It is known that the Schwartz class $\mathcal{S}(\mathbb{R}^n)$ is a Fréchet space and also that the space of test functions $\mathcal{D}(\mathbb{R}^n)$ is dense in $\mathcal{S}(\mathbb{R}^n)$. Let $\mathcal{S}'(\mathbb{R}^n)$ the topological dual space $\mathcal{S}(\mathbb{R}^n)$. Now we have that the mapping $u \in \mathcal{S}'(\mathbb{R}^n) \mapsto v=u_{|\mathcal{D}(\mathbb{R}^n)} \in \mathcal{D}'(\mathbb{R}^n)$ is linear and one-to-one because convergence in $\mathcal{D}(\mathbb{R}^n)$ implies convergence in $\mathcal{S}(\mathbb{R}^n)$, and $u_{|\mathcal{D}(\mathbb{R}^n)} \in \mathcal{D}'(\mathbb{R}^n)$ determines uniquely $u \in \mathcal{S}'(\mathbb{R}^n)$. Then a distribution $v \in \mathcal{D}'(\mathbb{R}^n)$ is the restriction of an element $u \in \mathcal{S}'(\mathbb{R}^n)$ if and only if there exist $N \in \mathbb{N}$ and a constant $C_N>0$ such that
- $\displaystyle |u(\varphi)| \leq C_N q_N(\varphi)=C_N \sup_{x \in \mathbb{R}^n; |\alpha| \leq N} (1+|x|^2)^N |D^\alpha \varphi(x)|$ for all $\varphi \in \mathcal{D}(\mathbb{R}^n)$
The elements of $\mathcal{S}'(\mathbb{R}^n)$ or their restriction to $\mathcal{D}(\mathbb{R}^n)$ are called tempered distributions. On $\mathcal{S}'(\mathbb{R}^n)$ we consider the topology $\sigma(\mathcal{S}'(\mathbb{R}^n), \mathcal{S}(\mathbb{R}^n))$, so that $u_k \rightarrow u$ in $\mathcal{S}'(\mathbb{R}^n)$ means that
- $\displaystyle \langle \varphi, u_k \rangle\rightarrow \langle \varphi, u \rangle$ for all $\varphi \in \mathcal{S}(\mathbb{R}^n)$.
But if (1) is true for every test function, then I can consider the convergence in the weak topology in (2) for all $\varphi \in \mathcal{D}(\mathbb{R}^n)$ and not for all $\varphi \in \mathcal{S}(\mathbb{R}^n) \supset \mathcal{D}(\mathbb{R}^n)$
It's correct? thank you
No, this is not sufficient. Consider the sequence of tempered (!) distributions on $\Bbb{R}^1$ (i.e. $n=1$) $(u_k)_{k \in \Bbb{N}}$ defined by
$$ \langle u_k , \varphi\rangle := e^{k^2} \cdot \varphi(k). $$ Furthermore, let $\langle u, \varphi\rangle :=0$.
It is not hard to see $\langle u_k, \varphi \rangle = 0$ for each $\varphi \in \mathcal{D}$ and $k \geq k_0 =k_0(\varphi)$, but we have $$ \langle u_k, x\mapsto e^{-x^2} \rangle = e^{k^2} \cdot e^{-k^2} = 1 \not\to 0 = \langle u, x \mapsto e^{-x^2} \rangle. $$
Your condition will be sufficient, however, if we have $$ |\langle u_k, \varphi \rangle| \leq C \cdot \sup_{x \in \Bbb{R}^n, |\alpha|\leq N} (1+|x|^2)^N \cdot |\partial^\alpha \varphi(x)| $$ for all $k$, where $C$ and $N$ do not depend on $k$.