PlanetMath: $T_f$ is a distribution of zeroth order

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is a distribution of zeroth order

(Proof)

To check that is a distribution of zeroth order, we shall use condition (3) on this page. First, it is clear that is a linear mapping. To see that is continuous, suppose is a compact set in and $u\in \mathcal{D}_K$ , i.e., is a smooth function with support in . We then have

$\displaystyle \vert T_f(u)\vert$	$\displaystyle =$	$\displaystyle \vert\int_K f(x) u(x) dx \vert$
	$\displaystyle \le$	$\displaystyle \int_K \vert f(x)\vert \,\, \vert u(x)\vert dx$
	$\displaystyle \le$	$\displaystyle \int_K \vert f(x)\vert dx \, \vert\vert u\vert\vert _\infty.$

Since

is locally integrable, it follows that $C=\int_K \vert f(x)\vert dx$ is finite, so

$\displaystyle \vert T_f(u)\vert \le C \vert\vert u\vert\vert _\infty.$

Thus

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

References

is a distribution of zeroth order" is owned by Koro. [ owner history (1) ]

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is a distribution of zeroth order, born on 2003-07-09, modified 2003-12-20.
Object id is 4434, canonical name is T_fIsADistributionOfZerothOrder.
Accessed 1924 times total.

Classification:

AMS MSC:	46F05 (Functional analysis :: Distributions, generalized functions, distribution spaces :: Topological linear spaces of test functions, distributions and ultradistributions)
	46-00 (Functional analysis :: General reference works )

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