1.\] The \(L^1\)-term tends to zero as \(x\to0\) by continuity of translations in \(L^1\); see [72], Proposition 8.5. For each \(|\alpha|\le k\), \(\partial^\alpha f\in C_0(\mathbb{R}^d;\mathbb{C})\), hence \(\partial^\alpha f\) is uniformly continuous. Therefore \[\|\tau_x(\partial^\alpha f)-\partial^\alpha f\|_\infty \to0 \qquad (x\to0).\] Thus \[\|\tau_x f-f\|_{\Xi^k}\to0 \qquad (x\to0).\] For any \(x_0\in\mathbb{R}^d\), \[\|\tau_x f-\tau_{x_0}f\|_{\Xi^k} = \|\tau_{x-x_0}f-f\|_{\Xi^k} \to0 \qquad (x\to x_0),\] because \(\tau_{x_0}\) is an isometry. This proves part (a).

We now prove part (b). Assumption 4 implies that \(\partial^\alpha f_V\in C_0(\mathbb{R}^d;\mathbb{C})\) for all \(|\alpha|\le k\). Since \(f_V\) is a density, \(f_V\in L^1(\mathbb{R}^d)\). Hence \[f_V\in\Xi^k(\mathbb{R}^d;\mathbb{C}).\] Define \[\Phi:\mathbb{R}^d\to\Xi^k(\mathbb{R}^d;\mathbb{C}), \qquad \Phi(x)\mathrel{\mathop{:}}=\tau_x f_V.\] By part (a), \(\Phi\) is bounded and continuous. Since \(\mathbb{R}^d\) is separable, \(\Phi\) has separable range and is strongly measurable with respect to \(|\mu|\). Also, \[\int_{\mathbb{R}^d}\|\Phi(x)\|_{\Xi^k}\,d|\mu|(x) = \|f_V\|_{\Xi^k}\,|\mu|(\mathbb{R}^d) <\infty.\] Thus \(\Phi\) is Bochner integrable with respect to \(|\mu|\).

Since \(\mu\) is a finite complex measure, there exists a measurable function \(\sigma_\mu\), with \(|\sigma_\mu|=1\) \(|\mu|\)-a.e., such that \[d\mu=\sigma_\mu\,d|\mu|;\] see [54], Theorem 6.12. We define \[\int_{\mathbb{R}^d}\Phi(x)\,d\mu(x) \mathrel{\mathop{:}}= \int_{\mathbb{R}^d}\sigma_\mu(x)\Phi(x)\,d|\mu|(x).\] Set \[F \mathrel{\mathop{:}}= \int_{\mathbb{R}^d}\tau_x f_V\,d\mu(x) \in \Xi^k(\mathbb{R}^d;\mathbb{C}).\]

We identify \(F\). For \(|\alpha|\le k\), the map \[J_\alpha:\Xi^k(\mathbb{R}^d;\mathbb{C})\to C_0(\mathbb{R}^d;\mathbb{C}), \qquad J_\alpha(h)\mathrel{\mathop{:}}=\partial^\alpha h,\] is bounded and linear. Bounded linear maps commute with Bochner integration; see [73], Theorem 3.1.7. Hence \[\partial^\alpha F = \int_{\mathbb{R}^d}\tau_x(\partial^\alpha f_V)\,d\mu(x) \qquad \text{in } C_0(\mathbb{R}^d;\mathbb{C}).\] Evaluating at \(t\in\mathbb{R}^d\) gives \[\partial^\alpha F(t) = \int_{\mathbb{R}^d} \partial^\alpha f_V(t-x)\,d\mu(x).\] Taking \(\alpha=0\), we obtain \[F(t) = \int_{\mathbb{R}^d}f_V(t-x)\,d\mu(x) = (f_V*\mu)(t).\] Therefore \[f_V*\mu = \int_{\mathbb{R}^d}\tau_x f_V\,d\mu(x) \qquad \text{in } \Xi^k(\mathbb{R}^d;\mathbb{C}).\] This proves part (b). ◻

Proof of Theorem 2. Set \[\lambda\mathrel{\mathop{:}}=\lambda_{g,V,y}, \qquad Q\mathrel{\mathop{:}}=\mathcal{Q}_{g,V,y}, \qquad L_Q\mathrel{\mathop{:}}=\mathcal{F}^{-1}\{\mathcal{I}_Q\}.\] By assumption (i) and Lemma 2, \[\mathcal{I}_Q\in\mathscr S'(\mathbb{R}^d).\] Hence \[L_Q\in\mathscr S'(\mathbb{R}^d).\] Assumption (ii) gives \[\label{eq:master-fourier-bound} |L_Q[\psi]| \le C\|\psi\|_{\Xi^k}, \qquad \psi\in\mathscr S(\mathbb{R}^d).\tag{4}\] Thus \(L_Q\) is continuous on \(\mathscr S(\mathbb{R}^d)\) when \(\mathscr S(\mathbb{R}^d)\) is equipped with the \(\Xi^k\)-norm.

By Lemma 1(c), \(\mathscr S(\mathbb{R}^d)\) is dense in \(\Xi^k(\mathbb{R}^d;\mathbb{C})\). Therefore 4 extends \(L_Q\) uniquely to a continuous linear functional \[\overline{L}_Q\in \bigl(\Xi^k(\mathbb{R}^d;\mathbb{C})\bigr)'.\] By Lemma 1(b), \[\mathcal{A}_V(\mathbb{R}^d) \subset \Xi^k(\mathbb{R}^d;\mathbb{C}),\] so the restriction \[\overline{L}_Q|_{\mathcal{A}_V(\mathbb{R}^d)}\] is well-defined. We prove that this restriction is \(\mathcal{T}_{g,V,y}\).

For \(\psi\in\mathscr S(\mathbb{R}^d)\), the definition of the inverse Fourier transform on tempered distributions gives \[\label{eq:master-fourier-action} \begin{align} L_Q[\psi] &= \mathcal{I}_Q[\psi^\sharp] \\ &= \int_{\mathbb{R}^d}Q(\omega)\psi^\sharp(\omega)\,d\omega \\ &= \frac{1}{(2\pi)^d} \int_{\mathbb{R}^d}Q(\omega)\widetilde{\psi}(-\omega)\,d\omega, \end{align}\tag{5}\] where \[\psi^\sharp(\omega) = (2\pi)^{-d}\widetilde{\psi}(-\omega).\]

Define \[\Psi(x) \mathrel{\mathop{:}}= \overline{L}_Q[\tau_x f_V], \qquad x\in\mathbb{R}^d.\] By Lemma 3(a), the map \(x\mapsto\tau_x f_V\) is bounded and continuous into \(\Xi^k(\mathbb{R}^d;\mathbb{C})\). Since \(\overline{L}_Q\) is continuous, \(\Psi\) is bounded and continuous.

We next identify \(\Psi\). Let \(\eta\in\mathscr S(\mathbb{R}^d)\), and let \(\mu_\eta\in\mathcal{M}(\mathbb{R}^d)\) be the finite complex measure \[d\mu_\eta(x)=\eta(x)\,dx.\] By Lemma 3(b), \[f_V*\eta = \int_{\mathbb{R}^d}\eta(x)\tau_x f_V\,dx \qquad \text{in } \Xi^k(\mathbb{R}^d;\mathbb{C}).\] Applying \(\overline{L}_Q\) and using again [73], Theorem 3.1.7, gives \[\label{eq:master-pairing-Psi} \int_{\mathbb{R}^d}\Psi(x)\eta(x)\,dx = \overline{L}_Q[f_V*\eta].\tag{6}\]

Choose \((u_n)_{n\ge1}\subset\mathscr S(\mathbb{R}^d)\) such that \[u_n\to f_V \qquad \text{in } \Xi^k(\mathbb{R}^d;\mathbb{C}),\] which is possible by Lemma 1(c). We use the elementary bound \[\label{eq:master-convolution-estimate} \|\zeta*\eta\|_{\Xi^k} \le \|\zeta\|_{\Xi^k}\|\eta\|_1, \qquad \zeta\in\Xi^k(\mathbb{R}^d;\mathbb{C}),\quad \eta\in L^1(\mathbb{R}^d;\mathbb{C}).\tag{7}\] Indeed, for every \(|\alpha|\le k\), \[\partial^\alpha(\zeta*\eta) = (\partial^\alpha\zeta)*\eta,\] and therefore \[\|\partial^\alpha(\zeta*\eta)\|_\infty \le \|\partial^\alpha\zeta\|_\infty\|\eta\|_1.\] Also, \[\|\zeta*\eta\|_1 \le \|\zeta\|_1\|\eta\|_1.\] Combining these inequalities proves 7 .

Applying 7 to \(\zeta=u_n-f_V\) gives \[u_n*\eta\to f_V*\eta \qquad \text{in } \Xi^k(\mathbb{R}^d;\mathbb{C}).\] Since \(\overline{L}_Q\) is continuous, \[\label{eq:master-limit-extension} \overline{L}_Q[f_V*\eta] = \lim_{n\to\infty}\overline{L}_Q[u_n*\eta].\tag{8}\] Moreover, \(u_n*\eta\in\mathscr S(\mathbb{R}^d)\), because the Schwartz space is closed under convolution; see [74], Chapter 5, Proposition 1.11. Hence \(\overline{L}_Q[u_n*\eta]=L_Q[u_n*\eta]\). By 5 , \[\label{eq:master-convolution-fourier} L_Q[u_n*\eta] = \frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} Q(\omega)\widetilde{u}_n(-\omega)\widetilde{\eta}(-\omega)\,d\omega.\tag{9}\]

Since \(u_n\to f_V\) in \(L^1(\mathbb{R}^d;\mathbb{C})\), \[\sup_{\omega\in\mathbb{R}^d} |\widetilde{u}_n(\omega)-\varphi_V(\omega)| \le \|u_n-f_V\|_1 \to0.\] Thus \(\widetilde{u}_n\to\varphi_V\) uniformly on \(\mathbb{R}^d\). Also, \[\sup_{n\ge1}\|\widetilde{u}_n\|_\infty \le \sup_{n\ge1}\|u_n\|_1 <\infty.\]

