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Let $C(R)$ denote the ring of all continuous real-valued functions on with the operations of pointwise addition and pointwise multiplication. Which of the following form an ideal in this ring?

  • The set of all $C^{\infty}$ functions with compact support.

  • The set of all continuous functions with compact support.

  • The set of all continuous functions which vanish at infinity, i.e. functions such that $$\lim_{|x|\to\infty}f(x) = 0$$

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    $\begingroup$ What are your thoughts on this? $\endgroup$ Commented Jan 20, 2014 at 2:38
  • $\begingroup$ "Set of all $C^\infty$" is a little weird to say, but I guess you mean the subset of smooth functions $C^\infty\subset C(R)$? $\endgroup$
    – rschwieb
    Commented Jan 20, 2014 at 13:42

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By the definition of a subring, an ideal of $C(\mathbb{R})$ would be a subset $I$ of $C(\mathbb{R})$ such that:

  • For $f_1,f_2 \in I$, $f_1 + f_2 \in I$
  • For any $f \in I$ and any $g \in C(\mathbb{R})$, $f\cdot g \in I$.

With that in mind, the set of continuous functions with compact support is the only set you've described satisfying all of these properties.

Why the other sets fail to be ideals:

The set $C^\infty(\mathbb{R})$:

Note that $f = 1$ given by $f(x) = 1$ is in $C^\infty(\mathbb{R})$ and $g$ given by $g(x) = |x|\in C(\mathbb{R})$, but $f\cdot g = g \not \in C^\infty(\mathbb{R})$. Incidentally, this is why no proper subset of a ring containing the multiplicative identity can be an ideal of that ring.

The set of functions with compact support:

Not all such functions are continuous, and so this fails to be a subset of $C(\mathbb{R})$ (I'll leave it to you to find a counterexample). Thus, it is not an ideal thereof.

The set $C_0(\mathbb{R})$ of continuous functions vanishing at $\pm \infty$

Note that $f$ given by $f(x) = e^{-|x|}$ is in $C_0(\mathbb{R})$, and $g$ given by $g(x) = e^{|x|}$ is continuous. However, $f \cdot g = 1$ is not in $C_0(\mathbb{R})$.

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  • $\begingroup$ It would be quite useful to see why the others dont satisfy this definition $\endgroup$ Commented Jan 20, 2014 at 16:58
  • $\begingroup$ how The set of all continuous functions with compact support becomes an ideal u did not explain...................... $\endgroup$
    – anonymous
    Commented Jan 17, 2017 at 18:53
  • $\begingroup$ It seems that the question has changed since I posted my answer, so the first two non-ideals no longer apply. $\endgroup$ Commented Jan 17, 2017 at 20:33
  • $\begingroup$ @sani you're right, and I don't intend to. If you have a question about it, perhaps you should post a new question. $\endgroup$ Commented Jan 17, 2017 at 20:34
  • $\begingroup$ so what will be the answer? How u prove ? $\endgroup$
    – anonymous
    Commented Jan 17, 2017 at 20:38

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