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Suppose that $\psi \in \mathscr{S}(\mathbb{R})^{*}$ is a tempered distribution and $\phi_n \in \mathscr{S}(\mathbb{R})$ are Schwartz functions so that $\phi_n(x) \rightarrow 1$ pointwise. For example $\phi_n(x) = e^{-x^2/n}$. Denote by $c$ any constant function, e.g. $c(x)=1$.

My questions:

1) What would be conditions that assure that $\langle \psi,c\rangle$ makes sense (exists)? I mean beyond the usual integral conditions for specific $\psi$.

2) Under which conditions does it hold that $\langle\psi, \phi_n\rangle \rightarrow \langle \psi,c\rangle$ as $n \rightarrow \infty$? Or more generally how to approximate a function $f$ not in $\mathscr{S}(\mathbb{R})$ so that $\langle\psi, \phi_n\rangle \rightarrow \langle \psi,f\rangle$?

My very preliminary thoughts:

Obviously the convergence in 2) is not in $\mathscr{S}(\mathbb{R})$ since the constant function is not Schwartz. So the continuity of $\psi$ in $\mathscr{S}(\mathbb{R})$ does not imply 2). I have not found a question like this treated anywhere. Maybe I can use Hermite expansion to show that $\langle\psi,\phi_n\rangle=\sum_m \langle\psi,p_m\rangle \langle p_m,\phi_n\rangle \rightarrow \sum_m \langle\psi,p_m\rangle \langle p_m,c\rangle$ where $p_m$ are the Hermite functions. But even then how to show that the last term equals $\langle\psi,c\rangle$?

Thank you for ideas, comments, suggestions! I don't really know how to approach this.

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Let $T$ be a tempered distribution. For a Schwartz function $\varphi$ then $T \ast \varphi \in C^\infty$.

If for some $\varphi$ Schwartz satisfying $\int_{-\infty}^\infty \varphi(x)dx = 1$ we have $T \ast \varphi \in L^1$ then let $$\langle T,1 \rangle = C, \qquad\qquad C = \int_{-\infty}^\infty T\ast \varphi(x)dx$$

$\langle T,1 \rangle = C$ makes sense in the sense that the value doesn't depend on the chosen $\varphi$.

Proof : the condition ensures that the tempered distribution $\widehat{T}$ is (represented by) a continuous function when restricted to $C^\infty_c(-r,r)$ with $r$ small enough such that $\widehat{\varphi}$ doesn't vanish there and $C = \widehat{T}(0)$.

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