[problem] **Problem 1.** Let $f, g \in \mathcal{S}(\mathbb{R}^n)$. Show that the convolution $f * g$ belongs to $\mathcal{S}(\mathbb{R}^n)$, where \begin{align*} f * g: \mathbb{R}^n &\to \mathbb{C} \\ x &\mapsto \int_{\mathbb{R}^n} f(x - y) \, g(y) \, d\mathcal{L}^n(y). \end{align*} [/problem]

[solution] **Step 1: Smoothness via differentiation under the integral.** For any multi-index $\beta$, we differentiate under the integral sign: $\partial^\beta(f * g)(x) = (\partial^\beta f * g)(x)$. To justify this, we verify the hypotheses of the Leibniz integral rule. The integrand $f(x - y) g(y)$ is smooth in $x$ for each $y$, and the derivative $\partial_x^\beta f(x - y) g(y)$ is dominated by $\|\partial^\beta f\|_{0,0} \cdot |g(y)|$, which is integrable since $g \in \mathcal{S} \subseteq L^1$. By induction on $|\beta|$, the Leibniz rule applies at each stage, giving $\partial^\beta(f * g) = (\partial^\beta f) * g$, and since $\partial^\beta f \in L^1$ and $g \in L^1$, this convolution is a well-defined continuous function. Hence $f * g \in C^\infty(\mathbb{R}^n)$. **Step 2: Semi-norm estimates.** Fix $\alpha, \beta \in \mathbb{N}_0^n$. By Step 1, \begin{align*} |x^\alpha \, \partial^\beta(f * g)(x)| &= \left|x^\alpha \int_{\mathbb{R}^n} (\partial^\beta f)(x - y) \, g(y) \, d\mathcal{L}^n(y)\right|. \end{align*} We use the binomial expansion for multi-indices: $x^\alpha = ((x-y) + y)^\alpha = \sum_{\gamma \leq \alpha} \binom{\alpha}{\gamma} (x-y)^\gamma \, y^{\alpha - \gamma}$. Substituting: \begin{align*} |x^\alpha \, \partial^\beta(f*g)(x)| &\leq \sum_{\gamma \leq \alpha} \binom{\alpha}{\gamma} \int_{\mathbb{R}^n} |(x-y)^\gamma \, (\partial^\beta f)(x-y)| \cdot |y^{\alpha-\gamma} \, g(y)| \, d\mathcal{L}^n(y). \end{align*} **Step 3: Bound each factor.** The first factor satisfies $|(x-y)^\gamma (\partial^\beta f)(x-y)| \leq \|f\|_{\gamma, \beta}$ by the definition of the Schwartz semi-norm, since $f \in \mathcal{S}(\mathbb{R}^n)$. For the second factor, we need $|y^{\alpha-\gamma} g(y)|$ to be integrable. Choose $N > n$. Since $g \in \mathcal{S}(\mathbb{R}^n)$, we have $|y^{\alpha-\gamma} g(y)| \leq \|g\|_{\alpha - \gamma + N\mathbf{e}, 0} \cdot (1 + |y|)^{-N}$ where $N\mathbf{e}$ is a multi-index of sufficient size, and $(1+|y|)^{-N}$ is integrable over $\mathbb{R}^n$ since $N > n$. Therefore \begin{align*} \int_{\mathbb{R}^n} |y^{\alpha-\gamma} g(y)| \, d\mathcal{L}^n(y) &\leq C_N \, \|g\|_{\alpha-\gamma + N\mathbf{e}, 0} < \infty. \end{align*} **Step 4: Conclude.** Combining: \begin{align*} \|f * g\|_{\alpha,\beta} &\leq \sum_{\gamma \leq \alpha} \binom{\alpha}{\gamma} \|f\|_{\gamma,\beta} \cdot C_N \, \|g\|_{\alpha - \gamma + N\mathbf{e}, 0} < \infty. \end{align*} Since $\alpha$ and $\beta$ were arbitrary, $f * g \in \mathcal{S}(\mathbb{R}^n)$. [/solution]

[problem] **Problem 2.** Give an example of a sequence $\{f_k\}_{k \geq 1} \subseteq \mathcal{S}(\mathbb{R})$ that converges to $0$ in $L^2(\mathbb{R})$ but does not converge to $0$ in $\mathcal{S}(\mathbb{R})$. [/problem]

