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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 ?

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    The function that is always $1$ doesn't have compact support, unless the domain is already compact, in which case all functions have compact support (but in your case the domain appears to be the real line or at least $[0,\infty)$, not compact). Perhaps you are confusing domain and range. – 2017-01-17
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    $g(x)=1$ for all $x \in \mathbb{R}$ has support $\mathbb{R}$ which is not compact. – 2017-01-17
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    I assume what is being said in your source is that if $R$ is the ring of continuous functions with pointwise addition and multiplication, and $I$ is the set of continuous functions with compact support, then $I$ is an ideal in $R$. And this is true. – 2017-01-17
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    sani, do you understand why your $f$ does not in fact have compact support? – 2017-01-17

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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 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.)

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.)

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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.

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 $S_{rf} \subset S_f$.
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    why can u please explain – 2017-01-17
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    What do you want me to explain, exactly? Do you agree that $\Bbb R$ is not compact? – 2017-01-17
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    Note that f=1f=1 given by f(x)=1f(x)=1 is in C∞(R)C∞(R) and gg given by g(x)=|x|∈C(R)g(x)=|x|∈C(R), but f⋅g=g∉C∞(R)f⋅g=g∉C∞(R). Incidentally, this is why no proper subset of a ring containing the multiplicative identity can be an ideal of that ring. How did u write this?/// – 2017-01-17
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    http://math.stackexchange.com/questions/644460/which-of-the-following-form-an-ideal-in-this-ring/645115?noredirect=1#comment4322268_645115 check this link – 2017-01-17
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    how did u prove that The set of all C∞ functions with compact support is not ideal – 2017-01-17
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    I didn't. I proved that the set **of all $C^\infty$ functions** (with or without compact support) is not an ideal. – 2017-01-17
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    ohhh sorry..That created confusion.. – 2017-01-17
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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.