# Why do we care whether the support of a function is compact or not?

This is a question for self-learning. I am too confused by the text to formulate a well-defined question now.

I am reading analysis of functions, and confused by the motivations of some theorems and definitions.

For example, we want to prove the following result:

where $$\phi$$ is any function $$\in C^0(\Omega)$$, and

Why do we care whether the support of a function is compact or not? (My thought: since the support is a closure, so it must be closed; then as long as the support is bounded, i.e. the function vanishes as $$x\to\infty$$, the support is compact.)

Why do we define a function $$\tau_x\phi$$ with $$y$$ as the independent variable? (My thought: both $$x, y \in \Omega$$, so here we define a function with arbitrarily chosen $$x$$ as a parameter, and maps a point in $$\phi$$'s domain $$\Omega$$, to the image of $$\phi$$ at the difference (distance) of the point from the parameter (reference point) $$x$$. $$\quad$$ The corollary basically says that a translated $$C^k_c(\Omega)$$ function is still $$C^k_c(\Omega)$$, at least if the translation $$x$$ is near $$0$$. In other words, a small perturbation does not change the property of the function. $$\quad$$ Nevertheless, it seems to me that any finite translation $$x < \infty$$ should not change the compactness of support and $$k$$-th differentiability, right?)

• I feel like the particular result you're citing is not going to give you any motivation for studying functions with compact support. It's more of a supporting result: once you are already studying such functions, it is a useful tool to use. May 14 at 17:46
• @MishaLavrov Thanks. I might read other parts and then return to this result later. May 14 at 17:50

There's lots of reasons. Functions which are compactly supported are zero outside some ball, so if $$f$$ is continuous with compact support, it's automatically integrable on $$\mathbb{R}^n$$ (and in every $$L^p$$ space). It's also the case that $$C^k_c$$ is dense in $$L^p$$, which is very helpful for proving a variety of facts about integrable functions which are more easily proved for continuous functions first. The same can't be said for continuous functions which have infinite support; constant functions are smooth but aren't integrable on $$\mathbb{R}^n$$.
• It makes sense to me that a compactly supported function is usually integrable. 1. What if, however, we define a function that is a sum of countably many Dirac delta functions? 2.What is 'dense' in $L^p$? It is mentioned here en.wikipedia.org/wiki/Lp_space#Dense_subspaces. Is 'dense' similar to 'complete' (except at the boundary)? May 14 at 18:02
• 1. The Dirac delta is not a function, it's a distribution - a continuous linear functional acting on $C^{\infty}_c$. 2. A set of functions $G$ is dense in $L^p$ if for any $\epsilon>0$, $f\in L^p$, there exists $g\in G$ such that $\|f-g\|_{L^p}<\epsilon$. Density just means that any element can be approximated by members of the dense set; for example the rationals are dense in $\mathbb{R}$. May 14 at 18:10