Let $X$ be a topological space and $C_c(X)$ be the set of all continuous complex functions on $X$ whose support is compact.
Let $f,g\in C_c(X)$. Trivially, $f+g$ are continuous, but how do i prove that supp$(f+g)$ is compact?
Asked
Active
Viewed 335 times
4 Answers
2
And to say why the inclusion holds, just note that if $f(x) = 0$ and $g(x) = 0$ then $(f+g)(x) = 0$.
SBF
- 36,041
1
Let me complete this answer. I will give the proof for real functions. First, let us recall the definition of support.
Given $\phi:X\subset \mathbb{R}^n\to \mathbb{R}$ a continuous function, the support of $\phi$, denoted by $\operatorname{supp} f$, is the closure (taken in $X$) of the subset of $X$ where f is non-zero, that is, if we write $S_\phi=\{x\in X;\ \phi(x)\neq 0\}$, $$\operatorname{supp}\phi=\overline{S_\phi}\cap X,$$ where $\overline{S_\phi}$ is the closure taken in $\mathbb{R}^n$.
Given $f,g:X\to \mathbb{R}$, define $h=f+g$.
Claim 1: supp$(f+g)\subset$ supp$(f)$+supp$(g)$
Set $h=f+g$ and note that
$x\not\in S_f\cup S_g \Rightarrow x\not\in S_f$ and $x\not\in S_g \Rightarrow f(x)=0$ and $g(x)=0\Rightarrow h(x)=0\Rightarrow x\not\in S_h$. As a csonequence,
$S_h\subset S_f\cup S_g\Rightarrow$ supp$(h)=\overline{S_h}\cap X\subset (\overline{S_f}\cup\overline{S_g})\cap X= (\overline{S_f}\cap X)\cup(\overline{S_g}\cap X)=$supp$(f)\cup$supp$(g)$
Claim 2: If supp$(f)$ and supp$(g)$ are compact in $\mathbb{R}^n$, then supp$(h)$ is also compact.
It suffices to prove that given $(x_n)$ a sequence in supp$(h)$, there exists a convergente subsequence in supp$(h)$. Indeed, from Claim 1, $(x_n)$ is a sequence in $K=$supp$(f)\cup$ supp$(g)$ that is compact. Hence, there exists a subsequence $(x_{n_k})$ such that $x_{n_k}\to x$ in $K$. It remais to prove that $x\in$ supp$(h)$.
In fact, $x\in K=$supp$(f)\cup$ supp$(g)\subset X$. Since $x_{n_k}\in$ supp$(h)\subset \overline{S_h}$, we have that $x\in \overline{S_h}$. Hence, $x\in \overline{S_h}\cap X=$supp$(h)$, as required.
