> [!theorem] > > The [[Schwartz Space|Schwartz space]] $\cs$ is a [[Fréchet Space|Fréchet space]] with its [[Topological Space|topology]] determined by the seminorms $\norm{\cdot}_{(N, \alpha)}$. > > *Proof*. > > # Hausdorff > > Since $\norm{\cdot}_{(0, 0)} = \norm{\cdot}_u$, for any $f \ne g$, $\norm{f - g}_u > 0$ and there exists open sets separating them. > > # Completeness > > ### Uniform Convergence of Partials > > Let $\seq{f_k} \subset \cs$ be a [[Cauchy Sequence|Cauchy sequence]] and $\alpha$ be a multi-index, then $\seq{\partial^\alpha f_k}$ converges [[Uniform Convergence|uniformly]] to a continuous function. > > *Proof*. Since $\seq{f_k}$ is Cauchy, $\norm{f_j - f_k}_{(N, \alpha)} \to 0$ for any $N \in \nat_0$ and $\alpha$. Fix $N = 0$, then > $ > \norm{f_j - f_k}_{(0, \alpha)} = \norm{\partial^\alpha f_j -\partial^\alpha f_k}_u \to 0 > $ > the sequence $\seq{\partial^\alpha f_n}$ is Cauchy with respect to the uniform norm. As $BC(\real^n)$ with uniform norm is complete, the sequence is convergent. > > ### Existence of Limit > > Denote $g_\alpha = \limv{n}\partial^\alpha$. Let $\seq{f_n} \subset \cs$ be a Cauchy sequence, then there exists $g = g_0 \in C^\infty$ such that > $ > \partial^\alpha g = \limv{n}\partial^\alpha f_n = g_\alpha \quad \forall \alpha > $ > where the sequence converges uniformly. > > ##### Induction Argument > > Let $k \in \nat$ and $\seq{f_n} \subset \cs$ be a Cauchy sequence such that > $ > \partial^\alpha g_0 = \partial^\alpha \lim_{n \to \infty}f_n = \limv{n}\partial^{\alpha}f_n = g_{\alpha} \quad \forall \alpha: \abs{\alpha} = k > $ > then > $ > \partial^\beta g_0 = g_{\beta} \quad \forall \beta: \abs{\beta} = k + 1 > $ > Since $\partial^0g_0 = g_0$, this applies to all multi-indices. > > *Proof*. Suppose that $g_0 \in C^{k}$ and that $\partial^\alpha g_0 = g_\alpha$ for all $\alpha: \abs{\alpha} = k$. Let $\beta$ be a multi-index with $\abs{\beta} = k + 1$, then there exists $\alpha: \abs{\alpha} = k$ and $j$ such that $\beta = \alpha + e_j$. > > Since $f_n \in C^\infty$, by the [[Mean Value Theorem|mean value theorem]], > $ > (\partial^\alpha f_n)(x + te_j) - (\partial^\alpha f_n)(x) = t\int_0^1(\partial^{\alpha + e_j}f_n)(x + ste_j)ds > $ > Taking the limit as $n \to \infty$ yields > $ > (g_\alpha)(x + te_j) - (g_\alpha)(x) = \limv{n}t\int_0^1(\partial^{\alpha + e_j}f_n)(x + ste_j)ds > $ > where the right-hand-side is dominated by $\sup_{n \in \nat}\norm{\partial^{\alpha + e_j}f_n}_u$, which thanks to being restricted to the interval $[0, 1]$, is in $L^1$. By the [[Dominated Convergence Theorem]], > $ > \limv{n}t\int_0^1(\partial^{\alpha + e_j}f_n)(x + ste_j)ds = t\int_0^1g_{\alpha + e_j}(x + ste_j)ds > $ > Push the $t$ in to obtain > $ > t\int_0^1(g_{\alpha + e_j})(x + ste_j)ds = \int_0^t g_{\alpha + e_j}(x + se_j)ds > $ > Since $s \mapsto g_{\alpha + e_j}(x + se_j)$ is continuous on $[-1, 1]$, and $0 \in [-1, 1]^o$, by FTC, > $ > \partial^{e_j}g_\alpha(x) = g_{\alpha + e_j}(x) > $ > we have $g_0 \in C^{k + 1}(\real^n, \complex)$. > > ### Convergence in Seminorms > > Let $\seq{f_k} \subset \cs$ be a Cauchy sequence and $g_0 = \limv{k}f_k$, then $\norm{f_k - g_0}_{(N, \alpha)} \to 0$ for all $N, \alpha$. > > *Proof*. Let $N$ be a non-negative integer and $\alpha$ be a multi-index. Since $\seq{f_k} \subset \cs$ is Cauchy, for any $\varepsilon > 0$, there exists $K \in \nat$ such that > $ > (1 + \abs{x})^N\abs{\partial^\alpha f_j(x) - \partial^\alpha f_k(x)} < \varepsilon \quad \forall j, k \ge K, x \in \real^n > $ > Since $\partial^\alpha f_k \to \partial^\alpha g_0$ uniformly as $k \to \infty$, > $ > (1 + \abs{x})^N\abs{\partial^\alpha f_j(x) - \partial^\alpha g_0(x)} \le \varepsilon \quad \forall j \ge K, x \in \real^n > $ > by sending $k \to \infty$. This places a bound on the supremum as > $ > \sup_{x \in \real^n}(1 + \abs{x})^N \abs{\partial^\alpha f_j(x) - \partial^\alpha g_0(x)} \le \varepsilon \quad \forall j \ge K > $ > which means that $\norm{f_k - g_0}_{(N, \alpha)} \to 0$. > > ### Membership in S > > Let $\seq{f_k} \subset \cs$ be a Cauchy sequence and $g_0 = \limv{k}f_k$, then $g_0 \in \cs$. > > *Proof*. Fix $(N, \alpha)$. Since $\seq{f_k}$ is Cauchy, there exists $K \in \nat$ such that > $ > \norm{f_K - f_k}_{(N, \alpha)} < 1 \quad \forall k \ge K > $ > and > $ > M = \sup_{k \in \nat}\norm{f_k}_{(N, \alpha)} \le \max_{k \le K}\norm{f_k}_{(N, \alpha)} + 1 < \infty > $ > which gives > $ > (1 + \abs{x})^N \cdot \abs{\partial^\alpha f_k(x)} \le M \quad \forall x \in \real^n > $ > As $\partial^\alpha f_k \to \partial^\alpha g_0$ uniformly, > $ > (1 + \abs{x})^N \cdot \abs{\partial^\alpha g_0(x)} \le M \quad \forall x \in \real^n > $ > Therefore $\norm{g_0}_{(N, \alpha)} < \infty$ and $g_0 \in \cs$.