Let \(N\) be the integer from assumption (i). Since \(\widetilde{\eta}\in\mathscr S(\mathbb{R}^d)\), there exists \(C_{\eta,N}>0\) such that \[|\widetilde{\eta}(-\omega)| \le C_{\eta,N}(1+\|\omega\|_2)^{-N}.\] Hence \[|Q(\omega)\widetilde{\eta}(-\omega)| \le C_{\eta,N} \frac{|Q(\omega)|}{(1+\|\omega\|_2)^N},\] and the right-hand side is integrable by assumption (i). Therefore \[|Q(\omega)\widetilde{u}_n(-\omega)\widetilde{\eta}(-\omega)| \le \left(\sup_{m\ge1}\|\widetilde{u}_m\|_\infty\right) |Q(\omega)\widetilde{\eta}(-\omega)|.\] Dominated convergence in 9 yields \[\label{eq:master-fVeta-fourier} \overline{L}_Q[f_V*\eta] = \frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} Q(\omega)\varphi_V(-\omega)\widetilde{\eta}(-\omega)\,d\omega.\tag{10}\]

Since \(|\varphi_V|\le1\), assumption (i) and Lemma 2 imply that \[\mathcal{I}_{\omega\mapsto Q(\omega)\varphi_V(-\omega)} \in \mathscr S'(\mathbb{R}^d).\] By 10 and the definition of the inverse Fourier transform on \(\mathscr S'(\mathbb{R}^d)\), \[\overline{L}_Q[f_V*\eta] = \mathcal{F}^{-1} \left\{ \mathcal{I}_{\omega\mapsto Q(\omega)\varphi_V(-\omega)} \right\}[\eta].\] Assumption (iii) gives \[\mathcal{F}^{-1} \left\{ \mathcal{I}_{\omega\mapsto Q(\omega)\varphi_V(-\omega)} \right\} = \mathcal{I}_\lambda \qquad \text{in } \mathscr S'(\mathbb{R}^d).\] Therefore \[\overline{L}_Q[f_V*\eta] = \mathcal{I}_\lambda[\eta] = \int_{\mathbb{R}^d}\lambda(x)\eta(x)\,dx, \qquad \eta\in\mathscr S(\mathbb{R}^d).\]

Combining this identity with 6 , we obtain \[\int_{\mathbb{R}^d}\Psi(x)\eta(x)\,dx = \int_{\mathbb{R}^d}\lambda(x)\eta(x)\,dx, \qquad \eta\in\mathscr S(\mathbb{R}^d).\] Thus \(\Psi-\lambda=0\) in \(\mathscr S'(\mathbb{R}^d)\). Since \(\Psi\) is continuous and \(\lambda\in C_0(\mathbb{R}^d;\mathbb{C})\), both functions are locally integrable. The distributional equality implies \(\Psi=\lambda\) Lebesgue-a.e. Since both functions are continuous, \[\Psi(x)=\lambda(x), \qquad x\in\mathbb{R}^d.\]

We now identify the restriction of \(\overline{L}_Q\) to \(\mathcal{A}_V(\mathbb{R}^d)\). Let \(f\in\mathcal{A}_V(\mathbb{R}^d)\), and set \[\mu_f\mathrel{\mathop{:}}=\mathcal{K}_V^{-1}[f].\] Then \[f=f_V*\mu_f.\] By Lemma 3(b), \[f = \int_{\mathbb{R}^d}\tau_x f_V\,d\mu_f(x) \qquad \text{in } \Xi^k(\mathbb{R}^d;\mathbb{C}).\] Applying \(\overline{L}_Q\) and using Bochner integration once more, \[\begin{align} \overline{L}_Q[f] &= \int_{\mathbb{R}^d} \overline{L}_Q[\tau_x f_V]\,d\mu_f(x) \\ &= \int_{\mathbb{R}^d}\Psi(x)\,d\mu_f(x) \\ &= \int_{\mathbb{R}^d}\lambda(x)\,d\mu_f(x) \\ &= \mathcal{T}_{g,V,y}[f]. \end{align}\] Hence \[\overline{L}_Q|_{\mathcal{A}_V(\mathbb{R}^d)} = \mathcal{T}_{g,V,y}.\]

Finally, let \(f\in\mathcal{A}_V(\mathbb{R}^d)\), and let \((f_n)_{n\ge1}\subset\mathscr S(\mathbb{R}^d)\) satisfy \[\|f_n-f\|_{\Xi^k}\to0.\] Since \(\overline{L}_Q\) extends \(L_Q\) and is continuous on \(\Xi^k\), \[L_Q[f_n] = \overline{L}_Q[f_n] \to \overline{L}_Q[f] = \mathcal{T}_{g,V,y}[f].\] Using 5 with \(\psi=f_n\), this is the same as \[\frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} Q(\omega)\widetilde{f}_n(-\omega)\,d\omega \to \mathcal{T}_{g,V,y}[f].\] The limit is therefore independent of the approximating sequence. ◻

Proof. Write \[\lambda_n\mathrel{\mathop{:}}=\lambda_{n,g,V,y}, \qquad \lambda\mathrel{\mathop{:}}=\lambda_{g,V,y}, \qquad \mathcal{T}_n\mathrel{\mathop{:}}=\mathcal{T}_{n,g,V,y}.\] For each \(n\), \(\lambda_n\in C_0(\mathbb{R}^d;\mathbb{C})\). Hence Theorem 1 gives a unique continuous linear functional \[\mathcal{T}_n\in\mathcal{A}_V'(\mathbb{R}^d)\] such that \[\mathcal{T}_n[f_V*\mu] = \int_{\mathbb{R}^d}\lambda_n(x)\,d\mu(x), \qquad \mu\in\mathcal{M}(\mathbb{R}^d).\]

We now construct the limiting functional. By assumption, \[\lambda_n(x)\to\lambda(x), \qquad x\in\mathbb{R}^d,\] and \[M_\lambda \mathrel{\mathop{:}}= \sup_{n\ge1}\|\lambda_n\|_\infty <\infty .\] Since \(\lambda\) is the pointwise limit of continuous functions, it is Borel measurable. Moreover, \[\|\lambda\|_\infty\le M_\lambda .\]

For \(f\in\mathcal{A}_V(\mathbb{R}^d)\), let \[\mu_f\mathrel{\mathop{:}}=\mathcal{K}_V^{-1}[f].\] The measure \(\mu_f\) is well-defined and unique by Proposition 2. Define \[\mathcal{T}_{g,V,y}[f] \mathrel{\mathop{:}}= \int_{\mathbb{R}^d}\lambda(x)\,d\mu_f(x).\] This map is linear because \(f\mapsto\mu_f\) is linear. It is also continuous: \[|\mathcal{T}_{g,V,y}[f]| \le \|\lambda\|_\infty\|\mu_f\|_{TV} = \|\lambda\|_\infty\|f\|_{\mathcal{A}_V}.\] Therefore \[\mathcal{T}_{g,V,y}\in\mathcal{A}_V'(\mathbb{R}^d).\]

This functional is the unique continuous linear functional with the displayed integral representation. Indeed, if \(S\in\mathcal{A}_V'(\mathbb{R}^d)\) satisfies \[S[f_V*\mu] = \int_{\mathbb{R}^d}\lambda(x)\,d\mu(x), \qquad \mu\in\mathcal{M}(\mathbb{R}^d),\] then, for any \(f\in\mathcal{A}_V(\mathbb{R}^d)\), \[S[f] = S[f_V*\mu_f] = \int_{\mathbb{R}^d}\lambda(x)\,d\mu_f(x) = \mathcal{T}_{g,V,y}[f].\] Hence \(S=\mathcal{T}_{g,V,y}\).

It remains to prove convergence. Fix \(f\in\mathcal{A}_V(\mathbb{R}^d)\), and write \(\mu_f=\mathcal{K}_V^{-1}[f]\). Then \[\mathcal{T}_n[f] = \int_{\mathbb{R}^d}\lambda_n(x)\,d\mu_f(x), \qquad \mathcal{T}_{g,V,y}[f] = \int_{\mathbb{R}^d}\lambda(x)\,d\mu_f(x).\] Since \[\lambda_n(x)-\lambda(x)\to0 \qquad \text{for every }x\in\mathbb{R}^d,\] and \[|\lambda_n(x)-\lambda(x)| \le 2M_\lambda, \qquad x\in\mathbb{R}^d,\] the dominated convergence theorem, applied with respect to the finite measure \(|\mu_f|\), gives \[\int_{\mathbb{R}^d} |\lambda_n(x)-\lambda(x)|\,d|\mu_f|(x) \to0.\] Consequently, \[\begin{align} |\mathcal{T}_n[f]-\mathcal{T}_{g,V,y}[f]| &= \left| \int_{\mathbb{R}^d} \{\lambda_n(x)-\lambda(x)\}\,d\mu_f(x) \right| \\ &\le \int_{\mathbb{R}^d} |\lambda_n(x)-\lambda(x)|\,d|\mu_f|(x) \to0 . \end{align}\] Thus \[\mathcal{T}_n[f]\to\mathcal{T}_{g,V,y}[f], \qquad f\in\mathcal{A}_V(\mathbb{R}^d).\]

Now suppose that \((Y,X,V)\) also satisfies Assumption 1. By Proposition 1, \[f_Y = f_V*P_X = \mathcal{K}_V[P_X],\] so \(f_Y\in\mathcal{A}_V(\mathbb{R}^d)\). Applying the preceding convergence result with \(f=f_Y\) yields \[\mathcal{T}_{g,V,y}[f_Y] = \lim_{n\to\infty}\mathcal{T}_{n,g,V,y}[f_Y].\]

Finally, by the definition of \(\lambda_{g,V,y}\), \[\mathcal{T}_{g,V,y}[f_Y] = \int_{\mathbb{R}^d}\lambda_{g,V,y}(x)\,dP_X(x) = \int_{\mathbb{R}^d}g(x)f_V(y-x)\,dP_X(x).\] Therefore, whenever \(f_Y(y)>0\), the ratio formula 3 gives \[\mathbb{E}[g(X)\mid Y=y] = \frac{ \int_{\mathbb{R}^d}g(x)f_V(y-x)\,dP_X(x) }{ f_Y(y) } = \frac{\mathcal{T}_{g,V,y}[f_Y]}{f_Y(y)} = \lim_{n\to\infty} \frac{\mathcal{T}_{n,g,V,y}[f_Y]}{f_Y(y)}.\] This proves the approximation formula. ◻

Proof. Set \[Q(\omega)\mathrel{\mathop{:}}=\mathcal{Q}_{g,V,y}(\omega), \qquad R(\omega)\mathrel{\mathop{:}}= \sum_{|\alpha|\le k}(i\omega)^\alpha \varphi_{\mu_\alpha}(\omega).\] By assumption, \(Q=R\) Lebesgue-a.e. Hence \[\mathcal{I}_Q=\mathcal{I}_R \qquad \text{on } \mathscr S(\mathbb{R}^d).\]

