functional analysis, Schwartz space, $L^p(\mathbb{R},\lambda)$

Question

Is $1\leq p\leq\infty$ then holds every $g\in L^p(\mathbb{R},\lambda)$ the condition here: linear, continuous functional, Schwartz space

As hint is given, to use, that $x\mapsto \frac{1}{1+x^2}$ is integrable over $\mathbb{R}$.

Can you give me a further hint? Thanks in advance.

Answer 1

Yes, the trick that I showed you in another post:

\begin{align*} \int|g(x)|^{p}dx&=\int|(1+|x|^{2})g(x)|^{p}\frac{1}{(1+|x|^{2})^{p}}dx\\ &\leq\left(\sup|(1+|x|^{2})g(x)|^{p}\right)\int\frac{1}{(1+|x|^{2})^{p}}dx\\ &\leq\left(\sup|(1+|x|^{2})g(x)|^{p}\right)\int\frac{1}{1+|x|^{2}}dx\\ &=\left(\sup|(1+|x|^{2})g(x)|\right)^{p}\int\frac{1}{1+|x|^{2}}dx \end{align*}

For $p=\infty$ is easy.