The set of all continuous functions with compact support is ideal??????

Question

AS far as ideal of a ring is concerned, it is not ideal. I am giving an counter example. f(x) = 1 for all x belongs to R. which is a continuous function with compact support. g(x) = x for all x belongs to closed interval 0 to infinity. = 0 otherwise Now f(x) . g(x) has not a compact support.

definition of compact support - A function has compact support if it is zero outside of a compact set. Alternatively, one can say that a function has compact support if its support is a compact set.(what I know)

Can anyone please help me out ?

Answer 1

The function $f(x) = 1$ (for all $x$) is not a function with compact support, so it does not belong to the ideal. In particular, $\overline{\{f \neq 0\}} = \Bbb R$ is not compact.

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To see that the continuous functions with compact support form an ideal, note the following:

• Suppose that $f,g$ have supports $S_f,S_g$. Then $S_{f+g} \subset S_f \cup S_g$, which is compact.
• Suppose that $f$ has compact support, and $r(x)$ is an arbitrary continuous function. Then $S_{rf} \subset S_f$.

Answer 2

The set of continuous functions from $\mathbb{R}$ to $\mathbb{R}$ with compact support is indeed an ideal, in the ring-theoretic sense, in the ring of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$.

I think the source of your confusion is the definition of a compact set.

The function $f:\mathbb{R}\rightarrow\mathbb{R}: x\mapsto 1$ that you mention does not have compact support - its support (the closure of the set of all values on which it is nonzero) is $\mathbb{R}$, and this is not compact.

(The function $f$ is constant on a compact set, and its range is compact - but that's not the same as having compact support.)

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Why is $\mathbb{R}$ not compact? Well, remember that a compact set is one with the property

every open cover has a finite subcover.

But $\{(-n, n): n\in\mathbb{N}\}$ is a cover of $\mathbb{R}$ by open sets, with no finite subcover

Alternatively, in a Euclidean metric space (i.e., $\mathbb{R}^n$ for some $n$), a set is compact iff it is closed and bounded. Well, $\mathbb{R}$ is certainly closed, but is it bounded?

(Note that above I am considering $\mathbb{R}$ with the usual topology and metric, but since you didn't say otherwise I assume this is the context you are working in.)

Answer 3

Let $g:\Bbb R\to\Bbb R$ be any continuous function and $f$ be continuous with compact support $K$. Then $fg=0$ outside $K$. Now $fg\ne 0$ on some subset $K_1\subset K$. The support of $fg$ is the closure of $K_1$. Since $\text{cl}K_1\subset K$ (as a compact set, $K$ is in particular closed), this closure is compact as a closed subset of a compact set, so $fg$ has a compact support.

In a similar way prove that the difference of two functions with compact support admits a compact support.