[proofplan] We define the convolution $u * \phi$ pointwise via duality and verify its properties. Smoothness follows from showing the difference quotient of the reflected-translated Schwartz function converges in $\mathcal{S}$, so continuity of $u$ gives differentiability. Polynomial growth follows from the seminorm bound combined with a binomial expansion of the translated seminorm. The Fourier exchange formula follows by transposing the Fourier transform through the duality definition. [/proofplan] [step:Define the convolution and verify well-definedness] For each $x \in \mathbb{R}^n$, define $\tilde\phi_x \in \mathcal{S}(\mathbb{R}^n)$ by $\tilde\phi_x(y) := \phi(x - y)$. Translation and reflection both preserve the Schwartz class: the Schwartz seminorms satisfy $\|\tilde\phi_x\|_{\alpha,\beta} \leq C_{\alpha,\beta}(1 + |x|)^{|\alpha|}$ for constants depending only on $\phi$. Set $F(x) := u(\tilde\phi_x) = (u * \phi)(x)$. [/step] [step:Prove $F \in C^\infty(\mathbb{R}^n)$ by showing difference quotients converge in $\mathcal{S}$] For $h \neq 0$, define the difference quotient \begin{align*} \psi_h(y) &:= \frac{\tilde\phi_{x + he_j}(y) - \tilde\phi_x(y)}{h} = \frac{\phi(x + he_j - y) - \phi(x - y)}{h}. \end{align*} By the [mean value theorem](/theorems/186), for each $y$ there exists $\theta = \theta(y, h) \in (0,1)$ with $\psi_h(y) = (\partial_j\phi)(x + \theta h e_j - y)$. Applying the mean value theorem again to $\partial^\gamma(\partial_j\phi)$ shows that each Schwartz seminorm satisfies \begin{align*} \|\psi_h - \widetilde{(\partial_j\phi)}_x\|_{\alpha,\gamma} &\leq |h| \cdot C(1 + |x| + |h|)^{|\alpha|} \|\phi\|_{\alpha, \gamma + 2e_j}, \end{align*} which tends to $0$ as $h \to 0$. Since $u \in \mathcal{S}'(\mathbb{R}^n)$ is continuous, \begin{align*} \frac{F(x + he_j) - F(x)}{h} &= u(\psi_h) \to u(\widetilde{(\partial_j\phi)}_x) \quad \text{as } h \to 0. \end{align*} Hence $\partial_{x_j} F(x) = (u * \partial_j\phi)(x)$. By induction, $\partial^\beta F(x) = u(\widetilde{(\partial^\beta\phi)}_x)$ for every multi-index $\beta$. The identity $u * (\partial^\beta\phi) = (\partial^\beta u) * \phi$ follows from the [distributional derivative](/page/Distributional%20Derivative) definition: $(\partial^\beta u)(\tilde\phi_x) = (-1)^{|\beta|} u(\partial_y^\beta \tilde\phi_x) = u(\widetilde{(\partial^\beta\phi)}_x)$. [/step] [step:Establish polynomial growth of $F$ and all its derivatives] By the [Semi-Norm Bound Characterisation](/theorems/456), there exist $C > 0$ and integers $N, M \geq 0$ such that $|u(\psi)| \leq C \sum_{|\alpha| \leq N, \, |\gamma| \leq M} \|\psi\|_{\alpha,\gamma}$ for every $\psi \in \mathcal{S}(\mathbb{R}^n)$. Applying this to $\psi = \widetilde{(\partial^\beta\phi)}_x$ and using the substitution $z = x - y$: \begin{align*} |\partial^\beta F(x)| &\leq C \sum_{|\alpha| \leq N, \, |\gamma| \leq M} \sup_{z \in \mathbb{R}^n} |(x - z)^\alpha \, (\partial^{\beta + \gamma}\phi)(z)|. \end{align*} Expanding $(x - z)^\alpha = \sum_{\delta \leq \alpha} \binom{\alpha}{\delta} x^{\alpha - \delta} (-z)^{\delta}$ and applying the triangle inequality: \begin{align*} \sup_{z} |(x - z)^\alpha (\partial^{\beta+\gamma}\phi)(z)| &\leq \sum_{\delta \leq \alpha} \binom{\alpha}{\delta} |x|^{|\alpha - \delta|} \|\phi\|_{\delta, \beta+\gamma}. \end{align*} Each $\|\phi\|_{\delta, \beta+\gamma}$ is finite, and the sum is a polynomial of degree at most $N$ in $|x|$. Summing over $\alpha, \gamma$ gives $|\partial^\beta F(x)| \leq C_\beta(1 + |x|)^{N_\beta}$ with $N_\beta = N$. [/step] [step:Derive the Fourier exchange formula $\widehat{u * \phi} = \hat{u} \cdot \hat{\phi}$] Since $F = u * \phi$ has polynomial growth, $T_F \in \mathcal{S}'(\mathbb{R}^n)$. For every $\psi \in \mathcal{S}(\mathbb{R}^n)$: \begin{align*} \widehat{T_F}(\psi) &= T_F(\hat\psi) = \int_{\mathbb{R}^n} F(x) \, \hat\psi(x) \, d\mathcal{L}^n(x). \end{align*} Substituting $F(x) = u_y(\phi(x - y))$ and interchanging $u_y$ with the $d\mathcal{L}^n(x)$ integral (justified by the seminorm bound on $u$ and the rapid decay of $\hat\psi$): \begin{align*} \widehat{T_F}(\psi) &= u_y\!\left(\int_{\mathbb{R}^n} \phi(x - y) \, \hat\psi(x) \, d\mathcal{L}^n(x)\right) = u_y\bigl((\phi * \hat\psi)(y)\bigr). \end{align*} The classical Fourier exchange for Schwartz functions gives $\phi * \hat\psi = \widehat{\hat\phi \cdot \psi}$. Therefore \begin{align*} \widehat{T_F}(\psi) &= u_y\bigl(\widehat{\hat\phi \cdot \psi}(y)\bigr) = \hat{u}(\hat\phi \cdot \psi) = (\hat{u} \cdot \hat\phi)(\psi), \end{align*} where the second equality uses the definition of $\hat{u}$ (the Fourier transform of a tempered distribution acts by $\hat{u}(\psi) = u(\hat\psi)$) and the last equality uses the definition of the product of a distribution with a smooth function. [/step]