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?)
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 $\mathbb{R}^n$.