linear, continuous functional, Schwartz space

Question

Is $g:\mathbb{R}\to\mathbb{C}$ measurable such that it exists $N\in\mathbb{N}$ with $x\mapsto \frac{g(x)}{(1+x^2)^n}\in L^1(\mathbb{R},\lambda)$, then defines $g$ a continuous, linear functional $S_g:\mathcal{S}(R)\to\mathbb{C}$, where $\mathcal{S}(\mathbb{R})$ is the Schwartz space.

$S_g(f)=\int_{\mathbb{R}} f(x)g(x)\, dx$

[This question is related to: Schwartz space, functional analysis ]

To show, that $S_g$ is linear, is easy. Let $\mu\in\mathbb{R}$, then

$(\mu S_g)(f)=\int_{\mathbb{R}} \mu f(x)g(x)\,dx=\mu \int_{\mathbb{R}} f(x)g(x)\, dx=\mu S_g(f)$

$(S_g+S_h)(f)=\int_{\mathbb{R}} f(x)(g(x)+h(x))\, dx=\dotso=S_g(f)+S_h(f)$

Is it correct, that I can show that $S_g$ is continuous by showing it is continuous in $0$?

Can you help me to show, that $S_g$ is continuous? Thanks in advance.

Answer 1

It suffices to prove at $0$ because it is translation invariant. Also, it is metrizable, it suffices to prove that, given a sequence $(f_{n})$ and $f$ of Schwartz functions with $f_{n}\rightarrow f$ in the Schwartz seminorms, then $S_{g}(f_{n})\rightarrow S_{g}(f)$. But $\sup|(1+|x|^{2})^{N}(f_{n}(x)-f(x))|\rightarrow 0$, so $\displaystyle\int|f_{n}(x)-f(x)||g(x)|dx\leq\sup|(1+|x|^{2})^{N}(f_{n}(x)-f(x))|\int\dfrac{|g(x)|}{(1+|x|^{2})^{N}}dx$, so the result follows.

And one thing, actually you need to show that $S_{g}(f)$ is defined, that is, the integral exists for Schwartz function $f$.

Answer 2

Continuity was shown. You need to show that the integral will always converge though.

If $f \in \mathbb{S}$, then there exists N sufficiently large such that outside of some compact interval $[-N,N]$ we have that $|f| < \frac{1}{(1+x^2)^n}$. Thus $$\int_{-\infty}^{\infty}|f(x) g(x)| dx < \int_{-N}^{N}|f(x) g(x)| dx + \int_{-\infty}^{-N} \frac{|g(x)|}{(1+x^2)^n} dx + \int_{N}^{\infty} \frac{|g(x)|}{(1+x^2)^n} < \infty $$