# $T_{f}$ 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 $u\in\mathcal{D}_{K}$, i.e., $u$ is a smooth function with support in $K$. We then have

 $\displaystyle|T_{f}(u)|$ $\displaystyle=$ $\displaystyle|\int_{K}f(x)u(x)dx|$ $\displaystyle\leq$ $\displaystyle\int_{K}|f(x)|\,\,|u(x)|dx$ $\displaystyle\leq$ $\displaystyle\int_{K}|f(x)|dx\,||u||_{\infty}.$

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

 $|T_{f}(u)|\leq C||u||_{\infty}.$

Thus $f$ is a distribution of zeroth order ([1], pp. 381). $\Box$

## References

• 1 S. Lang, Analysis II, Addison-Wesley Publishing Company Inc., 1969.
Title $T_{f}$ is a distribution of zeroth order TfIsADistributionOfZerothOrder 2013-03-22 13:44:28 2013-03-22 13:44:28 Koro (127) Koro (127) 6 Koro (127) Proof msc 46F05 msc 46-00