On a tempered distribution which comes from a locally integrable function

Let $$f\colon \mathbb{R}^n \to \mathbb{C}$$ be a measurable function. We say that $$f$$ is locally integrable if $$f|_K$$ is integrable for all compact subsets $$K \subseteq \mathbb{R}^n$$. A locally integrable function $$f$$ yields a distribution $$T_f$$, which is given by $$\langle T_f, \phi \rangle = \int_{\mathbb{R}^n} f(x) \phi(x) \,dx$$ for $$\phi \in \mathscr{D}(\mathbb{R}^n)$$ (test functions).

Question. Suppose that $$T_f$$ is tempered as a distribution. Then $$T_f$$ is extended so that $$\langle T_f, \phi \rangle$$ is defined for all $$\phi \in \mathscr{S}(\mathbb{R}^n)$$ (rapidly decreasing functions).

1. Is $$f \phi$$ integrable for all $$\phi \in \mathscr{S}(\mathbb{R}^n)$$?
2. For $$\phi \in \mathscr{S}(\mathbb{R}^n)$$ with $$f \phi$$ integrable, is it true that $$\langle T_f, \phi \rangle = \int_{\mathbb{R}^n} f(x) \phi(x) \,dx$$?

Is $$f\phi$$ integrable for all $$\phi \in \mathscr{S}(\mathbb{R}^n)$$?
Not necessarily. Since $$F(x) = \sin e^{x^2}$$ is tempered as a distribution, its differentiation $$f(x) = 2x e^{x^2} \cos e^{x^2}$$ is also tempered as a distribution. $$\phi(x) = e^{-x^2/2}$$ is rapidly decreasing, but $$f\phi$$ is not integrable.
For $$\phi \in \mathscr{S}(\mathbb{R}^n)$$ with $$f \phi$$ integrable, is it true that $$\langle T_f, \phi \rangle = \int_{\mathbb{R}^n} f(x)\phi(x) \,dx$$?
Yes. To prove this, we take a sequence $$(\phi_k)_{k \in \mathbb{N}}$$ in $$\mathscr{D}(\mathbb{R}^n)$$ such that $$\lvert \phi_k \rvert \leq \lvert \phi \rvert$$ and $$\phi_k \to \phi$$ in $$\mathscr{S}(\mathbb{R}^n)$$. Since $$T_f$$ is tempered, $$\langle T_f, \phi_k \rangle \to \langle T_f, \phi \rangle$$. On the other hand, since $$f\phi$$ is integrable, $$\int_{\mathbb{R}^n} f(x)\phi_k(x) \,dx \to \int_{\mathbb{R}^n} f(x)\phi(x) \,dx$$ by Lebesgue’s dominated convergence theorem. Therefore, we get $$\langle T_f, \phi \rangle = \lim_{k \to \infty} \langle T_f, \phi_k \rangle = \lim_{k \to \infty} \int_{\mathbb{R}^n} f(x)\phi_k(x) \,dx = \int_{\mathbb{R}^n} f(x)\phi(x) \,dx.$$
• So for 1, $f$ is $L^1_{loc}$ and so well defines a distribution on $\mathcal D$, but its extension to a tempered distribution is not $\phi \mapsto \int f\phi$. Instead it is defined with the distributional derivative $\phi\mapsto -\int F \phi'$. Jan 25, 2021 at 11:10