> [!definition] > > Denote $C^\infty = C^\infty(\real^n, \complex)$ as the [[Space of Smooth Functions|space of smooth functions]] on $R^n$. Let $N \in \nat$, $\alpha$ be a [[Multi-Index|multi-index]], and define > $ > \norm{f}_{(N, \alpha)} = \sup_{x \in \real^n}(1 + \abs{x})^N \cdot \abs{\partial^\alpha f(x)} > $ > where $\norm{f}_{(N, \alpha)} < \infty$ if $\partial^\alpha f$ vanishes faster than $(1 + \abs{x})^N$, then $\norm{\cdot}_{(N, \alpha)}$ is a [[Seminorm|seminorm]]. > > Since there are countably many of these seminorms, $\cs$ forms a [[Fréchet Space|Fréchet space]][^2]. > > *Proof*. Since for any $\lambda \in \complex$, > $ > \begin{align*} > &\bracs{x \in \real^n: (1 + \abs{x})^N \cdot \abs{\lambda \cdot \partial^\alpha f(x)}} \\ > &= \abs{\lambda} \cdot \bracs{x \in \real^n: (1 + \abs{x})^N \cdot \abs{\partial^\alpha f(x)}} > \end{align*} > $ > we have $\norm{\lambda f}_{(N, \alpha)} = \abs{\lambda}\norm{f}_{(N, \alpha)}$. Let $f, g \in C^\infty$, then > $ > \abs{\partial^\alpha (f(x) + g(x))} \le \abs{\partial^\alpha f(x)} + \abs{\partial^\alpha g(x)} > $ > so $\norm{f + g}_{(N, \alpha)} \le \norm{f}_{(N, \alpha)} + \norm{g}_{(N, \alpha)}$. > [!definition] > > The **Schwartz Space** is the space of all smooth functions, which together with all of their derivatives, vanish at infinity faster than any power of $\abs{x}$: > $ > \cs = \bracs{f \in C^\infty: \norm{f}_{(N, \alpha)} < \infty \quad \forall N \in \nat_0, \alpha} > $ > [!theorem]- > > If $f \in \cs$, then $\partial^\alpha f \in L^p$ for all $\alpha$ and all $p \in [0, \infty]$. > > *Proof*. Since > $ > \abs{\partial^\alpha f(x)} \le \norm{f}_{(N, \alpha)} \cdot (1 + \abs{x})^{-N} \quad \forall N \in \nat > $ > and $(1 + \abs{x})^{-N} \in L^p$ if $N > n/p$, $\partial^\alpha f \in L^p$. > [!theorem] > > Let $f \in C^\infty$, then the following are equivalent: > 1. $f \in \cs$. > 2. $x^\beta\partial^\alpha f$ is bounded for all multi-indices $\alpha, \beta$. > 3. $\partial^\alpha(x^\beta f)$ is bounded for all multi-indices $\alpha, \beta$. > > ### Partial First > > Let $f \in C^\infty$, then $f \in \cs$ if and only if $\norm{x^\beta \partial^\alpha f}_u < \infty$ for all $\alpha, \beta$. > > *Proof*. First note that $\abs{x^\beta} \le (1 + \abs{x})^N$ for all $\abs{\beta} \le N$, so > $ > \norm{x^\beta \partial^\alpha f}_u \le \norm{(1 + \abs{x})^N \partial^\alpha f}_u = \norm{f}_{(N, \alpha)} < \infty > $ > for any $f \in \cs$. > > On other other hand, $\sum_{j = 1}^n\abs{x_j}^N$ is strictly positive on the unit sphere, so it has a positive minimum[^1] $\delta$. Therefore $\sum_{j = 1}^n\abs{x_j}^N \ge \delta \abs{x}^N$ for all $\abs{x} = 1$. Since both sides are homogenous of degree $N$, this actually holds for all $x$. From here, > $ > \begin{align*} > (1 + \abs{x})^N &\le 2^N(1 + \abs{x}) \\ > &\le 2^n\braks{1 + \frac{1}{\delta}\sum_{j = 1}^n\abs{x_j^N}} \\ > &\le \frac{2^N}{\delta}\sum_{\abs{\beta} \le N}\abs{x^\beta} > \end{align*} > $ > Therefore > $ > \begin{align*} > \norm{f}_{(N, \alpha)} &= \norm{(1 + \abs{x})^N\partial^\alpha f}_u \\ > &\le C \norm{x^\beta \partial ^\alpha f}_u < \infty > \end{align*} > $ > and $f \in \cs$. > > ### Multiply First > > Let $f \in C^\infty$, then $f \in \cs$ if and only if $\partial^\alpha(x^\beta f)$ is bounded for all multi-indices $\alpha, \beta$. > > *Proof*. By the product rule, $\partial^\alpha(x^\beta f)$ is a linear combination of terms of the form $x^\gamma \partial^\delta f$. If $f \in \cs$, then each individual term is bounded, and $\partial^\alpha(x^\beta f)$ would be bounded. > > If $\partial^\alpha(x^\beta f)$ is bounded, for all $\alpha$ and $\beta$, then $x^\beta \partial^\alpha f$ is bounded for all $\alpha$ and $\beta$. Therefore $f \in \cs$. > [!theorem] > > Let $\phi \in C_c^\infty$ with $\phi(0) = 1$. Define $\phi^t(x) = \phi(t x)$, then for any $f \in \cs$, $\phi^t f \to f$ in $\cs$ as $t \to 0$. Therefore $\cd$ is [[Dense|dense]] in $\cs$. > > *Proof*. Let $N \in \nat$ and $\eps > 0$, then there exists $R \ge 0$ such that $(1 + \abs{x})^N \abs{f(x)} < \eps$ for all $x \in B(0, R)^c$. On $B(0, R)$, $\phi(tx) \to 1$ uniformly as $t \to 0$, so > $ > \norm{\phi^t f - f}_{(N, 0)} \le \norm{\paren{\phi^t f - f} \cdot \one_{B(0, R)}}_{(N, 0)} + \paren{1 +\norm{\phi}_u}\eps > $ > As $\eps$ is arbitrary, $\norm{\phi^t f - f}_{(N, 0)} \to 0$ as $t \to 0$. > > Let $\alpha$ be a [[Multi-Index|multi-index]], then > $ > \partial^\alpha (\phi^tf - f) = \phi^t \partial^\alpha f - \partial^\alpha f + E_t > $ > where $E_t$ is a sum of terms involving the derivatives of $\phi^t$. In particular, > $ > \abs{(\partial^\beta \phi^t)(x)} = t^{\abs \beta} \abs{(\partial^\beta \phi)(tx)} \le C_\beta t^{\abs \beta} > $ > so $\abs{E_t} \le C t$. By the same arguments, $\norm{\phi^tf - f}_{(N, \alpha)} \to 0$ as $t \to 0$. [^1]: [[Unit Circle is Compact]] [^2]: [[Schwartz Space is a Fréchet Space]]