Locally integrable distributions

Question

Is it possible that a locally integrable function which doesn't belong to any $\mathbb{L}^p$ space ($p\in [1,\infty]$) define a tempered distribution on the Schwartz space ?

Answer 1

If $f \in L^1_{loc}(\mathbb{R})$ and $\int_{-x}^x |f(t)|dt < C_1 |x|^k$ then $\varphi \mapsto \langle f, \varphi \rangle$ is a tempered distribution :

$$\int_{0}^A f(x) \varphi(x)dx = \varphi(A)\int_{0}^A f(t)dt-\int_{0}^A \left(\int_0^x f(t)dt\right) \varphi'(x)dx$$ with $\varphi \in S(\mathbb{R})$ : $\ \lim_{|A| \to \infty}\varphi(A)\int_{0}^A f(t)dt= 0$ so that $$\langle f,\varphi \rangle = \lim_{A \to \infty} \int_{0}^A f(x) \varphi(x)dx-\int_{0}^{-A} f(x) \varphi(x)dx = -\int_{-\infty}^\infty \left(\int_0^x f(t)dt\right) \varphi'(x)dx$$ And $$|\varphi'(x) (1+|x|^k)(1+x^2)| \le \sup_t |\varphi'(t)| (1+|t|^k)(1+t^2)$$ $$\le \|\varphi'\|_\infty+\|x^2\varphi'\|_\infty+\|x^k\varphi'\|_\infty+\|x^{k+2}\varphi'\|_\infty$$ i.e. $$|\langle f, \varphi \rangle| \le \int_{-\infty}^\infty \left(\int_0^x |f(t)|dt \right) |\varphi'(x)|dx \le \int_{-\infty}^\infty C_1 |x|^k \frac{|\varphi'(x)| (1+|x|^k)(1+x^2)}{(1+|x|^k)(1+x^2)}dx$$ $$ \le \int_{-\infty}^\infty C_1 \frac{\|\varphi'\|_\infty+\|x^2\varphi'\|_\infty+\|x^k\varphi'\|_\infty+\|x^{k+2}\varphi'\|_\infty }{1+x^2}dx \le \pi C_1 \sum_{\alpha+\beta \le k+3}\|x^\alpha\varphi^{(\beta)}\|_\infty$$

Showing the continuity of $\varphi \mapsto \langle f, \varphi \rangle$ in the Schwartz topology.