Recently, I read Methods of Modern Mathematical Physics by M. Reed and B. Simon to learn Fourier transform, there is a lemma before proving Fourier inversion theorem:
The maps $\hat{f}(\vec\lambda)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}e^{-i\vec x\cdot\vec \lambda}f(\vec x)d\vec x$ is continuous linear transformation of $\mathscr{S}(\mathbb{R}^n)$ into $\mathscr{S}(\mathbb{R}^n)$. Furthermore, if $\alpha$ and $\beta$ are multi-indices, then $$ ((i\vec\lambda)^\alpha D^\beta\hat{f})(\vec\lambda) = \widehat{D^\alpha((-i\vec x)^\beta f(\vec x))} $$
After I knew $\left \Vert \hat{f} \right \Vert_{\alpha,\beta}\equiv\underset{\lambda}{\text{sup}}\,\lvert \lambda^\alpha (D^\beta\hat{f})(\vec\lambda)\rvert\leq\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}\lvert D^\alpha_{\vec x}(\vec x^\beta f)\rvert d\vec x<\infty$ from the proof, it said
"Furthermore, if $k$ is large enough, $\int(1+x^2)^{-k}dx<\infty$ so that $$\begin{aligned} \left \Vert \hat{f} \right \Vert_{\alpha,\beta}&\leq\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}\frac{(1+x^2)^{-k}}{(1+x^2)^{-k}}\lvert D^\alpha_{\vec x}(-i\vec x)^\beta f(x)\rvert d\vec x\\ &\leq\frac{1}{(2\pi)^{n/2}}\left(\int_{\mathbb{R}^n}(1+x^2)^{-k}\right)\underset{\vec x}{\text{sup}}\{(1+x^2)^{+k}\lvert D^\alpha_{\vec x}(-i\vec x)^\beta f(x)\rvert\} \end{aligned}$$
Use Leibniz rule, we conclude that there exist multi-indices $\alpha_j$, $\beta_j$, and constants $c_j$ so that $$ \left \Vert \hat{f} \right \Vert_{\alpha,\beta}\leq\overset{M}{\underset{j=1}{\sum}}\,c_j\left \Vert \hat{f} \right \Vert_{\alpha_j,\beta_j} $$ Thus, $\hat{f}$ is bounded and by theorem is therefore continous."
I'm confused in regards to two questions. What is $x^2$ for $x\in\mathbb{R}^n$? Is it $(x^2_1,x^2_2,\cdots,x^2_n)$? And why does the book discuss "if $k$ is large enough, $\int(1+x^2)^{-k}dx<\infty \cdots$" after I already know Fourier transform is bounded.