Characterization of Sobolev space $H^m(\mathbb R^n)$ with $m\in\mathbb N_0$? I want to show that if $m\in\mathbb N_0:=\mathbb N\cup\{0\}$ then $$H^m(\mathbb R^n):=\{u\in\mathscr{S}^\prime(\mathbb R^n): \exists f\in L^2(\mathbb R^n); \partial^\alpha f\in L^2(\mathbb R^n)\ \forall\ |\alpha|\leq m\ \textrm{and}\ u=T_f\}.$$
Above $T_f$ stands for the tempered distribution $$\phi\longmapsto \int_{\mathbb R^n} f(x) \phi(x)\ dx,\ \forall \phi\in\mathscr{S}(\mathbb R^n).$$ I don't like simply writing $u\in L^2(\mathbb R^n)$.
I can prove the inclusion $\supseteq$ easily, however I'm stuck on the other direction, can anyone help me?
Obs: Indeed, which statement is corret: $$H^m(\mathbb R^n):=\{u\in\mathscr{S}^\prime(\mathbb R^n): \exists f\in L^2(\mathbb R^n); \partial^\alpha f\in L^2(\mathbb R^n)\ \forall\ |\alpha|\leq m\ \textrm{and}\ u=T_f\},$$ or: $$H^m(\mathbb R^n):=\{u\in \mathscr{S}^\prime(\mathbb R^n): \forall |\alpha|\leq m\ \exists f_\alpha\in L^2(\mathbb R^n); \partial^\alpha u=T_{f_\alpha}\}?$$
The strangest thing is that the inclusion $\supseteq$ holds in the first case and the inclusion $\subseteq$ holds in the second case, however I can't prove either converse.
Recall: For $s\in\mathbb R$ we define $$H^s(\mathbb R^n):=\{u\in \mathscr{S}^\prime(\mathbb R^n): \exists f\in L^1_{\textrm{loc}}(\mathbb R^n); \widehat{u}=T_f\ \textrm{and}\ (1+|\cdot|^2)^{s/2} f\in L^2(\mathbb R^n)\}$$  
In the definition above I also avoided any abuse of language (that bothers me indeed). 
 A: The valid statement is the first one. I'll add the proof just in case someone else needs it.
Start noticing the inequalities:
$$C\sum_{|\alpha|\leq k} |\xi^{\alpha}|^2\leq (1+|\xi|^2)^k\leq C^\prime\sum_{|\alpha|\leq k} |\xi^{\alpha}|^2$$
In fact, using that $|\xi^\alpha|\leq |\xi|^{|\alpha|}$ one gets:
\begin{align*}
\sum_{|\alpha|\leq k} |\xi^{\alpha}|^2&\leq \sum_{|\alpha|\leq k} |\xi|^{2k}\\
&=\binom{n+k}{k} |\xi|^{2k}\\
&\leq \binom{n+k}{k}\max\{1, |\xi|^{2k}\}\\
&=\binom{n+k}{k} (\max\{1, |\xi|^2\})^k\\
&\leq\binom{n+k}{k} (1+|\xi|^{2})^k\\
&=\frac{1}{C}(1+|\xi|^2)^k,
\end{align*}
so we get the first inequality. As to the second, I haven't proved it yet, but it surely holds. 
Using this let us show $$H^k(\mathbb R^n)=\{u\in\mathscr{S}^\prime(\mathbb R^n): \exists f\in L^2(\mathbb R^n); \partial^\alpha f\in L^2(\mathbb R^n)\ \forall \ |\alpha|\leq k\ \textrm{and}\ u=T_f\}.$$
Let $u\in H^k(\mathbb R^n)$. Since $k\geq 0$ one finds $$u\in H^0(\mathbb R^n)=\{T_f: f\in L^2(\mathbb R^n)\}.$$ 
Recall: $s\geq t$ implies $H^s(\mathbb R^n)\subseteq H^t(\mathbb R^n)$. 
Therefore, $u=T_f$ for some $f\in L^2(\mathbb R^n)$. Let us show $\partial^\alpha f\in L^2(\mathbb R^n)$ whenever $|\alpha|\leq k$. In fact, given $|\alpha|\leq k$:
\begin{align*}
\|\partial^\alpha f\|_{L^2(\mathbb R^n)}^2&\leq \sum_{|\alpha|\leq k} \|\partial^\alpha f\|_{L^2(\mathbb R^n)}^2\\
&=\sum_{|\alpha|\leq k} \|\widehat{\partial^\alpha f}\|_{L^2(\mathbb R^n)}\\
&=\sum_{|\alpha|\leq k} \int_{\mathbb R^n} |\xi^\alpha|^2|\widehat{f}(\xi)|^2\ d\xi\\
&\leq C\int_{\mathbb R^n} (1+|\xi|^2)^k |\widehat{f}(\xi)|^2\ d\xi<\infty.
\end{align*}
The another implication uses the same ideas.
Obs: 
$(i)$ I'm not very glad with the first equality where $\widehat{\partial^\alpha f}$ appears. Why is this defined? Why is $\partial^\alpha f$ integrable?
$(ii)$ In my definition $$H^s(\mathbb R^n):=\{u\in \mathscr{S}^\prime(\mathbb R^n):\ \exists f\in L^1_{\textrm{loc}}(\mathbb R^n); \widehat{u}=T_f\ \textrm{and}\ (1+|\cdot|^2)^{s/2} f\in L^2(\mathbb R^n)\},$$ the condition $f\in L^1_{\textrm{loc}}(\mathbb R^n)$ is superflous since $(1+|\cdot|^2)^{s/2} f\in L^2(\mathbb R^n)$ already implies $f\in L^2(\mathbb R^n)$ once $(1+|\xi|^2)^{-s/2}\leq 1$.
