# Is the set of all real valued continuous functions on $\mathbb R$ with compact support complete?

Is the set of all real valued continuous functions on $\mathbb R$ with compact support complete? the metric is the usual sup-norm metric

how to do it any hints .I cant prove it neither can find a counter example previous answers dont satisfy me

• What is your objection to the answer to math.stackexchange.com/questions/126797/…? Jan 3 '15 at 16:31
• i cant find the example Jan 3 '15 at 16:45
• can u please elaborate robert's answer @RobArthan Jan 3 '15 at 16:46
• You form a sequence of compactly supported functions $f_n$ that approximate the characteristic function of the closed interval $[-n, n]$, e.g., with $f_n(x) = 1$ for $x \in [-n, n]$ and with $f_n(x)$ decreasing linearly to zero on $[-n-1, -n]$ and $[n, n+1]$. The sequence of functions $f_n\cdot\phi$ will then tend uniformly to $\phi$ if $\phi(x)$ tends to $0$ as $x$ tends to $\pm \infty$. Jan 3 '15 at 16:59

Let $\phi(x)$ be a nonzero continuous function with support inside $[0, 1]$. Then let
$$f_1(x) = \phi(x), \ \ \ f_2 = \phi(x) + \frac{1}{2} \phi(x-1), \cdots, f_n = \sum_{i=0}^{n-1} \frac 1{i+1}\phi(x-i)$$
Then we have $f_i \in C_c(\mathbb R)$ and is a Cauchy sequence. The sequence $\{f_i\}$ converges uniformly to a continuous function
$$f = \sum_{i=0}^\infty \frac 1{i+1} \phi(x-i)$$
but $f$ is not of compact support. Thus $C_c(\mathbb R)$ is not complete with respect to the sup norm.