My attempt:
Let $H^2(\mathbb{R})=\left\{u\in L^2(\mathbb{R}): \left\|u\right\|_{H^2(\mathbb{R})}:=\left\|\mathcal{F}^{-1}((1+\xi^2)\widehat{u})\right\|_{L^2(\mathbb{R})}<\infty\right\}$ the Sobolev space
Let $T:H^2(\mathbb{R})\to L^2(\mathbb{R})$ with $Tu=\mathcal{F}^{-1}((1+\xi^2)\widehat{u})$ linear operator.
\begin{align} d_2(Tu,Tv)&=\left\|Tu-Tv\right\|_{L^2(\mathbb{R})}\\ &=\left\|T(u-v)\right\|_{L^2(\mathbb{R})}\\ &=\left\|\mathcal{F}^{-1}((1+\xi^2)\widehat{(u-v)})\right\|_{L^2(\mathbb{R})}\\ &=\left\|u-v\right\|_{\mathcal{H}^2(\mathbb{R})}\\ &=d_1(u,v) \end{align}
Therefore $H^2(\mathbb{R})$ is complete? (because $L^2(\mathbb{R})$ is complete)
Thanks.
Problem: Let $X$ be a complete normed space and assume the normed space $Y$ is isometric to Show that $Y$ is complete (Isometry of a complete normed space is also complete.)
Actualization: In my case, $T^{-1}:L^2(\mathbb{R})\to H^2(\mathbb{R})$ with $T^{-1}f=\mathcal{F}^{-1}(\frac{1}{1+\xi^2}\widehat{u}(\xi))$ satisfies $\left\|T^{-1}u\right\|_{H^2(\mathbb{R})}=\left\|u\right\|_{L^2(\mathbb{R})}$. Then $H^2(\mathbb{R})$ is complete? $