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.
2026-04-11 18:04:20.1775930660
Compactly supported test functions
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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)