Completeness of $C^1$ functions vanishing at infinity with sup-norm of derivatives I'm looking at
$$C_0^1(\mathbb{R}) := \{f \in C^1(\mathbb{R}) : \lim_\limits{|x|\rightarrow \infty}f(x) = \lim\limits_{|x|\rightarrow \infty} f'(x) = 0\},$$
along with the norm given by $||f|| := \sup_\limits{x\in\mathbb{R}} |f'(x)|$
If been struggling to figure out if this metric space with the given norm is complete.
I started considering the sequence $f_n(x) := \exp(-\sqrt{x^2+\frac{1}{n^2}})$, which can be nicely used to show that the same space along with the regular sup-norm isn't closed in $C_b(\mathbb{R}),$ i.e. complete. Yet, it doesn't seem to work out in this case, as proving that $(f_n)$ is Cauchy doesn't work as easily for $x^{-1}{\sqrt{x^2+1/n^2}}$'s discontinuity in $0$.
Yet, any Cauchy sequence in $(C_0^1,||\cdot||)$ should have a limit whose derivative is continuous and tends to zero for big $x$, as $(f_n)$ Cauchy in $C_0^1$ implies that $(f_n')$ is Cauchy in $(C_0,||\cdot||_{sup})$ (or do I miss something there?). But how would I then control the behavior at infinity of such a candidate limit function $f$?
So, I'm left puzzled if this space is complete.
 A: This space is not complete. Consider a continuous odd function $g:\mathbb{R} \rightarrow \mathbb{R}$ with $\lim_{|x| \rightarrow \infty} g = 0$ but $\int_{-\infty}^0 g(x) dx$ divergent.
Define $g_k = g \rho_k$ where $\rho_k$ is a smooth even function supported in $[-k-1,k+1]$, identically 1 on $[-k,k]$, and $0 \le \rho \le 1$ everywhere.
Our Cauchy sequence will be $f_k(x) = \int_{-\infty}^x g_k(s) ds$.
Note that $f_k$ is defined on $\mathbb{R}$ since $g_k$ is compactly supported, and that $f_k$ vanishes at infinity since $g_k$ is odd.
Moreover, $f_k' = g_k$ is continuous and vanishes at infinity, so $f_k \in C_0^1(\mathbb{R})$ for all $k$.
We also have that $f_k' = g_k$ converges uniformly to $g$ since $\lim_{|x|\rightarrow \infty} g(x) = 0$.
Therefore, the sequence $(f_k)$ is Cauchy in $C_0^1(\mathbb{R})$.
Suppose for contradiction that $(f_k)$ had a limit $f_\infty$ in $C_0^1(\mathbb{R})$.
Then $f_\infty' = g$ and the fundamental theorem of calculus would then yield that $f_\infty(x) = f_\infty(0) + \int_0^x g(s) ds$, which diverges as $|x| \rightarrow \infty$.
Therefore, this space is not complete.
A: As the previous answer shows that the space is not complete, I want to give some details of its completion as pointed out in a previous comment.
We claim that the completion of the space $C^1_0$ given the norm $||f||$ from above is $$\overline{C^1_0}(\mathbb{R}) = 
{\left\{u\in C^1(\mathbb{R}) \colon \lim\limits_{|x|\rightarrow \infty} u'(x) = 0\right\}}/{\mathbb{R}} =: X,$$
the space of all continuously differentiable functions whose derivatives vanish at infinity, modulo a constant/real number.
For the first inclusion, let $f\in X,$ and $\varepsilon > 0.$ Choose $c>0$ s.t. $|f'(x)| < \frac{\varepsilon}{2}$ for all $|x| \geq c$ and $sign(f'(c))=-sign(f(c)).$ Now let 
$ C:= \int_0^c f'(t) dt$ and $ D:=\int_{-c}^0 f'(t) dt,$
as well as $\tilde{c} = c+2$ and $d = c+\frac{-2D-2C-2f'(c)}{f'(-c)}.$
We define the continuous(!) function $g\colon \mathbb{R} \rightarrow \mathbb{R}$ as
$$ 
g(x) =
\begin{cases}
f'(x), & x\in[-c,c]\\
\frac{\tilde{c}-x}{\tilde{c}-c} f'(c), & x\in(c,\tilde{c})\\
\frac{x+d}{d-c} f'(c), & x\in(-d,-c)\\
0,& \text{else.}
\end{cases}$$
We can now define a second function as
$$\tilde{f}(x) := -C + \int_0^x g(t) dt - f'(c).$$
Using the Fundamental Theorem, we get that $\tilde{f}$ is continuously differentiable, for $g$ is continuous. Thus, $\tilde{f}'$ vanishes, too.
If $x>\tilde{c},$
\begin{align*}
\tilde{f}(x) = \int_c^{\tilde{c}} g(t) dt - f'(c) = 0.
\end{align*}
On the other hand, if $x<-d$
\begin{align*}
\tilde{f}(x) = -C -f'(c)-D - \int_{-d}^{-c} g(t) dt =0,
\end{align*}
thus, $\tilde{f}$ vanishes and we get $\tilde{f}\in C_0^1(\mathbb{R}).$
Moreover, 
\begin{align*} ||\tilde{f}-f|| &= \sup_{x\in\mathbb{R}} |\tilde{f}'(x)-f'(x)| \\
&= \max\left(
\sup_{c<x<\tilde{c}} \left|
\underbrace{\frac{\tilde{c}-x}{\tilde{c}-c}}_{\leq 1} \underbrace{f'(c)}_{|\cdot| \leq \frac{\varepsilon}{2}}
-\underbrace{f'(x)}_{|\cdot| \leq \frac{\varepsilon}{2}}\right|,
\sup_{-d\leq x\leq -c} \left|
\underbrace{\frac{x+d}{d-c}}_{\leq 1} \underbrace{f'(-c)}_{|\cdot| \leq \frac{\varepsilon}{2}}
-\underbrace{f'(x)}_{|\cdot| \leq \frac{\varepsilon}{2}}\right|,
 \sup_{x\notin [-d,\tilde{c}]} \underbrace{|f'(x)|}_{\leq \varepsilon}
\right)\\
& \leq \varepsilon.
\end{align*}
So $\tilde{f}$ converges to $f$ wrt. $||\cdot||.$
For the other inclusion, let $(f_n)$ be a CF in $C_0^1$ wrt. $||\cdot||$. Then $(f'_n)$ is a CF in $C_0^0$ wrt. its usual norm. Since this space is complete, there is a limit $f' \in C_0^0$.
If we let 
$$f(x):= \int_0^x f'(t) dt + \mathbb{R},$$
then $f \in X,$ for $f'$ has the property that $\lim_{|x|\rightarrow \infty} f'(x) = 0$ and is continuous, so $f$ is continuously differentiable. That $f_n\rightarrow f$ wrt. $||\cdot||$ is clear by construction.
