${T}_{f}$ 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
$|{T}_{f}(u)|$ | $=$ | $|{\displaystyle {\int}_{K}}f(x)u(x)\mathit{d}x|$ | ||
$\le $ | ${\int}_{K}}|f(x)||u(x)|\mathit{d}x$ | |||
$\le $ | ${\int}_{K}}|f(x)|\mathit{d}x{||u||}_{\mathrm{\infty}}.$ |
Since $f$ is locally integrable, it follows that $C={\int}_{K}|f(x)|\mathit{d}x$ is finite, so
$$|{T}_{f}(u)|\le C{||u||}_{\mathrm{\infty}}.$$ |
Thus $f$ is a distribution of zeroth order ([1], pp. 381). $\mathrm{\square}$
References
- 1 S. Lang, Analysis II, Addison-Wesley Publishing Company Inc., 1969.
Title | ${T}_{f}$ is a distribution of zeroth order |
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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 |