Step 1: temperedness. For each \(|\alpha|\le k\), the Fourier–Stieltjes transform of \(\mu_\alpha\) is bounded: \[|\varphi_{\mu_\alpha}(\omega)| = \left| \int_{\mathbb{R}^d}e^{i\omega^\top z}\,d\mu_\alpha(z) \right| \le \|\mu_\alpha\|_{TV}, \qquad \omega\in\mathbb{R}^d.\] Therefore there is a finite constant \(C>0\) such that \[|R(\omega)| \le \sum_{|\alpha|\le k} |\omega^\alpha|\|\mu_\alpha\|_{TV} \le C(1+\|\omega\|_2)^k .\] Since \(Q=R\) Lebesgue-a.e., the same polynomial bound holds for \(Q\) after changing \(Q\) on a null set. Thus, for any integer \(N>k+d\), \[\int_{\mathbb{R}^d} \frac{|Q(\omega)|}{(1+\|\omega\|_2)^N}\,d\omega \le C \int_{\mathbb{R}^d} (1+\|\omega\|_2)^{k-N}\,d\omega <\infty.\] Lemma 2 gives \[\mathcal{I}_Q\in\mathscr S'(\mathbb{R}^d).\]

Step 2: inverse Fourier transform on the Schwartz class. Let \(\psi\in\mathscr S(\mathbb{R}^d)\). Since \[\psi^\sharp(\omega) = (2\pi)^{-d}\widetilde{\psi}(-\omega)\] belongs to \(\mathscr S(\mathbb{R}^d)\), we have \[\int_{\mathbb{R}^d} |\omega^\alpha\psi^\sharp(\omega)|\,d\omega <\infty \qquad (|\alpha|\le k).\] Because each \(\mu_\alpha\) is finite, Fubini’s theorem applies: \[\int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |\omega^\alpha\psi^\sharp(\omega)| \,d\omega\,d|\mu_\alpha|(z) <\infty .\] Using \(Q=R\) a.e., \[\begin{align} \mathcal{F}^{-1}\{\mathcal{I}_Q\}[\psi] &= \mathcal{I}_Q[\psi^\sharp] \\ &= \int_{\mathbb{R}^d}Q(\omega)\psi^\sharp(\omega)\,d\omega \\ &= \sum_{|\alpha|\le k} \int_{\mathbb{R}^d} (i\omega)^\alpha \varphi_{\mu_\alpha}(\omega) \psi^\sharp(\omega)\,d\omega \\ &= \sum_{|\alpha|\le k} \int_{\mathbb{R}^d} \left[ \int_{\mathbb{R}^d} (i\omega)^\alpha e^{i\omega^\top z} \psi^\sharp(\omega)\,d\omega \right]d\mu_\alpha(z). \end{align}\] By Fourier inversion under our convention, \[\psi(z) = \int_{\mathbb{R}^d} e^{i\omega^\top z}\psi^\sharp(\omega)\,d\omega .\] Differentiating under the integral sign gives, for every \(|\alpha|\le k\), \[\partial^\alpha\psi(z) = \int_{\mathbb{R}^d} (i\omega)^\alpha e^{i\omega^\top z} \psi^\sharp(\omega)\,d\omega .\] Therefore \[\mathcal{F}^{-1}\{\mathcal{I}_Q\}[\psi] = \sum_{|\alpha|\le k} \int_{\mathbb{R}^d} \partial^\alpha\psi(z)\,d\mu_\alpha(z), \qquad \psi\in\mathscr S(\mathbb{R}^d).\]

Step 3: \(\Xi^k\)-continuity. Define, for \(\psi\in\Xi^k(\mathbb{R}^d;\mathbb{C})\), \[L[\psi] \mathrel{\mathop{:}}= \sum_{|\alpha|\le k} \int_{\mathbb{R}^d} \partial^\alpha\psi(z)\,d\mu_\alpha(z).\] This is well-defined because \(\partial^\alpha\psi\in C_0(\mathbb{R}^d;\mathbb{C})\) for every \(|\alpha|\le k\). Moreover, \[\begin{align} |L[\psi]| &\le \sum_{|\alpha|\le k} \left| \int_{\mathbb{R}^d} \partial^\alpha\psi(z)\,d\mu_\alpha(z) \right| \\ &\le \sum_{|\alpha|\le k} \|\partial^\alpha\psi\|_\infty \|\mu_\alpha\|_{TV} \\ &\le \left( \sum_{|\alpha|\le k} \|\mu_\alpha\|_{TV} \right) \|\psi\|_{\Xi^k}. \end{align}\] Thus \(L\) is a continuous linear functional on \(\Xi^k(\mathbb{R}^d;\mathbb{C})\). Step 2 shows that \[\mathcal{F}^{-1}\{\mathcal{I}_Q\}[\psi] = L[\psi], \qquad \psi\in\mathscr S(\mathbb{R}^d).\] Hence condition (ii) of Theorem 2 holds.

The assumptions of the proposition give the standing hypotheses of Theorem 2. Moreover, Step 1 verifies condition (i), Step 3 verifies condition (ii), and the assumption that \(Q\) is a Fourier-domain representer for \(\lambda_{g,V,y}\) gives condition (iii). Applying Theorem 2, we obtain, for every \(f\in\mathcal{A}_V(\mathbb{R}^d)\), \[\mathcal{T}_{g,V,y}[f] = \lim_{n\to\infty} \mathcal{F}^{-1}\{\mathcal{I}_Q\}[f_n],\] whenever \((f_n)_{n\ge1}\subset\mathscr S(\mathbb{R}^d)\) satisfies \[\|f_n-f\|_{\Xi^k}\to0.\] Since \(L\) is continuous on \(\Xi^k(\mathbb{R}^d;\mathbb{C})\), Step 2 gives \[\lim_{n\to\infty} \mathcal{F}^{-1}\{\mathcal{I}_Q\}[f_n] = \lim_{n\to\infty}L[f_n] = L[f].\] Therefore \[\mathcal{T}_{g,V,y}[f] = \sum_{|\alpha|\le k} \int_{\mathbb{R}^d} \partial^\alpha f(z)\,d\mu_\alpha(z), \qquad f\in\mathcal{A}_V(\mathbb{R}^d).\]

Finally, under Assumption 1, Proposition 1 gives \[f_Y = f_V*P_X \in \mathcal{A}_V(\mathbb{R}^d).\] Taking \(f=f_Y\) in the preceding display yields \[\mathcal{T}_{g,V,y}[f_Y] = \sum_{|\alpha|\le k} \int_{\mathbb{R}^d} \partial^\alpha f_Y(z)\,d\mu_\alpha(z).\] By the ratio formula 3 , whenever \(f_Y(y)>0\), \[\mathbb{E}[g(X)\mid Y=y] = \frac{\mathcal{T}_{g,V,y}[f_Y]}{f_Y(y)} = \frac{ \displaystyle \sum_{|\alpha|\le k} \int_{\mathbb{R}^d} \partial^\alpha f_Y(z)\,d\mu_\alpha(z) }{ f_Y(y) }.\] This is the stated posterior representation. ◻

Proof. We prove the three assertions separately. Throughout, \(\phi\) denotes the standard \(d\)-dimensional Gaussian density.

Part (a): conditional Bayes formula. Since \(\mathbb{R}^d\) and \(\mathbb{S}_{++}^d\) are standard Borel spaces, there exists a regular conditional law \[K(\Sigma_0,A) \mathrel{\mathop{:}}= P_{X\mid\Sigma}(A\mid \Sigma_0), \qquad A\in\mathcal{B}(\mathbb{R}^d).\] Thus \(K\) is a probability kernel from \(\mathbb{S}_{++}^d\) to \(\mathbb{R}^d\), and \[\mathbb{P}(X\in A,\Sigma\in C) = \int_C K(\Sigma_0,A)\,dP_\Sigma(\Sigma_0)\] for all \(A\in\mathcal{B}(\mathbb{R}^d)\) and \(C\in\mathcal{B}(\mathbb{S}_{++}^d)\).

Because \(V\perp\!\!\!\!\perp(X,\Sigma)\), the product kernel \[K(\Sigma_0,\cdot)\otimes P_V\] is a version of \(P_{(X,V)\mid\Sigma}(\cdot\mid\Sigma_0)\). Indeed, for \(A,B\in\mathcal{B}(\mathbb{R}^d)\) and \(C\in\mathcal{B}(\mathbb{S}_{++}^d)\), \[\begin{align} \mathbb{P}(X\in A,V\in B,\Sigma\in C) &= P_V(B)\mathbb{P}(X\in A,\Sigma\in C) \\ &= \int_C K(\Sigma_0,A)P_V(B)\,dP_\Sigma(\Sigma_0). \end{align}\] The equality on rectangles determines the product kernel. For \(\Sigma_0\in\mathbb{S}_{++}^d\), define \[P^W_{\Sigma_0}(B) \mathrel{\mathop{:}}= P_{W\mid\Sigma}(B\mid\Sigma_0) \mathrel{\mathop{:}}= K(\Sigma_0,\Sigma_0^{1/2}B), \qquad B\in\mathcal{B}(\mathbb{R}^d),\] where \[\Sigma_0^{1/2}B = \{\Sigma_0^{1/2}w:w\in B\}.\] This is well-defined because \(w\mapsto\Sigma_0^{1/2}w\) is a homeomorphism of \(\mathbb{R}^d\). It is also a probability kernel. For fixed \(B\), \[P^W_{\Sigma_0}(B) = \int_{\mathbb{R}^d} \mathbf{1}_B(\Sigma_0^{-1/2}x)\,K(\Sigma_0,dx),\] and the integrand is Borel in \((\Sigma_0,x)\).

Under the chosen conditional law given \(\Sigma=\Sigma_0\), \[W=\Sigma_0^{-1/2}X, \qquad Z=W+V, \qquad W\perp\!\!\!\!\perp V, \qquad V\sim\mathcal{N}(0,I_d).\] Hence the conditional density of \(Z\) given \(\Sigma=\Sigma_0\) is \[f_{Z\mid\Sigma}(z\mid\Sigma_0) = \int_{\mathbb{R}^d} \phi(z-w)\,dP^W_{\Sigma_0}(w).\] This density is strictly positive for every \(z\in\mathbb{R}^d\).

