Does Weak Convergence Hold for Sequences Approaching Infinity?

[tom_rylex]

Homework Statement

Show that if {x_k} is any sequence of points in space R^n with |{x_k}| \rightarrow \infty, then \delta(x-x_k) \rightarrow 0 weakly

Homework Equations

The Attempt at a Solution

I'm still trying to grasp the concept of weak convergence for distributions. It would appear that this function doesn't converge pointwise. The distribution on a test function is
\int \delta(x-x_k)\theta(x)dx = \theta(x_k) Does the function converge weakly to zero because x_k approaches infinity, and therefore would be outside of the region of support of any locally integrable test function?

 

[Dick]

I think that's correct. But what exactly is your set of test functions?

 

[tom_rylex]

My set of test functions meet the following criteria:
* function has a finite region of support, inside of which \theta(x) \neq 0, outside of which \theta(x)=0
* \theta(x) has derivatives of all orders.

 

[Dick]

Then you are right. Finite support is enough.