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I'm asked to show the following:

Show that the class $C_c(\mathbb{R^n})$ of continuous functions with compact support on $\mathbb{R^n}$ with the sup-norm metric $d(f,g):= \text{sup}_{x\in \mathbb{R^n}} |f(x) - g(x)|$is not a complete metric space. In other words, show that $C_c(\mathbb{R^n})$ is not closed within the class of bounded continuous functions on $\mathbb{R^n}$

I'm not really completely sure even how to approach this problem. To sort of get some intuition on this, I first tried working only with the class $C_c(\mathbb{R})$ of continuous functions with compact support on $\mathbb{R}$. So I need to construct a sequence $\{f_j\}$ of functions in $C_c(\mathbb{R})$ that converge to a function that's either unbounded, discontinuous or both.

I will define a function

$$f_j(x):= x^j \quad \text{if} \quad 0 \leq x < 1$$ $$\quad \quad \quad \quad \quad -x+2 \quad \text{if} \quad 1 \leq x \leq 2$$ $$0, \text{else}$$

Clearly this $f_j(x)$ will be $0$ for all $x \in \mathbb{R}\setminus[0,2]$, so $f_j$ has compact support. To be brief, I will just say that $f_j$ is bounded and continuous (this can easily be made rigorous). Now, consider the sequence of numbers $\{f_1(x), f_2(x), \ldots \}$. We see that this sequence converges to the function

$$f(x) = 0 \quad \text{if} \quad x<1 \quad \text{or} \quad x>2$$ $$-x+2 \quad \text{if} \quad 1 \leq x \leq 2$$

In particular, $f$ will be discontinuous at $x =1 $, so $f$ is not in the class of bounded continuous functions on $\mathbb{R}$. Thus, $C_c(\mathbb{R})$ is not a complete metric space.

Question 1: Is this even a correct way to show that $C_c(\mathbb{R})$ is not a complete metric space? I'm pretty new to these types of questions, so perhaps I'm not even quite sure what I need to show in my proof.

Question 2: If it is, is there any way to generalize this argument to show that $C_c(\mathbb{R^n})$ is not complete?

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  • $\begingroup$ Your example does not work, since your sequence does not converge to its point-wise limit uniformly (in the sup - norm). Just consider points to the left of $x=1$. You cannot make the difference $|f_j(x)-0|$ uniformly small there. In other words, your sequence is not a Cauchy sequence in the ambient space of continuous, bounded functions. The incompleteness of $C_c$ lies at infinity. $\endgroup$ – GReyes Jan 29 at 7:18
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Since we are working in metric spaces, it is important to always consider the metric when talking about convergencies.

The way the question is asked it sounds like you already know that the class of bounded continuous functions on $\mathbb R^n$ with the metric $d$ is a complete metric space. Lets call it $C_b(\mathbb R^n)$ for now.

So I need to construct a sequence $\{f_j\}$ of functions in $C_c(\mathbb{R})$ that converge to a function that's either unbounded, discontinuous or both.

This is not correct, because then $\{f_j\}$ would also converge in $C_b(\mathbb R)$ (with respect to $d$) and the limit is always continuous and bounded.

Instead, what you have to do is to find a sequence $\{f_j\}$ in $C_c(\mathbb R^n)$ that converges (with respect to $d$!) to a function $f$ in $C_b(\mathbb R^n)$ such that $f$ is not in $C_c(\mathbb R^n)$. This is only possible if $f$ has no compact support.

I suggest you try to find examples where possibly $f_j\to f$ and $f$ has no compact support. Don't forget to verify that the converges is in the correct metric, i.e. you have to show that $d(f,f_n)\to 0$.

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  • $\begingroup$ why the downvote? It would be nice if you could comment what I did wrong $\endgroup$ – supinf Jan 29 at 7:51
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Let $f$ be any continuous function such that $f(x) \to 0$ as $|x| \to \infty$ but $f$ does not have compact support. Let $g$ be any continuous function such that $0\leq g \leq 1$, $g(x)=1$ for $|x| \leq 1$ and $0$ for $|x| >2$. Consider the sequence $g(\frac x n)f(x)$. This sequence converges uniformly to $f$. It is a Cauchy sequence in $C_c(\mathbb R^{n})$ but it is not convergent because its pointwise limit $f$ is not in this space.

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  • $\begingroup$ Hi! I'm facing the same problem but I don't undeestand how $\|f_n-f\|_\infty$ tends to zero. Could you be more specific in this step, please? $\endgroup$ – user326159 Feb 24 at 1:52
  • $\begingroup$ Choose $N$ such that $|f(x)| <\epsilon $ for $|x| >N$. Then $|g(\frac x n)f(x)-f(x)| <\epsilon $ for $|x| >N$. For any $n>N$ note that $|g(\frac x n)f(x)-f(x)|=0$ for $|x| \leq N$. $\endgroup$ – Kavi Rama Murthy Feb 24 at 5:11

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