Existence of Certain Locally Integrable Function Defining a Tempered Distribution

Question

We say that a distribution $T$ is tempered if for every sequence $\left\{\phi_{n}\right\}$ in $C_{c}^{\infty}(\mathbb{R}^{n})$ tending to $0$ in the topology of the Schwartz space $\mathcal{S}(\mathbb{R}^{n})$, $\langle{T,\phi_{n}}\rangle\rightarrow 0$. In this case, the density of $C_{c}^{\infty}(\mathbb{R}^{n})$ in $\mathcal{S}(\mathbb{R}^{n})$ implies that $T$ has a unique extension to a continuous linear functional on $\mathcal{S}(\mathbb{R}^{n})$, which we denote by $T\in\mathcal{S}'(\mathbb{R}^{n})$.

Given a locally integrable function $f\in L_{loc}^{1}(\mathbb{R}^{n})$, let $T_{f}\in\mathcal{D}'(\mathbb{R}^{n})$ be the distribution defined by

$$\langle{T_{f},\phi}\rangle,\qquad\forall\phi\in C_{c}^{\infty}(\mathbb{R}^{n})$$

It is known that $L_{loc}^{1}(\mathbb{R}^{n})\not\subset\mathcal{S}'(\mathbb{R}^{n})$. As noted in the linked question, the Lebesgue integral

$$\int_{\mathbb{R}^{n}}f\phi$$

may not exist for some $\phi\in\mathcal{S}(\mathbb{R}^{n})$, which gives some indication that $T_{f}$ is not tempered if $f$ grows too fast at $\infty$. My question is whether this condition is sufficient. My intuition is that it's not sufficient, but I don't have an example to show this at this moment.

Question. Does there exist an $f\in L_{loc}^{1}(\mathbb{R}^{n})$ such that the distribution $T_{f}$ is tempered, but $f\phi\notin L^{1}(\mathbb{R}^{n})$ for some $\phi\in\mathcal{S}(\mathbb{R}^{n})$?

Answer 1

It helps to recall Which functions are tempered distributions? — exactly the distributional derivatives of continuous functions of polynomial growth. The full statement of this theorem isn't needed here, but it gives an idea where to look for weird tempered distributions that are represented by a locally integrable function: take the derivative of a (locally) absolutely continuous function of polynomial growth.

For example, $f(x) = \sin(\exp(x^2))$ represents a tempered distribution on $\mathbb{R}$. Therefore, so does its derivative $$f'(x) = 2xe^{x^2}\cos(\exp(x^2))$$ Multiplying $f'$ by $\phi(x)=\exp( -x^2)$, which is in $\mathcal{S}(\mathbb{R}^{n})$, we get $$f(x)\phi(x) = 2x\cos(\exp(x^2))$$ which is pretty far from $L^1(\mathbb{R})$.