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.
Best Answer
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.