$T_f$ is a distribution of zeroth order

To check that $T_f$ is a distribution of zeroth order (http://planetmath.org/Distribution4),we shall usecondition (3) on this page (http://planetmath.org/Distribution4).First, it is clear that $T_f$ is a linear mapping.To see that $T_f$ is continuous, suppose $K$ is a compact set in$U$ and $u\in {𝒟}_K$, i.e., $u$ is a smooth function with support in $K$.We then have

$|T_f(u)|$$=$$|{\displaystyle {\int }_K}f(x)u(x)𝑑x|$

$\le$${\displaystyle {\int }_K}|f(x)||u(x)|𝑑x$

$\le$${\displaystyle {\int }_K}|f(x)|𝑑x{||u||}_{\mathrm{\infty }}.$

Since $f$ is locally integrable, it follows that $C={\int }_K|f(x)|𝑑x$is finite, so

$$|T_f(u)|\le C{||u||}_{\mathrm{\infty }}.$$

Thus $f$ is a distribution of zeroth order ([1], pp. 381). $\mathrm{\square }$