Schwartz space is a Fréchet space

I have a question about the proof of the completeness of the Schwartz space in Folland, Proposition 8.2.

Take $(f_k)$ to be a Cauchy sequence in the Schwartz space $S$.

I understand that in the proof he constructed $g_0$ which satisfies

$$\partial^\alpha f_k \to \partial^\alpha g_0$$

uniformly. But in the definition of the norm $\|\dot\|_{(N,\alpha)}$, there is a factor $(1+|x|)^N$ and after taking the sup, how can we guarantee the uniform convergence of

$$(1+|x|)^N(\partial^\alpha f_k) \to (1+|x|)^N(\partial^\alpha g_0)?$$

faster than any power of $|x| .$ More precisely, for any nonnegative integer $N$ and any multi-index $\alpha$ we define $$ \|f\|_{(N, \alpha)}=\sup _{x \in \mathbb{R}^{n}}(1+|x|)^{N}\left|\partial^{\alpha} f(x)\right| $$ then $$ \mathcal{S}=\left\{f \in C^{\infty}:\|f\|_{(N, \alpha)}<\infty \text { for all } N, \alpha\right\} $$ Examples of functions in $\mathcal{S}$ are easy to find: for instance, $f_{\alpha}(x)=x^{\alpha} e^{-|x|^{2}}$ where $\alpha$ is any multi-index. Also, clearly $C_{c}^{\infty} \subset \mathcal{S}$.

It is an important observation that if $f \in \mathcal{S}$, then $\partial^{\alpha} f \in L^{p}$ for all $\alpha$ and all $p \in[1, \infty] .$ Indeed, $\left|\partial^{\alpha} f(x)\right| \leq C_{N}(1+|x|)^{-N}$ for all $N$, and $(1+|x|)^{-N} \in L^{p}$ for $N>n / p$ by Corollary $2.52$. 8.2 Proposition. $\mathcal{S}$ is a Fréchet space with the topology defined by the norms $\|\cdot\|_{(N, \alpha)}$.

Proof. The only nontrivial point is completeness. If $\left\{f_{k}\right\}$ is a Cauchy sequence in $\mathcal{S}$, then $\left\|f_{j}-f_{k}\right\|_{(N, \alpha)} \rightarrow 0$ for all $N, \alpha .$ In particular, for each $\alpha$ the sequence $\left\{\partial^{\alpha} f_{k}\right\}$ converges uniformly to a function $g_{\alpha}$. Denoting by $e_{j}$ the vector $(0, \ldots, 1, \ldots, 0)$ with the 1 in the $j$ th position, we have $$ f_{k}\left(x+t e_{j}\right)-f_{k}(x)=\int_{0}^{t} \partial_{j} f_{k}\left(x+s e_{j}\right) d s . $$ Letting $k \rightarrow \infty$, we obtain $$ g_{0}\left(x+t e_{j}\right)-g_{0}(x)=\int_{0}^{t} g_{e_{j}}\left(x+s e_{j}\right) d s $$ The fundamental theorem of calculus implies that $g_{e_{j}}=\partial_{j} g_{0}$, and an induction on $|\alpha|$ then yields $g_{\alpha}=\partial^{\alpha} g_{0}$ for all $\alpha .$ It is then easy to check that $\left\|f_{k}-g_{0}\right\|_{(N, \alpha)} \rightarrow 0$ for all $\alpha$. QED Transcribed from screenshot

Solution 1:

The key point is to remember we have control of $\Vert f_n - f_m\Vert_{N,\alpha}$ from the Cauchy hypothesis. Let $\epsilon >0$ and pick $L$ such that $m,n > L$ implies that $$\sup_{x} (1+ \vert x \vert)^N \vert \partial^\alpha f_n(x) - \partial^\alpha f_m(x) \vert = \Vert f_n - f_m\Vert_{N,\alpha} < \epsilon . $$ For any $x$ we then have that $m,n > L$ implies that $$ (1+ \vert x \vert)^N \vert \partial^\alpha f_n(x) - \partial^\alpha f_m(x) \vert < \epsilon, $$ but we know that $\partial^\alpha f_m(x) \to \partial^\alpha g_0(x)$ as $m \to \infty$, so we can send $m\to \infty$ in the previous inequality to get $$ (1+ \vert x \vert)^N \vert \partial^\alpha f_n(x) - \partial^\alpha g_0(x) \vert \le \epsilon. $$ Since $x$ was arbitrary, we can take the supremum and deduce that $$ n \ge L \Rightarrow \Vert f_n - g_0 \Vert_{N,\alpha} \le \epsilon, $$ and we're done.