[solution] **Step 1: Construction.** Let $\psi \in C_c^\infty(\mathbb{R})$ satisfy $\psi \geq 0$, $\mathrm{supp}(\psi) \subseteq [0, 1]$, and $\psi(1/2) > 0$. Define \begin{align*} f_k: \mathbb{R} &\to \mathbb{R} \\ x &\mapsto k^{-1/2} \, \psi(x - k). \end{align*} Each $f_k$ is a translate of a smooth compactly supported function, scaled by $k^{-1/2}$, so $f_k \in C_c^\infty(\mathbb{R}) \subseteq \mathcal{S}(\mathbb{R})$. **Step 2: $L^2$ convergence.** The $L^2$ norm is \begin{align*} \|f_k\|_{L^2}^2 &= k^{-1} \int_{\mathbb{R}} |\psi(x - k)|^2 \, d\mathcal{L}^1(x) = k^{-1} \|\psi\|_{L^2}^2 \to 0 \end{align*} as $k \to \infty$, using the translation-invariance of Lebesgue measure. **Step 3: Failure of Schwartz convergence.** Consider the semi-norm with $\alpha = (2)$ and $\beta = (0)$: \begin{align*} \|f_k\|_{(2),(0)} &= \sup_{x \in \mathbb{R}} |x^2 \, f_k(x)| \geq |{(k + 1/2)}^2 \cdot k^{-1/2} \, \psi(1/2)|. \end{align*} Since $\psi(1/2) > 0$, this lower bound is at least $c \, k^{3/2}$ for some constant $c > 0$ and all sufficiently large $k$. Therefore $\|f_k\|_{(2),(0)} \to \infty$, and the sequence does not converge to $0$ in $\mathcal{S}(\mathbb{R})$. The geometric intuition is clear: the bump slides to infinity without shrinking, so the polynomial weight $x^2$ amplifies it by a factor that grows like $k^2$, overwhelming the $k^{-1/2}$ amplitude decay. The $L^2$ norm sees only amplitude, not location, so it detects only the shrinkage. [/solution]

[problem] **Problem 3.** Let $f \in \mathcal{S}(\mathbb{R}^n)$ and let $g \in \mathcal{S}(\mathbb{R}^n)$. Verify the identity \begin{align*} \int_{\mathbb{R}^n} \hat{f}(\xi) \, g(\xi) \, d\mathcal{L}^n(\xi) &= \int_{\mathbb{R}^n} f(x) \, \hat{g}(x) \, d\mathcal{L}^n(x) \end{align*} directly from the definition of the Fourier transform. [/problem]

[solution] **Step 1: Write out the left-hand side.** Substituting $\hat{f}(\xi) = \int_{\mathbb{R}^n} f(x) e^{-ix \cdot \xi} \, d\mathcal{L}^n(x)$: \begin{align*} \int_{\mathbb{R}^n} \hat{f}(\xi) \, g(\xi) \, d\mathcal{L}^n(\xi) &= \int_{\mathbb{R}^n} \left(\int_{\mathbb{R}^n} f(x) \, e^{-ix \cdot \xi} \, d\mathcal{L}^n(x)\right) g(\xi) \, d\mathcal{L}^n(\xi). \end{align*} **Step 2: Justify Fubini.** The double integral is absolutely convergent: since $f, g \in \mathcal{S}(\mathbb{R}^n) \subseteq L^1(\mathbb{R}^n)$, \begin{align*} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} |f(x)| \, |g(\xi)| \, d\mathcal{L}^n(x) \, d\mathcal{L}^n(\xi) &= \|f\|_{L^1} \, \|g\|_{L^1} < \infty. \end{align*} By Fubini's theorem, we may exchange the order of integration. **Step 3: Exchange and recognise.** After exchanging: \begin{align*} \int_{\mathbb{R}^n} \hat{f}(\xi) \, g(\xi) \, d\mathcal{L}^n(\xi) &= \int_{\mathbb{R}^n} f(x) \left(\int_{\mathbb{R}^n} g(\xi) \, e^{-ix \cdot \xi} \, d\mathcal{L}^n(\xi)\right) d\mathcal{L}^n(x). \end{align*} The inner integral is $\int g(\xi) e^{-i\xi \cdot x} \, d\mathcal{L}^n(\xi) = \hat{g}(x)$, since $x \cdot \xi = \xi \cdot x$. Therefore \begin{align*} \int_{\mathbb{R}^n} \hat{f}(\xi) \, g(\xi) \, d\mathcal{L}^n(\xi) &= \int_{\mathbb{R}^n} f(x) \, \hat{g}(x) \, d\mathcal{L}^n(x). \end{align*} This identity — sometimes called the *transpose formula* — is the starting point for defining the Fourier transform of tempered distributions by duality. [/solution]