$T_f$ is a distribution of zeroth order

To check that $T_f$ is a distribution of zeroth order (http://planetmath.org/Distribution4), we shall use condition (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). $\mathrm{\square }$

Title	$T_f$ is a distribution of zeroth order
Canonical name	TfIsADistributionOfZerothOrder
Date of creation	2013-03-22 13:44:28
Last modified on	2013-03-22 13:44:28
Owner	Koro (127)
Last modified by	Koro (127)
Numerical id	6
Author	Koro (127)
Entry type	Proof
Classification	msc 46F05
Classification	msc 46-00