# Existence of Certain Locally Integrable Function Defining a Tempered Distribution

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 $\phi\in\mathcal{S}(\mathbb{R}^{n})$?

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 $L^1(\mathbb{R})$.
• Many Thanks!. The example I worked out is $f(x)=e^{x}cos(e^{x})$, which one can integrate against something like a $C^{\infty}$ function equal to $0$ for $x\leq 0$ and equal to $e^{-x}$ for $x\geq 1$. But the idea is the same as your example. – Matt Rosenzweig May 22 '15 at 19:05