is a distribution of first order
Now it is clear that the integrand is continuous for all . What is more, the integrand approaches for , so the integrand has a removable discontinuity at . That is, by assigning the value to the integrand at , the integrand becomes continuous in . This means that the integrand is Lebesgue measurable on . Then, by defining (where is the characteristic function), and applying the Lebesgue dominated convergence theorem, we have
It follows that is finite, i.e., takes values in . Since is a vector space, if follows easily from the above expression that is linear.
To prove that is continuous, we shall use condition (3) on this page (http://planetmath.org/Distribution4). For this, suppose is a compact subset of and . Again, we can assume that for some . For , we have
|Title||is a distribution of first order|
|Date of creation||2013-03-22 13:46:07|
|Last modified on||2013-03-22 13:46:07|
|Last modified by||Koro (127)|