Call a function $f: \mathbb{R} \to \mathbb{C}$ a $\textit{tempered}$ function if there exists $N \in \mathbb{N}$ such that $$\int_{\mathbb{R}} |f(x)| \, (1+|x|)^{-N} < \infty.$$ Then it is known that the $\textit{tempered distribution}$ induced by $f$, $$T_f: \mathcal{S}(\mathbb{R}) \to \mathbb{C}$$ $$T_f(\phi) = \int_{\mathbb{R}} f(x) \, \phi(x) \, dx$$ is indeed a tempered distribution, where $\mathcal{S}(\mathbb{R})$ is the space of Schwartz functions and a tempered distribution has the property that if $\phi_j \to 0$ in $\mathcal{S}(\mathbb{R})$, then $T_f(\phi_j) \to 0$ . I would like to show that the (a priori) tempered distribution $$\delta: \mathcal{S}(\mathbb{R}) \to \mathbb{C}$$ $$\delta(\phi) = \phi(0)$$ is not induced by any tempered function $f$.
My argument is intuitively clear to me, but the rigor in some steps, I am not sure.
On the contrary, suppose that we have some $f$ so that $T_f = \delta$. Let $\varphi$ be some symmetric bump function such that $\mathrm{Supp}(\varphi) \subseteq [-1,1]$ and $\varphi(0) = 1$. Define a sequence $\varphi_j = \varphi(jx)$. Then for all $j \in \mathbb{N}$ we have $$\delta(\varphi_j)=\delta(\varphi(jx)) = \varphi(0) = 1.$$ Thus $\delta(\phi_j) \to 1$. On then other hand, since $\mathrm{Supp}(\varphi_j) = [-1/j, 1/j]$ we have $$T_f(\varphi_j) = \int_{-1/j}^{1/j} f(x) \varphi_j(x) dx$$ so that $T_f(\varphi_j) \to 0$.
By uniqueness of limits we must have $T_f \neq \delta$.