[proofplan] We show that every [Cauchy sequence](/page/Cauchy%20Sequence) in $\mathcal{S}(\mathbb{R}^n)$ converges to a Schwartz function. The Cauchy condition in the semi-norms $\|\cdot\|_{0,\beta}$ gives [uniform convergence](/page/Uniform%20Convergence) of all derivatives $\partial^\beta f_k$ to [continuous](/page/Continuity) limits $g_\beta$. An induction on $|\beta|$ using the classical theorem on uniform limits of derivatives identifies $g_\beta = \partial^\beta g_0$, producing a smooth limit $f = g_0$. Passing to the limit in the Cauchy condition for general $\|\cdot\|_{\alpha,\beta}$ shows $f$ has finite Schwartz semi-norms and $f_k \to f$ in every semi-norm. [/proofplan] [step:Construct a smooth pointwise limit via uniform convergence of derivatives] Let $\{f_k\}_{k \geq 1}$ be a [sequence](/page/Sequence) in $\mathcal{S}(\mathbb{R}^n)$ that is Cauchy in every semi-norm. The Cauchy condition in $\|\cdot\|_{0,\beta}$ states: for every $\varepsilon > 0$, there exists $K$ such that \begin{align*} \sup_{x \in \mathbb{R}^n} |\partial^\beta f_k(x) - \partial^\beta f_m(x)| &= \|f_k - f_m\|_{0,\beta} < \varepsilon \end{align*} for all $k, m \geq K$. Hence $\{\partial^\beta f_k\}$ is a uniformly Cauchy sequence of [continuous](/page/Continuity) [functions](/page/Function) on $\mathbb{R}^n$. By completeness of $(C_b(\mathbb{R}^n), \|\cdot\|_\infty)$, there exists $g_\beta \in C_b(\mathbb{R}^n)$ with $\partial^\beta f_k \to g_\beta$ uniformly. It remains to show $g_0 \in C^\infty(\mathbb{R}^n)$ with $\partial^\beta g_0 = g_\beta$ for every $\beta$. Proceed by induction on $|\beta|$. For $|\beta| = 0$ there is nothing to prove. For the inductive step, suppose $\partial^\gamma g_0 = g_\gamma$ for all $|\gamma| < |\beta|$, and write $\beta = \gamma + e_j$ for some coordinate direction $j$ with $|\gamma| = |\beta| - 1$. We have $\partial^\gamma f_k \to g_\gamma$ uniformly and $\partial_j(\partial^\gamma f_k) = \partial^\beta f_k \to g_\beta$ uniformly. By the classical theorem on uniform convergence of derivatives (if $h_k \to h$ pointwise and $h_k' \to \ell$ uniformly, then $h' = \ell$, applied coordinate-by-coordinate), $g_\gamma$ is [differentiable](/page/Derivative) in $x_j$ with $\partial_j g_\gamma = g_\beta$. This gives $\partial^\beta g_0 = g_\beta$ and completes the induction. Setting $f = g_0$ yields $f \in C^\infty(\mathbb{R}^n)$ with $\partial^\beta f_k \to \partial^\beta f$ uniformly for every $\beta$. [guided] We need to find a candidate [limit](/page/Limit) for the Cauchy sequence $(f_k)$. The Cauchy condition in the semi-norm $\|\cdot\|_{0,\beta} = \sup_x |\partial^\beta(\cdot)(x)|$ means the sequence of $\beta$-th derivatives $(\partial^\beta f_k)$ is uniformly Cauchy on $\mathbb{R}^n$: \begin{align*} \sup_{x \in \mathbb{R}^n} |\partial^\beta f_k(x) - \partial^\beta f_m(x)| &= \|f_k - f_m\|_{0,\beta} < \varepsilon \quad \text{for all } k, m \geq K. \end{align*} Since $(C_b(\mathbb{R}^n), \|\cdot\|_\infty)$ is complete, there exists a continuous bounded function $g_\beta$ with $\partial^\beta f_k \to g_\beta$ uniformly. But do these limits fit together? That is, is $g_\beta$ really the $\beta$-th derivative of $g_0$? We verify this by induction on $|\beta|$. The base case $|\beta| = 0$ is vacuous. For the inductive step, write $\beta = \gamma + e_j$ where $|\gamma| = |\beta| - 1$. By the