Here is a question I was trying to solve:

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My attempt:

Let $(f_n)$ be a Cauchy sequence in $(\mathcal C_0(\mathbb R),||.||_\infty).$ Then for each $x\in\mathbb R$ $(f_n(x))$ is a Cauchy sequence in $\mathbb R$ and hence convergent.

Define $f:\mathbb R\to\mathbb R$ by $f(x)=\lim_{n\to\infty}f_n(x).$ To show that $f\in\mathcal C_0(\mathbb R)$ and $||f_n-f||_\infty\to0.$

1 $f\in\mathcal C(\mathbb R):$ Choose $\epsilon>0.$ Then $\exists~k\in\mathbb N$ such that $||f_m-f_n||_\infty<\epsilon/2~\forall~m,n\ge k.$

That is, $\forall~x\in\mathbb R,~|f_m(x)-f_n(x)|<\epsilon/2~\forall~m,n\ge k.$

That is, $\forall~x\in\mathbb R,~\lim_{n\to\infty}|f_m(x)-f_n(x)|\le\epsilon/2~\forall~m\ge k.$

That is, $\forall~x\in\mathbb R,|f_m(x)-f(x)|\le\epsilon/2<\epsilon~\forall~m\ge k.$

So $(f_n)$ converges uniformly to $f$ on $\mathbb R\implies f\in\mathcal C(\mathbb R).$

[2] $f$ vanishes at infinity: Choose $\epsilon>0.$ Then $\exists~k\in\mathbb N,$ such that $\forall~x\in\mathbb R,~|f_n(x)-f(x)|<\epsilon/2~\forall~n\ge k.$

That is $\forall~x\in\mathbb R,~|f_k(x)-f(x)|<\epsilon/2.$

Also there exists a compact subset $C$ of $\mathbb R$ such that $|f_k(x)|<\epsilon/2~\forall~x\notin C.$

So $|f(x)|\le|f_k(x)-f(x)|+|f_k(x)|<\epsilon~\forall~x\notin C.$

Thus $f\in\mathcal C_0(\mathbb R).$

[3] Choose $\epsilon>0.$ Then there exists $k\in\mathbb N$ such that $\forall~x\in\mathbb R,|f_n(x)-f(x)|<\epsilon/2~\forall~n\ge k.$

That is $\sup_{x\in\mathbb R}|f_n(x)-f(x)|\le\epsilon/2<\epsilon~\forall~n\ge k.$

So $||f_n-f||_\infty\to0.$ Hence $f_n\to f$ in $\mathcal C_0(\mathbb R).$

Please tell me whether my attempt is correct.

  • 1
    $\begingroup$ "That is, $\forall~x\in\mathbb R,~\lim_{n\to\infty}|f_m(x)-f(x)|\le\epsilon/2<\epsilon~\forall~m\ge k.$" You mean "That is, $\forall~x\in\mathbb R. ||f_m(x)-f(x)|\le\epsilon/2<\epsilon~\forall~m\ge k.$" $\endgroup$ – Matematleta Nov 15 '17 at 14:18

Another way to do it is to note that it suffices to show that $C_0(\mathbb R)\subset C_b(\mathbb R)$ is closed.

If $C_0(\mathbb R)\ni f_n\to f,$ then

$\tag1 |f(x)|\le |f(x)-f_n(x)|+|f_n(x)||.$

Now choose an integer $N$ such that $ \sup_\left \{ x\in \mathbb R \right \}\left \{ |f(x)-f_N(x)| \right \}<\frac{\epsilon }{2}$ and for this $N$ a compact set $K$ for which $|f_N(x)|<\frac{\epsilon }{2}$ whenever $x\notin K.$ Then, $(1)$ with $n=N$ is less than $\epsilon$ as soon as $x\notin K\Rightarrow f\in C_0(\mathbb R).$

  • $\begingroup$ This would work if $C(\mathbb{R})$ were a Banach space, but it is not a normed space under $\|\cdot\|_\infty$ as functions can be unbounded. It seems it isn't even a topological vector space with uniform convergence topology, see here. $\endgroup$ – mechanodroid Nov 15 '17 at 20:42
  • 1
    $\begingroup$ Should have been $C_b(\mathbb R)$ with the sup norm. $\endgroup$ – Matematleta Nov 16 '17 at 1:36

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