Now define the Gaussian density with covariance \(\Sigma_0\): \[\phi_{\Sigma_0}(u) \mathrel{\mathop{:}}= \det(\Sigma_0)^{-1/2} \phi(\Sigma_0^{-1/2}u).\] Then the conditional density of \(Y\) given \(\Sigma=\Sigma_0\) is \[f_{Y\mid\Sigma}(y\mid\Sigma_0) = \int_{\mathbb{R}^d} \phi_{\Sigma_0}(y-x)\,K(\Sigma_0,dx).\] Equivalently, with \[z_0\mathrel{\mathop{:}}=\Sigma_0^{-1/2}y,\] we have \[f_{Y\mid\Sigma}(y\mid\Sigma_0) = \det(\Sigma_0)^{-1/2} f_{Z\mid\Sigma}(z_0\mid\Sigma_0).\] Therefore \[f_{Y\mid\Sigma}(y\mid\Sigma_0)>0 \qquad \text{for all }(y,\Sigma_0)\in\mathbb{R}^d\times\mathbb{S}_{++}^d .\]

For \(A\in\mathcal{B}(\mathbb{R}^d)\), define \[\Pi_{\Sigma_0,y}(A) \mathrel{\mathop{:}}= \frac{ \displaystyle \int_A \phi_{\Sigma_0}(y-x)\,K(\Sigma_0,dx) }{ f_{Y\mid\Sigma}(y\mid\Sigma_0) }.\] Then \(\Pi_{\Sigma_0,y}\) is a probability measure. For fixed \(A\), the map \[(\Sigma_0,y,x) \mapsto \mathbf{1}_A(x)\phi_{\Sigma_0}(y-x)\] is nonnegative and Borel. Hence integration against the kernel \(K(\Sigma_0,dx)\) shows that \[(\Sigma_0,y)\mapsto\Pi_{\Sigma_0,y}(A)\] is Borel measurable. Thus \(\Pi\) is a probability kernel from \(\mathbb{S}_{++}^d\times\mathbb{R}^d\) to \(\mathbb{R}^d\).

We now verify that \(\Pi\) is a version of \(P_{X\mid Y,\Sigma}\). Let \(A,B\in\mathcal{B}(\mathbb{R}^d)\) and \(C\in\mathcal{B}(\mathbb{S}_{++}^d)\). Then \[\begin{align} &\int_C\int_B \Pi_{\Sigma_0,y}(A) f_{Y\mid\Sigma}(y\mid\Sigma_0)\,dy\,dP_\Sigma(\Sigma_0) \\ &\qquad = \int_C\int_B \int_A \phi_{\Sigma_0}(y-x)\,K(\Sigma_0,dx)\,dy\,dP_\Sigma(\Sigma_0) \\ &\qquad = \int_C \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \mathbf{1}_A(x)\mathbf{1}_B(y) \phi_{\Sigma_0}(y-x)\,dy\, K(\Sigma_0,dx)\,dP_\Sigma(\Sigma_0) \\ &\qquad = \mathbb{P}(X\in A,Y\in B,\Sigma\in C). \end{align}\] Therefore \(\Pi_{\Sigma_0,y}\) is a Bayes version of \(P_{X\mid Y,\Sigma}(\cdot\mid y,\Sigma_0)\).

In the rest of the proof, the expression \[\mathbb{E}[g(X)\mid Y=y,\Sigma=\Sigma_0]\] is understood with respect to this version. If \[\int_{\mathbb{R}^d} |g(x)|\phi_{\Sigma_0}(y-x)\,K(\Sigma_0,dx)<\infty,\] then \[\mathbb{E}[g(X)\mid Y=y,\Sigma=\Sigma_0] = \frac{ \displaystyle \int_{\mathbb{R}^d} g(x)\phi_{\Sigma_0}(y-x)\,K(\Sigma_0,dx) }{ f_{Y\mid\Sigma}(y\mid\Sigma_0) }.\] Using the definition of \(P^W_{\Sigma_0}\), the numerator becomes \[\int_{\mathbb{R}^d} g(x)\phi_{\Sigma_0}(y-x)\,K(\Sigma_0,dx) = \det(\Sigma_0)^{-1/2} \int_{\mathbb{R}^d} g(\Sigma_0^{1/2}w)\phi(z_0-w)\,dP^W_{\Sigma_0}(w).\] The same determinant factor appears in \[f_{Y\mid\Sigma}(y\mid\Sigma_0) = \det(\Sigma_0)^{-1/2} f_{Z\mid\Sigma}(z_0\mid\Sigma_0).\] After cancellation, \[\mathbb{E}[g(X)\mid Y=y,\Sigma=\Sigma_0] = \frac{ \displaystyle \int_{\mathbb{R}^d} g(\Sigma_0^{1/2}w)\phi(z_0-w)\,dP^W_{\Sigma_0}(w) }{ f_{Z\mid\Sigma}(z_0\mid\Sigma_0) }.\] This proves part (a).

Part (b): standardized Tweedie functional. Fix \(\Sigma_0\in\mathbb{S}_{++}^d\) and \(y\in\mathbb{R}^d\), and set \[z_0=\Sigma_0^{-1/2}y.\] Define \[\lambda_{g,y,\Sigma_0}(w) \mathrel{\mathop{:}}= g(\Sigma_0^{1/2}w)\phi(z_0-w).\] Assume that \[\lambda_{g,y,\Sigma_0}\in C_0(\mathbb{R}^d;\mathbb{C}).\]

For the standardized Gaussian noise, \(f_V=\phi\) and \[\varphi_V(\omega)=e^{-\|\omega\|_2^2/2}.\] The characteristic function is nowhere zero, so the convolution operator \(\mu\mapsto \phi*\mu\) is injective on \(\mathcal{M}(\mathbb{R}^d)\). Therefore, for \(f=\phi*\mu\in\mathcal{A}_V(\mathbb{R}^d)\), the measure \(\mu\) is unique.

Define \[\mathcal{T}_{g,y,\Sigma_0}[f] \mathrel{\mathop{:}}= \int_{\mathbb{R}^d} \lambda_{g,y,\Sigma_0}(w)\,d\mu(w), \qquad f=\phi*\mu.\] This is well-defined and linear. Moreover, \[\begin{align} |\mathcal{T}_{g,y,\Sigma_0}[\phi*\mu]| &\le \|\lambda_{g,y,\Sigma_0}\|_\infty \|\mu\|_{TV} \\ &= \|\lambda_{g,y,\Sigma_0}\|_\infty \|\phi*\mu\|_{\mathcal{A}_V}. \end{align}\] Thus \[\mathcal{T}_{g,y,\Sigma_0}\in\mathcal{A}_V'(\mathbb{R}^d).\] Uniqueness follows from the same injectivity argument: any continuous linear functional with this integral representation must agree with \(\mathcal{T}_{g,y,\Sigma_0}\) on every element of \(\mathcal{A}_V(\mathbb{R}^d)\).

Since \[f_{Z\mid\Sigma}(\cdot\mid\Sigma_0) = \phi*P^W_{\Sigma_0},\] we have \[\mathcal{T}_{g,y,\Sigma_0} \bigl[f_{Z\mid\Sigma}(\cdot\mid\Sigma_0)\bigr] = \int_{\mathbb{R}^d} g(\Sigma_0^{1/2}w)\phi(z_0-w)\,dP^W_{\Sigma_0}(w).\] Combining this identity with part (a) gives \[\mathbb{E}[g(X)\mid Y=y,\Sigma=\Sigma_0] = \frac{ \mathcal{T}_{g,y,\Sigma_0} \bigl[f_{Z\mid\Sigma}(\cdot\mid\Sigma_0)\bigr] }{ f_{Z\mid\Sigma}(z_0\mid\Sigma_0) }.\] This proves part (b).

Part (c): Fourier representation. Fix \(\Sigma_0\in\mathbb{S}_{++}^d\) and \(y\in\mathbb{R}^d\), and keep \[z_0=\Sigma_0^{-1/2}y.\] Consider the standardized model \[W\sim P^W_{\Sigma_0}, \qquad V\sim\mathcal{N}(0,I_d), \qquad W\perp\!\!\!\!\perp V, \qquad Z=W+V.\] In this model, the noise density is \(\phi\), the characteristic function \[\varphi_V(\omega)=e^{-\|\omega\|_2^2/2}\] is nowhere zero, and \(\phi\in C_0^k(\mathbb{R}^d)\) for every \(k\ge0\).

Assume the hypotheses in part (c). In particular, \[\lambda_{g,y,\Sigma_0}\in C_0(\mathbb{R}^d;\mathbb{C}),\] conditions (i) and (ii) of Theorem 2 hold for \(\mathcal{Q}_{g,y,\Sigma_0}\), and \(\mathcal{Q}_{g,y,\Sigma_0}\) is a Fourier-domain representer for \(\lambda_{g,y,\Sigma_0}\). Therefore Theorem 2 applies to the standardized Gaussian model.

Set \[Q_{\Sigma_0,y} \mathrel{\mathop{:}}= \mathcal{Q}_{g,y,\Sigma_0}.\] The theorem implies that \[\mathcal{F}^{-1}\{\mathcal{I}_{Q_{\Sigma_0,y}}\}\] extends continuously to \(\Xi^k(\mathbb{R}^d;\mathbb{C})\), and that its restriction to \(\mathcal{A}_V(\mathbb{R}^d)\) equals \(\mathcal{T}_{g,y,\Sigma_0}\). Hence, for every \(f\in\mathcal{A}_V(\mathbb{R}^d)\) and every sequence \((f_n)_{n\ge1}\subset\mathscr S(\mathbb{R}^d)\) such that \[\|f_n-f\|_{\Xi^k}\to0,\] we have \[\mathcal{T}_{g,y,\Sigma_0}[f] = \lim_{n\to\infty} \mathcal{F}^{-1}\{\mathcal{I}_{Q_{\Sigma_0,y}}\}[f_n].\] Equivalently, using the Fourier convention fixed above, \[\mathcal{T}_{g,y,\Sigma_0}[f] = \lim_{n\to\infty} \frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} Q_{\Sigma_0,y}(\omega)\widetilde{f}_n(-\omega)\,d\omega .\] The limit is independent of the approximating sequence.