inductive hypothesis, $\partial^\gamma g_0 = g_\gamma$, and we know $\partial^\gamma f_k \to g_\gamma$ uniformly and $\partial_j(\partial^\gamma f_k) = \partial^\beta f_k \to g_\beta$ uniformly. The classical theorem on interchanging uniform limits and derivatives (applied in the single coordinate $x_j$, with all other coordinates fixed) gives $\partial_j g_\gamma = g_\beta$, i.e., $\partial^\beta g_0 = g_\beta$. Setting $f = g_0$, we obtain $f \in C^\infty(\mathbb{R}^n)$ with $\partial^\beta f_k \to \partial^\beta f$ uniformly on $\mathbb{R}^n$ for every multi-index $\beta \in \mathbb{N}_0^n$. [/guided] [/step] [step:Verify rapid decay and semi-norm convergence $f_k \to f$ in $\mathcal{S}(\mathbb{R}^n)$] Fix $\alpha, \beta \in \mathbb{N}_0^n$ and $\varepsilon > 0$. By the Cauchy condition, there exists $K$ such that \begin{align*} \|f_k - f_m\|_{\alpha,\beta} &= \sup_{x \in \mathbb{R}^n} |x^\alpha \partial^\beta(f_k - f_m)(x)| < \varepsilon \end{align*} for all $k, m \geq K$. For each fixed $x \in \mathbb{R}^n$, the pointwise convergence $\partial^\beta f_m(x) \to \partial^\beta f(x)$ (from the previous step) gives \begin{align*} |x^\alpha \partial^\beta(f_k - f)(x)| &= \lim_{m \to \infty} |x^\alpha \partial^\beta(f_k - f_m)(x)| \leq \varepsilon. \end{align*} Taking the supremum over $x$ yields $\|f_k - f\|_{\alpha,\beta} \leq \varepsilon$ for all $k \geq K$, so $f_k \to f$ in the semi-norm $\|\cdot\|_{\alpha,\beta}$. In particular, $\|f\|_{\alpha,\beta} \leq \|f_K\|_{\alpha,\beta} + \|f_K - f\|_{\alpha,\beta} \leq \|f_K\|_{\alpha,\beta} + \varepsilon < \infty$. Since $\alpha$ and $\beta$ were arbitrary, every semi-norm of $f$ is finite, so $f \in \mathcal{S}(\mathbb{R}^n)$. [guided] We now show two things: that $f$ belongs to $\mathcal{S}(\mathbb{R}^n)$ (i.e., all its Schwartz semi-norms are finite), and that $f_k \to f$ in every semi-norm. Fix $\alpha, \beta \in \mathbb{N}_0^n$ and $\varepsilon > 0$. The Cauchy condition provides $K$ such that $\|f_k - f_m\|_{\alpha,\beta} < \varepsilon$ for all $k, m \geq K$. This means \begin{align*} |x^\alpha \partial^\beta(f_k - f_m)(x)| &< \varepsilon \quad \text{for all } x \in \mathbb{R}^n, \; k, m \geq K. \end{align*} How do we pass from the Cauchy condition (involving $f_m$) to a bound involving the limit $f$? Fix $x$ and $k \geq K$, and send $m \to \infty$. By the previous step, $\partial^\beta f_m(x) \to \partial^\beta f(x)$ pointwise, so: \begin{align*} |x^\alpha \partial^\beta(f_k - f)(x)| &= \lim_{m \to \infty} |x^\alpha \partial^\beta(f_k - f_m)(x)| \leq \varepsilon. \end{align*} The inequality $\leq \varepsilon$ (rather than $< \varepsilon$) comes from the fact that the limit of quantities bounded by $\varepsilon$ is bounded by $\varepsilon$. Taking the supremum over $x \in \mathbb{R}^n$: \begin{align*} \|f_k - f\|_{\alpha,\beta} &\leq \varepsilon \quad \text{for all } k \geq K. \end{align*} This establishes $f_k \to f$ in the semi-norm $\|\cdot\|_{\alpha,\beta}$. For membership in $\mathcal{S}(\mathbb{R}^n)$, we need $\|f\|_{\alpha,\beta} < \infty$. The triangle inequality gives $\|f\|_{\alpha,\beta} \leq \|f - f_K\|_{\alpha,\beta} + \|f_K\|_{\alpha,\beta} \leq \varepsilon + \|f_K\|_{\alpha,\beta} < \infty$, since $f_K \in \mathcal{S}(\mathbb{R}^n)$. Since $\alpha$ and $\beta$ were arbitrary, $f \in \mathcal{S}(\mathbb{R}^n)$. [/guided] [/step]