Completion of the continuous real valued functions with compact support is the continuous real valued functions vanishing at infinity

analysisfunctional-analysisreal-analysis

All functions here are from $\Bbb{R}$ to $\Bbb{R}$ and I use the supremum metric $d$.

$C_c=$ continuous functions with compact support and $C_0 = \{f: \forall \epsilon > 0 \ \exists K \subseteq X \text{ compact} : \vert f(x) \vert \lt \epsilon \text{ on } X \setminus K\}$, i.e, continuous functions vanishing at infinity.

I thought that I could prove this in 2 steps:

a) $C_c$ is dense in $C_0$.

b) $(C_0,d)$ is complete.

For part a), I argued as follows:

Let $f\in{C_0}$ and $\epsilon \gt 0$. Then by definition, there exists $K\subset{\Bbb{R}}$ compact such that $\vert f(x) \vert \lt \epsilon/2 \text{ on } \Bbb{R} \setminus K$. Take the function $g$ such that $g=f$ on $K$, and $g(x)=0$ outside of $K$. Then it is clear that $g$ belongs to $C_c$, and $d(f,g)\lt \epsilon$. For, on $K$ functions are equal so $d(f,g)=0$ and on $\Bbb{R} \setminus K$ we have $d(f,g)= \text{sup}\vert f \vert\le \epsilon/2 \lt \epsilon$.

My problem is that I am not sure whether the function $g$ is continuous with this construction, and if it is how to show it. Also, does this proves part a) correctly?

A much more important problem for me is part b), because I have no idea how to approach this.

Lastly, can we show the completion by using both parts a) and b), or do we need an extra condition? I do not really want to get into too much with the technical approach of completion (such as Cauchy construction).

Best Answer

The completion of a metric space $X$ is a complete metric space $X'$ such that:

  • There is an isometry $X\hookrightarrow X'$
  • If $Y$ is a complete metric space and there is an isometry $X\hookrightarrow Y$, then this isometry factors as $X\hookrightarrow X'\to Y$ for a unique continuous $X'\to Y$

In other words, $X'$ is the 'minimal' complete container of $X$. Theorem:

  • If $X'$ is complete and there is an isometry $X\hookrightarrow X'$ with dense image, $X'$ is a completion of $X$.

So, you might as well take this to be the definition too.

So, you need to check $a),b)$ and you also need to check that the inclusion $C_c\hookrightarrow C_0$ is an isometry. That's true since you are using the 'same' metric.

Your proof of $a)$ is not quite right since the $g$ thus defined is not continuous, necessarily. One easy way round this:

Let $M>0$ be such that $[-M,M]$ contains $K$. Define $g\equiv f$ on $[-M,M]$. Let $\alpha=f(-M),\beta=f(M)$. Define $g$ on $[-M-1,-M]$ to be the linear extension (or whatever else, many functions would do) $0\to\alpha$, and on $[M,M+1]$ to be the linear extension $\beta\to0$.

Finally, set $g\equiv0$ on $(-\infty,-M-1)\cup(M+1,\infty)$. Now $g$ is continuous, and $\|f-g\|_\infty\le\epsilon$ (the presence of a weak inequality sign is not important, but I think it should be this way: both $f$ and $g$ are bounded by $\le\epsilon/2$ outside of $[-M,M]$ so the overall variation is $\le\epsilon$).

Let's check the completeness of $C_0$, I'll leave some hints.

It is very important that the uniform metric is used here. Suppose there is a sequence $(f_n)_{n\in\Bbb N}\subseteq C_0$ which is Cauchy in the uniform metric.

  • Check $(f_n(x))_{n\in\Bbb N}\subseteq\Bbb R$ is Cauchy for all $x\in\Bbb R$, so convergent. Define $f:\Bbb R\to\Bbb R$ by $x\mapsto\lim_{n\to\infty}f_n(x)$.
  • Is $f$ continuous? Think about the mode of convergence.
  • Is it possible for $f$ to not vanish at infinity? Approximate $f$ by $f_n$ for large $n$, and use the vanishing property of that particular $f_n$ to bound $f$ on some $\Bbb R\setminus K$. Again, think about the mode of convergence

Conclude $C_0$ is complete!

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