Taking \[f=f_{Z\mid\Sigma}(\cdot\mid\Sigma_0) = \phi*P^W_{\Sigma_0}\] gives \[\mathcal{T}_{g,y,\Sigma_0} \bigl[f_{Z\mid\Sigma}(\cdot\mid\Sigma_0)\bigr] = \lim_{n\to\infty} \frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} Q_{\Sigma_0,y}(\omega)\widetilde{f}_n(-\omega)\,d\omega ,\] for any \((f_n)_{n\ge1}\subset\mathscr S(\mathbb{R}^d)\) satisfying \[\|f_n-f_{Z\mid\Sigma}(\cdot\mid\Sigma_0)\|_{\Xi^k}\to0.\] Combining this with the ratio from part (b) yields \[\mathbb{E}[g(X)\mid Y=y,\Sigma=\Sigma_0] = \frac{1}{f_{Z\mid\Sigma}(z_0\mid\Sigma_0)} \lim_{n\to\infty} \frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} Q_{\Sigma_0,y}(\omega)\widetilde{f}_n(-\omega)\,d\omega .\] This proves part (c). ◻

Proof. Write \[Q\mathrel{\mathop{:}}=\mathcal{Q}_{g,V,y}.\]

Step 1: temperedness. Since \(Q\in L^1(\mathbb{R}^d;\mathbb{C})\), the functional \[\mathcal{I}_Q[\psi] = \int_{\mathbb{R}^d}Q(\omega)\psi(\omega)\,d\omega, \qquad \psi\in\mathscr S(\mathbb{R}^d),\] is well-defined. Moreover, \[|\mathcal{I}_Q[\psi]| \le \|Q\|_1\|\psi\|_\infty .\] Thus \[\mathcal{I}_Q\in\mathscr S'(\mathbb{R}^d).\]

Step 2: inverse Fourier transform on the Schwartz class. Define \[Q^\sharp(z) \mathrel{\mathop{:}}= \frac{1}{(2\pi)^d} \int_{\mathbb{R}^d}e^{-i\omega^\top z}Q(\omega)\,d\omega .\] By the Riemann–Lebesgue lemma, \[Q^\sharp\in C_0(\mathbb{R}^d;\mathbb{C}),\] and \[\|Q^\sharp\|_\infty \le (2\pi)^{-d}\|Q\|_1 .\]

Let \(\psi\in\mathscr S(\mathbb{R}^d)\). Since \[\int_{\mathbb{R}^d}\int_{\mathbb{R}^d} |Q(\omega)\psi(z)|\,d\omega\,dz = \|Q\|_1\|\psi\|_1 <\infty,\] Fubini’s theorem applies. Therefore \[\begin{align} \int_{\mathbb{R}^d}Q^\sharp(z)\psi(z)\,dz &= \frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} Q(\omega) \left[ \int_{\mathbb{R}^d}e^{-i\omega^\top z}\psi(z)\,dz \right]d\omega \\ &= \frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} Q(\omega)\widetilde{\psi}(-\omega)\,d\omega \\ &= \int_{\mathbb{R}^d} Q(\omega)\psi^\sharp(\omega)\,d\omega \\ &= \mathcal{I}_Q[\psi^\sharp] \\ &= \mathcal{F}^{-1}\{\mathcal{I}_Q\}[\psi]. \end{align}\] Hence \[\mathcal{F}^{-1}\{\mathcal{I}_Q\}[\psi] = \int_{\mathbb{R}^d}Q^\sharp(z)\psi(z)\,dz, \qquad \psi\in\mathscr S(\mathbb{R}^d).\]

Step 3: extension to \(\Xi^0\) and application of the master theorem. Define \[L[f] \mathrel{\mathop{:}}= \int_{\mathbb{R}^d}Q^\sharp(z)f(z)\,dz, \qquad f\in L^1(\mathbb{R}^d;\mathbb{C}).\] Since \(Q^\sharp\in L^\infty(\mathbb{R}^d;\mathbb{C})\), \[|L[f]| \le \|Q^\sharp\|_\infty\|f\|_1.\] In particular, for \(f\in\Xi^0(\mathbb{R}^d;\mathbb{C})\), \[|L[f]| \le \|Q^\sharp\|_\infty\|f\|_{\Xi^0}.\] Thus \(L\) is continuous on \(\Xi^0(\mathbb{R}^d;\mathbb{C})\). Step 2 shows that \(L\) extends \(\mathcal{F}^{-1}\{\mathcal{I}_Q\}\) from \(\mathscr S(\mathbb{R}^d)\) to \(\Xi^0(\mathbb{R}^d;\mathbb{C})\).

The assumptions of the proposition give the standing hypotheses of Theorem 2. Moreover, Step 1 verifies condition (i), the preceding bound verifies condition (ii) with \(k=0\), and the assumption that \(Q\) is a Fourier-domain representer for \(\lambda_{g,V,y}\) gives condition (iii). Therefore Theorem 2 gives, for every \(f\in\mathcal{A}_V(\mathbb{R}^d)\),

\[\mathcal{T}_{g,V,y}[f] = L[f] = \int_{\mathbb{R}^d}Q^\sharp(z)f(z)\,dz .\]

Now assume Assumptions 1 and 3. By Proposition 1, \[f_Y = f_V*P_X \in \mathcal{A}_V(\mathbb{R}^d).\] Taking \(f=f_Y\) in the preceding display gives \[\mathcal{T}_{g,V,y}[f_Y] = \int_{\mathbb{R}^d}Q^\sharp(z)f_Y(z)\,dz.\] Since \(Q^\sharp\) is bounded, the right-hand side equals \[\mathbb{E}[Q^\sharp(Y)].\] Finally, the ratio formula 3 yields, whenever \(f_Y(y)>0\), \[\mathbb{E}[g(X)\mid Y=y] = \frac{\mathcal{T}_{g,V,y}[f_Y]}{f_Y(y)} = \frac{\mathbb{E}[Q^\sharp(Y)]}{f_Y(y)}.\] ◻

Proof. Set \[Z\mathrel{\mathop{:}}=g(X).\] The conditional expectation \(\mathbb{E}[Z\mid Y]\) exists as an integrable \(\sigma(Y)\)-measurable random variable if and only if \(Z\in L^1\). Since the marginal law of \(X\) is \(P_X\), \[\mathbb{E}|Z| = \int_{\mathbb{R}^d}|g(x)|\,dP_X(x).\] Thus \(Z\in L^1\) if and only if \[\int_{\mathbb{R}^d}|g(x)|\,dP_X(x)<\infty .\] Because \(P_X\) is a probability measure, \(|P_X|=P_X\), so this condition is equivalent to \[P_X\in\mathcal{M}_g(\mathbb{R}^d).\] ◻

Proof. Let \[f=\mathcal{K}_V[\mu]\in\mathcal{A}_{V,g}(\mathbb{R}^d).\] By Assumption 2, the representing measure \(\mu\) is unique.

We first show that the Bochner integral is well-defined. By Assumption 4, \[f_V\in\Xi^k(\mathbb{R}^d;\mathbb{C}).\] By Lemma 3, the map \[x\longmapsto \tau_x f_V\] is continuous from \(\mathbb{R}^d\) into \(\Xi^k(\mathbb{R}^d;\mathbb{C})\). Since \(g\) is measurable and scalar multiplication is continuous, the map \[x\longmapsto g(x)\tau_x f_V\] is strongly measurable. Translations are isometries on \(\Xi^k\), so \[\|g(x)\tau_x f_V\|_{\Xi^k} = |g(x)|\|f_V\|_{\Xi^k}.\] Therefore \[\int_{\mathbb{R}^d} \|g(x)\tau_x f_V\|_{\Xi^k}\,d|\mu|(x) = \|f_V\|_{\Xi^k} \int_{\mathbb{R}^d}|g(x)|\,d|\mu|(x) <\infty,\] because \(\mu\in\mathcal{M}_g(\mathbb{R}^d)\). Hence \(x\mapsto g(x)\tau_x f_V\) is Bochner integrable with respect to \(\mu\).

Define \[\mathscr T_{g,V}[f] \mathrel{\mathop{:}}= \int_{\mathbb{R}^d}g(x)\tau_x f_V\,d\mu(x).\] The definition is unambiguous because \(\mu=\mathcal{K}_V^{-1}[f]\) is unique. Linearity follows from the linearity of \(\mathcal{K}_V^{-1}\) and of the Bochner integral. This proves existence and uniqueness of the operator with the displayed representation.

We now identify its pointwise values. For each \(y\in\mathbb{R}^d\), evaluation at \(y\) is a continuous linear functional on \(\Xi^k(\mathbb{R}^d;\mathbb{C})\), because \[|h(y)|\le \|h\|_\infty\le \|h\|_{\Xi^k}.\] Thus evaluation commutes with the Bochner integral: \[\begin{align} (\mathscr T_{g,V}[f])(y) &= \int_{\mathbb{R}^d} g(x)(\tau_x f_V)(y)\,d\mu(x) \\ &= \int_{\mathbb{R}^d} g(x)f_V(y-x)\,d\mu(x). \end{align}\] If \(\lambda_{g,V,y}\in C_0(\mathbb{R}^d;\mathbb{C})\), then Theorem 1 gives \[\mathcal{T}_{g,V,y}[f] = \int_{\mathbb{R}^d} g(x)f_V(y-x)\,d\mu(x).\] Hence \[(\mathscr T_{g,V}[f])(y) = \mathcal{T}_{g,V,y}[f].\]

It remains to prove the almost-sure posterior representation. Suppose that \(P_X\in\mathcal{M}_g(\mathbb{R}^d)\) and that Assumptions 1 and 3 hold. By Proposition 1, \[f_Y = f_V*P_X = \mathcal{K}_V[P_X],\] so \[f_Y\in\mathcal{A}_{V,g}(\mathbb{R}^d).\] Moreover, since \(f_V\) is bounded under Assumption 4, \[\int_{\mathbb{R}^d} |g(x)|f_V(y-x)\,dP_X(x) \le \|f_V\|_\infty \int_{\mathbb{R}^d}|g(x)|\,dP_X(x) <\infty.\] Thus the pointwise ratio formula applies for every \(y\) such that \(f_Y(y)>0\): \[\mathbb{E}[g(X)\mid Y=y] = \frac{ \displaystyle \int_{\mathbb{R}^d}g(x)f_V(y-x)\,dP_X(x) }{ f_Y(y) }.\] Using the pointwise identity above with \(\mu=P_X\), we obtain \[\mathbb{E}[g(X)\mid Y=y] = \frac{(\mathscr T_{g,V}[f_Y])(y)}{f_Y(y)}, \qquad f_Y(y)>0.\]

Finally, \[\mathbb{P}\{f_Y(Y)=0\} = P_Y\{y:f_Y(y)=0\} = \int_{\{y:f_Y(y)=0\}}f_Y(y)\,dy = 0.\] Hence the ratio is defined \(\mathbb{P}\)-almost surely, with arbitrary values on \(\{f_Y(Y)=0\}\). Therefore \[\mathbb{E}[g(X)\mid Y] = \frac{\mathscr T_{g,V}[f_Y](Y)}{f_Y(Y)} \qquad \mathbb{P}\text{-almost surely}.\] ◻

Proof. By Proposition 7, \[N_g \mathrel{\mathop{:}}= \mathscr T_{g,V}[f_Y] \in \Xi^k(\mathbb{R}^d;\mathbb{C}).\] By Lemma 1(b), \[f_Y=\mathcal{K}_V[P_X]\in\Xi^k(\mathbb{R}^d;\mathbb{C}).\] Hence \(N_g\) and \(f_Y\) both belong to \(C^k(\mathbb{R}^d;\mathbb{C})\).

