The Schwartz space $\mathcal S(\mathbb R^d)$ is the set of all complex-valued function $f \in C^{\infty}(\mathbb R^d)$ such that $\sup_{x\in \mathbb R^d}|x^{\alpha}D^{\beta}f(x)|<\infty$ where $\alpha , \beta \in \mathbb N_{0}^{d}, D^{\beta}=\frac{\partial^{|\beta|}}{\partial^{\beta_{1}} ...\partial^{\beta_{d}}},|\beta|=\beta_{1}+ ... +\beta_{d} $.
$\mathcal S'(\mathbb R^d)$ is the collection of all complex-valued linear continuous functions over $\mathcal S(\mathbb R^d)$ (tempered distribution).
Sobolev space $H^{s}(\mathbb R^d)=\left\{f\in \mathcal S'(\mathbb R^d): (1+|x|^{2})^{\frac{s}{2}}\hat{f}\in L^{2}(\mathbb R^d)\right\}, s\in \mathbb R$, where $\hat{f}$ is the fourier transform of f.
My question is f is in $H^{1}(\mathbb R)$ if and only if it is a tempered distribution such as $\hat{f}(1+|xi|)$ belongs to $L^{2}$.
First a comment on the definition you provide. Explicitly, the the definition is that $f$ is in $H^1$ if and only if all of the following conditions hold true:
Again, let me stress the fact that the condition $(1+|x|^2)^{1/2}\hat f\in L^2$ requires $f$ to be a function!
Observe now that $(1+|x|^2)^{1/2}\leq 1+|xi| \leq 2^{1/2}(1+|x|^2)^{1/2}$. It follows for any measurable function $\hat f$ that $$ (1+|x|^2)^{1/2}|\hat f|\leq ´(1+|xi|)|\hat f| \leq 2^{1/2}(1+|x|^2)^{1/2}|\hat f|. $$
The point of this computation is that if $f$ is a distribution such that condition 1 and 2 above hold, then condition $3$ is equivalent to the condition
$\hspace{0.1cm}$ 3'. The Fourier transform is in a certain weighted $L^2$ space, $(1+|xi|)\hat f\in L^2$