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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: enter image description here

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

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?)

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    $\begingroup$ 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. $\endgroup$
    – Misha Lavrov
    May 14 at 17:46
  • $\begingroup$ @MishaLavrov Thanks. I might read other parts and then return to this result later. $\endgroup$
    – Charlie Chang
    May 14 at 17:50

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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 $\mathbb{R}^n$.

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  • $\begingroup$ 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)? $\endgroup$
    – Charlie Chang
    May 14 at 18:02
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    $\begingroup$ 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 $\mathbb{R}$. $\endgroup$
    – kieransquared
    May 14 at 18:10
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    $\begingroup$ By the way, I'm not sure what you're currently reading, but it'll all make much more sense if you start from the basics - introductory measure and integration theory, for example. Density is a common introductory analysis concept. $\endgroup$
    – kieransquared
    May 14 at 18:12
  • $\begingroup$ Thanks very much. So density implies approximability by the (function) set's elements. I'm reading a lecture note about functional analysis and distributions. I will go back to the measure and Lebesgue integration. $\endgroup$
    – Charlie Chang
    May 14 at 18:22

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