Since \(f_Y\) is continuous, \[\mathcal{O}_Y = \{y\in\mathbb{R}^d:f_Y(y)>0\}\] is open. On \(\mathcal{O}_Y\), the denominator does not vanish. Therefore the quotient rule gives \[\frac{N_g}{f_Y} \in C^k(\mathcal{O}_Y;\mathbb{C}).\] By Proposition 7, this quotient is a version of \(\mathbb{E}[g(X)\mid Y]\) on \(\mathcal{O}_Y\). ◻

Proof. By definition, \[\widetilde{\lambda}_{g,V,y}(\omega) = \int_{\mathbb{R}} e^{i\omega x}x f_V(y-x)\,dx .\] Make the change of variables \(u=y-x\). Then \(x=y-u\), and \[\widetilde{\lambda}_{g,V,y}(\omega) = \int_{\mathbb{R}} e^{i\omega(y-u)}(y-u)f_V(u)\,du .\] Therefore \[\widetilde{\lambda}_{g,V,y}(\omega) = e^{i\omega y} \left[ y \int_{\mathbb{R}}e^{-i\omega u}f_V(u)\,du - \int_{\mathbb{R}}u e^{-i\omega u}f_V(u)\,du \right].\] The first integral is \(\varphi_V(-\omega)\).

It remains to identify the second integral. Since \(\lambda_{g,V,y}\in L^1(\mathbb{R})\), the same change of variables gives \[\int_{\mathbb{R}}|y-u|f_V(u)\,du<\infty .\] Because \(f_V\) is a probability density, \[\int_{\mathbb{R}}|u|f_V(u)\,du \le \int_{\mathbb{R}}|y-u|f_V(u)\,du + |y|\int_{\mathbb{R}}f_V(u)\,du <\infty .\] Thus \(u\mapsto u f_V(u)\) is integrable. By dominated convergence, \[\frac{d}{d\omega}\varphi_V(-\omega) = \frac{d}{d\omega} \int_{\mathbb{R}}e^{-i\omega u}f_V(u)\,du = \int_{\mathbb{R}}(-iu)e^{-i\omega u}f_V(u)\,du .\] Hence \[\int_{\mathbb{R}}u e^{-i\omega u}f_V(u)\,du = i\frac{d}{d\omega}\varphi_V(-\omega).\] Substituting this identity into the expression for \(\widetilde{\lambda}_{g,V,y}\) gives \[\widetilde{\lambda}_{g,V,y}(\omega) = e^{i\omega y} \left[ y\varphi_V(-\omega) - i\frac{d}{d\omega}\varphi_V(-\omega) \right].\]

On \(Z_V^c\), division by \(\varphi_V(-\omega)\) gives \[\frac{\widetilde{\lambda}_{g,V,y}(\omega)}{\varphi_V(-\omega)} = e^{i\omega y} \left[ y - i \frac{\frac{d}{d\omega}\varphi_V(-\omega)}{\varphi_V(-\omega)} \right].\] On \(Z_V\), we set \(\mathcal{Q}_{g,V,y}=0\). Since \(Z_V\) is Lebesgue null, \[\mathcal{Q}_{g,V,y}(\omega)\varphi_V(-\omega) = \widetilde{\lambda}_{g,V,y}(\omega) \qquad \text{for Lebesgue-a.e. }\omega .\] Thus \(\mathcal{Q}_{g,V,y}\) is a Fourier-domain representer. ◻

Proof. Write \[c\mathrel{\mathop{:}}=c_{g,V,y}, \qquad \kappa\mathrel{\mathop{:}}=k_{g,V,y}, \qquad Q\mathrel{\mathop{:}}=\mathcal{Q}_{g,V,y}.\]

Step 1: temperedness. By assumption, \(Q\) agrees Lebesgue-a.e. with \[Q_*(\omega) \mathrel{\mathop{:}}= e^{i\omega y} \left[ -i\pi c\,\operatorname{sgn}(\omega) + q_\kappa(\omega) \right],\] where \[q_\kappa(\omega) \mathrel{\mathop{:}}= \int_0^\infty \bigl(e^{-i\omega t}-e^{i\omega t}\bigr)\kappa(t)\,dt .\] The integrability assumptions on \(\kappa\) imply that \(q_\kappa\) is well-defined. Indeed, \[|e^{-i\omega t}-e^{i\omega t}| = 2|\sin(\omega t)| \le 2\min\{|\omega|t,1\}.\] Therefore \[|q_\kappa(\omega)| \le 2|\omega|\int_0^1 t|\kappa(t)|\,dt + 2\int_1^\infty |\kappa(t)|\,dt .\] Hence there is a finite constant \(C>0\) such that \[|Q_*(\omega)| \le C(1+|\omega|), \qquad \omega\in\mathbb{R} .\] Since \(Q=Q_*\) Lebesgue-a.e., the function \(Q\) may be replaced by \(Q_*\) for distributional purposes. For any integer \(N>2\), \[\int_{\mathbb{R}} \frac{|Q(\omega)|}{(1+|\omega|)^N}\,d\omega <\infty .\] Lemma 2 gives \[\mathcal{I}_Q\in\mathscr S'(\mathbb{R}).\]

Step 2: inverse Fourier transform on the Schwartz class. Let \(\psi\in\mathscr S(\mathbb{R})\), and write \[\psi^\sharp(\omega) = (2\pi)^{-1}\widetilde{\psi}(-\omega).\] Since \(\psi^\sharp\in\mathscr S(\mathbb{R})\), \[\int_{\mathbb{R}}|\psi^\sharp(\omega)|\,d\omega<\infty, \qquad \int_{\mathbb{R}}|\omega|\,|\psi^\sharp(\omega)|\,d\omega<\infty .\] Moreover, \[\begin{align} &\int_0^\infty \int_{\mathbb{R}} \left| e^{i\omega y} \bigl(e^{-i\omega t}-e^{i\omega t}\bigr) \kappa(t)\psi^\sharp(\omega) \right| \,d\omega\,dt \\ &\qquad\le 2\int_0^1 t|\kappa(t)|\,dt \int_{\mathbb{R}}|\omega|\,|\psi^\sharp(\omega)|\,d\omega + 2\int_1^\infty |\kappa(t)|\,dt \int_{\mathbb{R}}|\psi^\sharp(\omega)|\,d\omega <\infty . \end{align}\] Thus Fubini’s theorem applies to the \(q_\kappa\)-term. Fourier inversion gives \[\int_{\mathbb{R}} e^{i\omega(y-t)}\psi^\sharp(\omega)\,d\omega = \psi(y-t), \qquad \int_{\mathbb{R}} e^{i\omega(y+t)}\psi^\sharp(\omega)\,d\omega = \psi(y+t).\] Hence \[\int_{\mathbb{R}} e^{i\omega y}q_\kappa(\omega)\psi^\sharp(\omega)\,d\omega = \int_0^\infty \bigl(\psi(y-t)-\psi(y+t)\bigr)\kappa(t)\,dt .\]

It remains to handle the singular term. Under the Fourier convention used in this paper, the identity in [56], Chapter 5, Equation 5.1.12, gives \[\mathcal{F}^{-1} \left\{ \mathcal{I}_{-i\pi\operatorname{sgn}} \right\} = -\operatorname{p.v.}\!\left(\frac{1}{\,\cdot\,}\right).\] Multiplication by \(e^{i\omega y}\) shifts the inverse transform. Therefore \[\begin{align} \mathcal{F}^{-1} \left\{ \mathcal{I}_{e^{i\omega y}(-i\pi c\operatorname{sgn}(\omega))} \right\}[\psi] &= c\operatorname{p.v.} \int_{\mathbb{R}} \frac{\psi(z)}{y-z}\,dz \\ &= \pi c\,\mathcal{H}[\psi](y). \end{align}\] Combining the regular and singular parts gives, for every \(\psi\in\mathscr S(\mathbb{R})\), \[\mathcal{F}^{-1}\{\mathcal{I}_Q\}[\psi] = \pi c\,\mathcal{H}[\psi](y) + \int_0^\infty \bigl(\psi(y-t)-\psi(y+t)\bigr)\kappa(t)\,dt .\]

Step 3: \(\Xi^1\)-continuity. For \(\psi\in\Xi^1(\mathbb{R};\mathbb{C})\), define \[L[\psi] \mathrel{\mathop{:}}= \pi c\,\mathcal{H}[\psi](y) + \int_0^\infty \bigl(\psi(y-t)-\psi(y+t)\bigr)\kappa(t)\,dt .\] We show that \(L\) is well-defined and continuous on \(\Xi^1(\mathbb{R};\mathbb{C})\).

