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).
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 $(1+|\xi|^2)^{-s/2}\leq 1$.