Suppose that a distribution $R\in D'(\Bbb R)$ satisfies the following estimation for an independent constant $c$:
$$\forall \phi\in D(\Bbb R)\quad |\langle R,\phi\rangle|\le c\|\phi, \,L^1(\Bbb R)\|.$$ It is easy to show that $R$ is a tempered distribution; this follows form the inequality $$\int_{\Bbb R}|\phi(x)|dx\le \sup_{x\in\Bbb R}\left((1+x^2)|\phi(x)|\right)\int_{\Bbb R}\frac{dx}{1+x^2}.$$
Take now a distribution $T\in D'(\Bbb R)$ such that $$\forall \phi\in D(\Bbb R)\quad |\langle T,\phi\rangle|\le c\|\hat \phi, \,L^1(\Bbb R)\|,$$where $\hat \phi$ is the Fourier transform of $\phi$ (we can even show that $T$ is of order at most $1$).
Can we say that there exists a tempered distribution $R\in S'(\Bbb R)$ such that $$\forall \phi\in D(\Bbb R)\quad \langle T,\phi\rangle=\langle R,\hat\phi\rangle? $$
Yes, there is such a tempered distribution. In fact, a unique such tempered distribution. Recall that the Fourier transform is an automorphism of $S(\mathbb{R})$, and the inclusion $\iota \colon D(\mathbb{R}) \hookrightarrow S(\mathbb{R})$ has dense image. Therefore $\hat{D}(\mathbb{R}) = \mathscr{F}(D(\mathbb{R}))$ is a dense subspace of $S(\mathbb{R})$. Since the inclusion $S(\mathbb{R}) \hookrightarrow L^1(\mathbb{R})$ is continuous, the given estimate $\lvert \langle T,\phi\rangle\rvert \leqslant c\lVert \hat{\phi}, L^1(\mathbb{R})\rVert$ says that $U = T \circ \mathscr{F}^{-1}$ is a continuous linear functional on $\hat{D}(\mathbb{R}) \subset S(\mathbb{R})$. By denseness, $U$ has a unique continuous extension $R$ to $S(\mathbb{R})$.