First consider the regular integral. For \(0<t\le1\), \[\psi(y-t)-\psi(y+t) = -\int_{-t}^t \psi'(y+s)\,ds,\] so \[|\psi(y-t)-\psi(y+t)| \le 2t\|\psi'\|_\infty .\] For \(t\ge1\), \[|\psi(y-t)-\psi(y+t)| \le 2\|\psi\|_\infty .\] Therefore \[\int_0^\infty |\psi(y-t)-\psi(y+t)|\,|\kappa(t)|\,dt \le 2\|\psi'\|_\infty\int_0^1 t|\kappa(t)|\,dt + 2\|\psi\|_\infty\int_1^\infty|\kappa(t)|\,dt .\]

Now consider the Hilbert-transform term. For \(\varepsilon>0\), \[\int_{|t|>\varepsilon} \frac{\psi(y-t)}{t}\,dt = \int_\varepsilon^\infty \frac{\psi(y-t)-\psi(y+t)}{t}\,dt .\] On \(0<t\le1\), \[\frac{|\psi(y-t)-\psi(y+t)|}{t} \le 2\|\psi'\|_\infty .\] On \(t\ge1\), \[\begin{align} \int_1^\infty \frac{|\psi(y-t)|}{t}\,dt &= \int_{-\infty}^{y-1} \frac{|\psi(u)|}{y-u}\,du \le \|\psi\|_1, \\ \int_1^\infty \frac{|\psi(y+t)|}{t}\,dt &= \int_{y+1}^{\infty} \frac{|\psi(u)|}{u-y}\,du \le \|\psi\|_1 . \end{align}\] Thus the principal value defining \(\mathcal{H}[\psi](y)\) exists, and \[|\mathcal{H}[\psi](y)| \le \frac{1}{\pi} \left( 2\|\psi'\|_\infty+2\|\psi\|_1 \right).\] Combining the bounds gives a finite constant \(C_{g,V,y}\) such that \[|L[\psi]| \le C_{g,V,y}\|\psi\|_{\Xi^1}, \qquad \psi\in\Xi^1(\mathbb{R};\mathbb{C}).\] Hence \[L\in\bigl(\Xi^1(\mathbb{R};\mathbb{C})\bigr)'.\] Step 2 shows that \[L[\psi] = \mathcal{F}^{-1}\{\mathcal{I}_Q\}[\psi], \qquad \psi\in\mathscr S(\mathbb{R}).\]

Step 4: passage to mixtures. The assumptions of the proposition give the standing hypotheses of Theorem 2. Moreover, Step 1 verifies condition (i), Step 3 verifies condition (ii) with smoothness order \(1\), and the assumption that \(Q\) is a Fourier-domain representer for \(\lambda_{g,V,y}\) gives condition (iii). Therefore Theorem 2 applies.

Let \(f\in\mathcal{A}_V(\mathbb{R})\), and choose \((f_n)_{n\ge1}\subset\mathscr S(\mathbb{R})\) such that \[\|f_n-f\|_{\Xi^1}\to0 .\] Then \[\mathcal{T}_{g,V,y}[f] = \lim_{n\to\infty} \mathcal{F}^{-1}\{\mathcal{I}_Q\}[f_n].\] Using Step 2 and the continuity of \(L\), \[\lim_{n\to\infty} \mathcal{F}^{-1}\{\mathcal{I}_Q\}[f_n] = \lim_{n\to\infty}L[f_n] = L[f].\] Thus \[\mathcal{T}_{g,V,y}[f] = \pi c\,\mathcal{H}[f](y) + \int_0^\infty \bigl(f(y-t)-f(y+t)\bigr)\kappa(t)\,dt .\] Restoring \(c=c_{g,V,y}\) and \(\kappa=k_{g,V,y}\) proves the functional identity.

Finally, assume Assumptions 1 and 3. By Proposition 1, \[f_Y = f_V*P_X \in \mathcal{A}_V(\mathbb{R}).\] Taking \(f=f_Y\) in the functional identity and applying the ratio formula 3 gives, whenever \(f_Y(y)>0\), \[\mathbb{E}[g(X)\mid Y=y] = \frac{1}{f_Y(y)} \left\{ \pi c_{g,V,y}\mathcal{H}[f_Y](y) + \int_0^\infty \bigl(f_Y(y-t)-f_Y(y+t)\bigr)k_{g,V,y}(t)\,dt \right\}.\] ◻

Proof. Write \[c\mathrel{\mathop{:}}=c_{g,V,y}, \qquad \kappa\mathrel{\mathop{:}}=k_{g,V,y}, \qquad Q\mathrel{\mathop{:}}=\mathcal{Q}_{g,V,y}.\]

Step 1: temperedness. By assumption, \(Q\) agrees Lebesgue-a.e. with \[Q_*(\omega) \mathrel{\mathop{:}}= e^{i\omega y} \bigl(c+q_\kappa(\omega)\bigr),\] where \[q_\kappa(\omega) \mathrel{\mathop{:}}= \int_0^\infty \bigl(1-e^{-i\omega t}\bigr)\kappa(t)\,dt .\] The integrability assumptions on \(\kappa\) imply that \(q_\kappa\) is well-defined. Indeed, \[|1-e^{-i\omega t}| \le \min\{2,|\omega|t\}.\] Therefore \[|q_\kappa(\omega)| \le |\omega|\int_0^1 t|\kappa(t)|\,dt + 2\int_1^\infty |\kappa(t)|\,dt .\] Hence there is a finite constant \(C>0\) such that \[|Q_*(\omega)| \le C(1+|\omega|), \qquad \omega\in\mathbb{R} .\] Since \(Q=Q_*\) Lebesgue-a.e., the two functions induce the same distribution on \(\mathscr S(\mathbb{R})\). For any integer \(N>2\), \[\int_{\mathbb{R}} \frac{|Q(\omega)|}{(1+|\omega|)^N}\,d\omega <\infty,\] after replacing \(Q\) by \(Q_*\) on a null set. Lemma 2 gives \[\mathcal{I}_Q\in\mathscr S'(\mathbb{R}).\]

Step 2: inverse Fourier transform on the Schwartz class. Let \(\psi\in\mathscr S(\mathbb{R})\), and write \[\psi^\sharp(\omega) = (2\pi)^{-1}\widetilde{\psi}(-\omega).\] Since \(\psi^\sharp\in\mathscr S(\mathbb{R})\), \[\int_{\mathbb{R}} (1+|\omega|)|\psi^\sharp(\omega)|\,d\omega <\infty .\] Moreover, \[\begin{align} &\int_0^\infty \int_{\mathbb{R}} \left| e^{i\omega y} \bigl(1-e^{-i\omega t}\bigr) \kappa(t)\psi^\sharp(\omega) \right| \,d\omega\,dt \\ &\qquad\le \int_0^1 t|\kappa(t)|\,dt \int_{\mathbb{R}}|\omega|\,|\psi^\sharp(\omega)|\,d\omega + 2\int_1^\infty |\kappa(t)|\,dt \int_{\mathbb{R}}|\psi^\sharp(\omega)|\,d\omega <\infty . \end{align}\] Thus Fubini’s theorem applies. Using \(Q=Q_*\) a.e., \[\begin{align} \mathcal{F}^{-1}\{\mathcal{I}_Q\}[\psi] &= \mathcal{I}_Q[\psi^\sharp] \\ &= \int_{\mathbb{R}}Q(\omega)\psi^\sharp(\omega)\,d\omega \\ &= c \int_{\mathbb{R}}e^{i\omega y}\psi^\sharp(\omega)\,d\omega \\ &\qquad + \int_0^\infty \left[ \int_{\mathbb{R}} e^{i\omega y} \bigl(1-e^{-i\omega t}\bigr) \psi^\sharp(\omega)\,d\omega \right]\kappa(t)\,dt . \end{align}\] Fourier inversion on \(\mathscr S(\mathbb{R})\) gives \[\int_{\mathbb{R}}e^{i\omega y}\psi^\sharp(\omega)\,d\omega = \psi(y), \qquad \int_{\mathbb{R}}e^{i\omega(y-t)}\psi^\sharp(\omega)\,d\omega = \psi(y-t).\] Therefore \[\mathcal{F}^{-1}\{\mathcal{I}_Q\}[\psi] = c\psi(y) + \int_0^\infty \bigl(\psi(y)-\psi(y-t)\bigr)\kappa(t)\,dt .\]

Step 3: \(\Xi^1\)-continuity. For \(\psi\in\Xi^1(\mathbb{R};\mathbb{C})\), define \[L[\psi] \mathrel{\mathop{:}}= c\psi(y) + \int_0^\infty \bigl(\psi(y)-\psi(y-t)\bigr)\kappa(t)\,dt .\] This is well-defined. For \(0<t\le1\), \[|\psi(y)-\psi(y-t)| \le t\|\psi'\|_\infty,\] while for \(t\ge1\), \[|\psi(y)-\psi(y-t)| \le 2\|\psi\|_\infty .\] Hence \[\begin{align} |L[\psi]| &\le |c|\,\|\psi\|_\infty + \|\psi'\|_\infty\int_0^1 t|\kappa(t)|\,dt + 2\|\psi\|_\infty\int_1^\infty |\kappa(t)|\,dt \\ &\le C_{g,V,y}\|\psi\|_{\Xi^1} \end{align}\] for some finite constant \(C_{g,V,y}\). Thus \[L\in \bigl(\Xi^1(\mathbb{R};\mathbb{C})\bigr)'.\] Step 2 shows that \[L[\psi] = \mathcal{F}^{-1}\{\mathcal{I}_Q\}[\psi], \qquad \psi\in\mathscr S(\mathbb{R}).\] Therefore condition (ii) of Theorem 2 holds with smoothness order \(1\).

Step 4: passage to mixtures. The assumptions of the proposition give the standing hypotheses of Theorem 2. Moreover, Step 1 verifies condition (i), Step 3 verifies condition (ii) with smoothness order \(1\), and the assumption that \(Q\) is a Fourier-domain representer for \(\lambda_{g,V,y}\) gives condition (iii). Therefore Theorem 2 applies.

Let \(f\in\mathcal{A}_V(\mathbb{R})\), and choose \((f_n)_{n\ge1}\subset\mathscr S(\mathbb{R})\) such that \[\|f_n-f\|_{\Xi^1}\to0 .\] Then \[\mathcal{T}_{g,V,y}[f] = \lim_{n\to\infty} \mathcal{F}^{-1}\{\mathcal{I}_Q\}[f_n].\] Using Step 2 and the continuity of \(L\), \[\lim_{n\to\infty} \mathcal{F}^{-1}\{\mathcal{I}_Q\}[f_n] = \lim_{n\to\infty}L[f_n] = L[f].\] Therefore \[\mathcal{T}_{g,V,y}[f] = c f(y) + \int_0^\infty \bigl(f(y)-f(y-t)\bigr)\kappa(t)\,dt .\] Restoring \(c=c_{g,V,y}\) and \(\kappa=k_{g,V,y}\) proves the functional identity.

