I am considering the test function:
$$f(x)=\frac{1}{\cos(\pi x)},$$
My question is whether the test function is compactly supported? I am aware of the definition of compact support, namely:
"A function has compact support if it is zero outside of a compact set".
I believe the domain of this function is: $$x\in \mathbb{R}\setminus \left\{\pm\frac{n}{2},n\in\mathbb{N}\right\},$$
This is a closed and bounded set, so also compact. But how do I show that this function is zero outside of this set? as it surely takes no value outside of this set? I'm fairly sure it is compactly supported, but I don't know how to show it.
Compactly supported test functions
compactnessdistribution-theoryfunctions
Best Answer
As suggested by @Fred , @OlivierMoschetta and @Chrystomath this test function is not compactly supported. However, this function can be used in the context of the composition integral, as per @md2perpe 's comment as it is smooth on a neighbourhood of the support of $\delta \circ g$ (I posted this as answer just so the thread can be closed)