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).
- Is $ f \phi $ integrable for all $ \phi \in \mathscr{S}(\mathbb{R}^n) $?
- 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 $?
Best Answer
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.
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 $$ \langle T_f, \phi \rangle = \lim_{k \to \infty} \langle T_f, \phi_k \rangle = \lim_{k \to \infty} \int_{\mathbb{R}^n} f(x)\phi_k(x) \,dx = \int_{\mathbb{R}^n} f(x)\phi(x) \,dx. $$