Finally, assume Assumptions 1 and 3. By Proposition 1, \[f_Y = f_V*P_X \in \mathcal{A}_V(\mathbb{R}).\] Taking \(f=f_Y\) in the functional identity and applying the ratio formula 3 gives, whenever \(f_Y(y)>0\), \[\mathbb{E}[g(X)\mid Y=y] = \frac{1}{f_Y(y)} \left\{ c_{g,V,y}f_Y(y) + \int_0^\infty \bigl(f_Y(y)-f_Y(y-t)\bigr)k_{g,V,y}(t)\,dt \right\}.\] ◻

Proof. Let \[\rho_n\mathrel{\mathop{:}}=\frac{\sigma}{s_n}.\] Since \(n\ge1\), we have \(0<\rho_n<1\). We use the probabilists’ Hermite polynomials \[\operatorname{He}_k(z) = (-1)^k e^{z^2/2} \frac{d^k}{dz^k}e^{-z^2/2}, \qquad z\in\mathbb{R}.\]

Step 1: Hermite expansion of the smooth cdf. For \(u\in\mathbb{R}\), set \[z=\frac{y-u}{\sigma}.\] Then \[\frac{x+n^{-1/2}-u}{n^{-1}} = \frac{a_n+\rho_n z}{\sqrt{1-\rho_n^2}}.\] Mehler’s formula gives, for \(|\rho|<1\), \[\frac{\phi_{-\rho}(w,z)}{\phi(w)\phi(z)} = \sum_{k=0}^{\infty} \frac{(-\rho)^k}{k!} \operatorname{He}_k(w)\operatorname{He}_k(z),\] where \(\phi_{-\rho}\) is the standard bivariate normal density with correlation \(-\rho\); see [77], Theorem 4.7.2. Therefore \[\int_{-\infty}^{a} \frac{\phi_{-\rho}(w,z)}{\phi(z)}\,dw = \Phi\left(\frac{a+\rho z}{\sqrt{1-\rho^2}}\right).\]

We justify termwise integration. By Cramer’s bound for Hermite polynomials, there is a constant \(C>0\) such that \[|\operatorname{He}_k(r)| \le C e^{r^2/4}\sqrt{k!}, \qquad r\in\mathbb{R},\quad k\ge0;\] see [76], Formula 22.14.17. Hence, for fixed \(z\in\mathbb{R}\), \[\phi(w) \left| \frac{(-\rho)^k}{k!} \operatorname{He}_k(w)\operatorname{He}_k(z) \right| \le C^2 e^{z^2/4}\rho^k e^{-w^2/4}.\] Since \(\sum_{k\ge0}\rho^k<\infty\) and \(e^{-w^2/4}\) is integrable on \((-\infty,a]\), Fubini’s theorem gives \[\int_{-\infty}^{a} \frac{\phi_{-\rho}(w,z)}{\phi(z)}\,dw = \sum_{k=0}^{\infty} \frac{(-\rho)^k}{k!} \operatorname{He}_k(z) \int_{-\infty}^{a} \operatorname{He}_k(w)\phi(w)\,dw .\]

For \(k\ge1\), Rodrigues’ formula implies \[\operatorname{He}_k(w)\phi(w) = (-1)^k\phi^{(k)}(w).\] Thus \[\int_{-\infty}^{a} \operatorname{He}_k(w)\phi(w)\,dw = (-1)^k\phi^{(k-1)}(a) = (-1)^k\Phi^{(k)}(a).\] The same identity holds for \(k=0\), with \(\Phi^{(0)}=\Phi\). Therefore \[\Phi\left(\frac{a+\rho z}{\sqrt{1-\rho^2}}\right) = \sum_{k=0}^{\infty} \frac{\rho^k}{k!} \Phi^{(k)}(a)\operatorname{He}_k(z).\] Taking \(a=a_n\), \(\rho=\rho_n\), and \(z=(y-u)/\sigma\), we obtain \[g_{n,x}(u) = \sum_{k=0}^{\infty} \frac{\rho_n^k}{k!} \Phi^{(k)}(a_n) \operatorname{He}_k\!\left(\frac{y-u}{\sigma}\right).\]

Step 2: finite approximants. For \(M\ge0\), define \[g_{n,M,x,y}(u) \mathrel{\mathop{:}}= \sum_{k=0}^{M} \frac{\rho_n^k}{k!} \Phi^{(k)}(a_n) \operatorname{He}_k\!\left(\frac{y-u}{\sigma}\right).\] Let \[\lambda_{n,M,x,y}(u) \mathrel{\mathop{:}}= g_{n,M,x,y}(u)f_V(y-u), \qquad \lambda_{n,x,y}(u) \mathrel{\mathop{:}}= g_{n,x}(u)f_V(y-u).\] We now show that \[\lambda_{n,M,x,y}\to \lambda_{n,x,y} \qquad \text{uniformly on }\mathbb{R} .\]

Since \[f_V(y-u) = \frac{1}{\sigma}\phi\!\left(\frac{y-u}{\sigma}\right),\] it is enough to bound the terms \[\frac{1}{\sigma}\phi(z) \frac{\rho_n^k}{k!} \Phi^{(k)}(a_n)\operatorname{He}_k(z).\] For \(k\ge1\), \[\Phi^{(k)}(a_n) = \phi^{(k-1)}(a_n) = (-1)^{k-1} \operatorname{He}_{k-1}(a_n)\phi(a_n).\] Cramer’s bound gives \[|\Phi^{(k)}(a_n)| \le \frac{C}{\sqrt{2\pi}} e^{-a_n^2/4}\sqrt{(k-1)!}.\] Hence, uniformly in \(z\in\mathbb{R}\), \[\frac{1}{\sigma}\phi(z) \frac{\rho_n^k}{k!} |\Phi^{(k)}(a_n)| |\operatorname{He}_k(z)| \le \frac{C^2}{2\pi\sigma} e^{-a_n^2/4} \frac{\rho_n^k}{\sqrt{k}}.\] Since \[\sum_{k=1}^{\infty}\frac{\rho_n^k}{\sqrt{k}}<\infty,\] the Weierstrass \(M\)-test gives uniform absolute convergence; see [78], Theorem 7.10. Thus \[\lambda_{n,M,x,y}\to \lambda_{n,x,y} \quad \text{uniformly,}\] and \[\sup_{M\ge0} \|\lambda_{n,M,x,y}\|_\infty <\infty .\] Moreover, each \(\lambda_{n,M,x,y}\) is a finite sum of polynomials times a Gaussian density, and therefore belongs to \(C_0(\mathbb{R};\mathbb{C})\).

By Proposition 3, applied with \(M\) as the approximation index, \[\mathcal{T}_{g_{n,x},V,y}[f] = \lim_{M\to\infty} \mathcal{T}_{n,M,x,y}[f], \qquad f\in\mathcal{A}_V(\mathbb{R}),\] where \(\mathcal{T}_{n,M,x,y}\) denotes the Tweedie functional generated by \(g_{n,M,x,y}\) at the point \(y\).

Step 3: explicit calculation for the finite approximants. For the Gaussian density, \[\frac{d^k}{du^k}f_V(y-u) = \frac{1}{\sigma^k} \operatorname{He}_k\!\left(\frac{y-u}{\sigma}\right) f_V(y-u).\] Therefore \[\operatorname{He}_k\!\left(\frac{y-u}{\sigma}\right)f_V(y-u) = \sigma^k\frac{d^k}{du^k}f_V(y-u).\] Using the Fourier convention of the paper, \[\widetilde{\frac{d^k}{du^k}f_V(y-\cdot)}(\omega) = (-i\omega)^k e^{i\omega y}e^{-\sigma^2\omega^2/2}.\] Since \(\varphi_V(-\omega)=e^{-\sigma^2\omega^2/2}\), the Fourier-domain representer for \(\lambda_{n,M,x,y}\) is \[\mathcal{Q}_{n,M,x,y}(\omega) = \sum_{k=0}^{M} \frac{1}{k!} \left( -\frac{\sigma^2}{s_n} \right)^k \Phi^{(k)}(a_n) (i\omega)^k e^{i\omega y}.\] Thus Proposition 4, applied with smoothness order \(M\), gives \[\mathcal{T}_{n,M,x,y}[f] = \sum_{k=0}^{M} \frac{1}{k!} \left( -\frac{\sigma^2}{s_n} \right)^k \Phi^{(k)}(a_n) f^{(k)}(y).\]

Letting \(M\to\infty\) yields \[\mathcal{T}_{g_{n,x},V,y}[f] = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \left(\frac{\sigma^2}{s_n}\right)^k \Phi^{(k)}(a_n) f^{(k)}(y),\] provided the series is absolutely convergent. We prove this next.

Step 4: absolute convergence. Write \[f=f_V*\mu, \qquad \mu\in\mathcal{M}(\mathbb{R}).\] Since all derivatives of the Gaussian density are bounded and continuous, differentiation under the integral gives \[f^{(k)}(y) = \int_{\mathbb{R}} f_V^{(k)}(y-u)\,d\mu(u).\] By Rodrigues’ formula, \[f_V^{(k)}(y-u) = \frac{(-1)^k}{\sigma^k} \operatorname{He}_k\!\left(\frac{y-u}{\sigma}\right) f_V(y-u).\] Thus \[f^{(k)}(y) = \frac{(-1)^k}{\sigma^k} \int_{\mathbb{R}} \operatorname{He}_k\!\left(\frac{y-u}{\sigma}\right) f_V(y-u)\,d\mu(u).\] Consequently, \[\begin{align} &\left| \frac{(-1)^k}{k!} \left(\frac{\sigma^2}{s_n}\right)^k \Phi^{(k)}(a_n)f^{(k)}(y) \right| \\ &\qquad\le \int_{\mathbb{R}} \frac{1}{\sigma} \phi\!\left(\frac{y-u}{\sigma}\right) \frac{\rho_n^k}{k!} |\Phi^{(k)}(a_n)| \left| \operatorname{He}_k\!\left(\frac{y-u}{\sigma}\right) \right| \,d|\mu|(u). \end{align}\] The uniform bound from Step 2 gives, for \(k\ge1\), \[\left| \frac{(-1)^k}{k!} \left(\frac{\sigma^2}{s_n}\right)^k \Phi^{(k)}(a_n)f^{(k)}(y) \right| \le C_n\frac{\rho_n^k}{\sqrt{k}}\|\mu\|_{TV},\] where \(C_n<\infty\) does not depend on \(k\). Since \[\sum_{k=1}^{\infty}\frac{\rho_n^k}{\sqrt{k}}<\infty,\] the derivative series converges absolutely for each fixed \(n\). The \(k=0\) term is finite, so the proof is complete. ◻