Let C denote continuous bounded functions from R -> R which are identically zero outside some closed bounded interval. Is C complete in the sup norm?
I think it is not because there will be some problems if we pick one function from the zero range, and one which is not. However, I'm not sure if this is right or how to prove this formally.
As stated, define $\displaystyle d(f,g):= \sup_\mathbb{R} |f-g|$. We will build a Cauchy sequence $\{f_n\}$ such that does not converge in $C$ because its pointwise limit is actually outside $C$. (In fact, we'll have that $\displaystyle d(f_n, f_{n-1}) \leq \frac{1}{2^n}$. You should check that this is condition implies that $\{f_n\}$ is actually Cauchy).
Let $$g_n(x) = \left\{\begin{array}{rl} \frac{1}{2^n} & x\in [-n, n]\\ 0 & x\notin [-n-1, n+1] \end{array}\right.$$ And for $x\in [-n-1, -n] \cup [n, n+1]$ just connect by linear line so that $g_n(x)$ is continuous. Now, it is clear that $g_n$ are all in $C$. Moreover, any finite sum $\sum_{i=1}^n g_i$ is in $C$.
Hence, consider $f_n := \sum_{i=1}^n g_i$. You can see that $\{f_n\}$ is strongly Cauchy, and pointwise converges to a function that is asymptotically zero as $x\to\pm \infty$ but not identically zero outside a closed bounded interval.