Is this a separable Banach space? Define $X=\{ f\in C^1 ( \mathbb{R} ):~\int_{-\infty}^{\infty} \left| f'(x) \right| \mbox{d}x < \infty \}$ with norm $\| f \| = \left| f(0)\right| + \int_{-\infty}^{\infty} \left| f'(x) \right| \mbox{d}x$.
Is this a Banach space? Is it a separable space?
 A: No, this is not a Banach space. But its completion is a separable Banach space.
First. Let $\|f_m-f_n\|\to 0$ implies that $f_n(0)\to a\in\mathbb R$ and $$\|f_n'-g\|_{L^1(\mathbb R)}\to 0,$$ 
for some $g\in L^1(\mathbb R)$. But does not belong to $C(\mathbb R)$ in general. For example let
$$
g_n(x)=\left\{\begin{array}{lll}
0 & \text{if $x\le -1-\frac{1}{n}$},\\
n(x+1)+1 & \text{if $x\in(-1-\frac{1}{n},1)$}, \\
1 & \text{if $x\in(-1,1)$}, \\
-n(x-1)+1 & \text{if $x\in(1,1+\frac{1}{n})$}, \\
0 & \text{if $x\ge 1\frac{1}{n}$}.
\end{array}
\right.
$$
Then $g_n$ continuous, $f_n(x)=\int_{-\infty}^x g(t)\,dt\in C^1(\mathbb R)$, and $\{f_n\}$ Cauchy sequence in $X$ which does not converge is $X$, as $\|g_n-g\|_{L^1(\mathbb R)}\to 0$,
where
$$
g(x)=\left\{\begin{array}{lll}
0 & \text{if $x\le -1$},\\
1 & \text{if $x\in(-1,1)$}, \\
0 & \text{if $x\ge 1$}.
\end{array}
\right.
$$
Its completion is a complete normed space, i.e., a Banach space, and it is separable, as $L^1(\mathbb R)$ is separable.
