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Let $X$ be a topological space. Recall that $C_c (X)$ is the space of continuous complex-valued functions with compact support on $X$.

Is $C_c (X)$ (with the supremum norm) a Banach space?

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closed as off-topic by user99914, JonMark Perry, Claude Leibovici, erfink, Ove Ahlman Dec 14 '17 at 12:06

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It is obviously a normed space but it is not complete. [ In fact, when $X=\mathbb R$ its completion is the space $C_0 (\mathbb R)$ of continuous functions that vanish at $\infty$ and $-\infty$. To show that the space is not complete when $X=\mathbb R$ take any function g in $C_0 (\mathbb R)$. There exists a sequence $\{f_n\}$ in $C_c (\mathbb R)$ such that $f_n =1$ on $[-n,n]$ and $0 \leq f_n \leq 1$ everywhere. The sequence $\{gf_n\}$ is a Cauchy sequence which is not convergent.

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  • $\begingroup$ Thank you for your answer. I can take $g:\mathbb{R}\longrightarrow \mathbb{C}$ by $g(x)=\frac{1}{x^2 +1}$ which satisfies in the above condition. But I don't know what I can take for $f_n$. Can you help me? $\endgroup$ – user481657 Dec 14 '17 at 11:19
  • $\begingroup$ Take $f_n=1$ on $[-n,n]$, 0 on $(n+1,\infty)$ as well as $(-\infty,-n-1)$ and make f 'linear' in $[n,n+1]$ and $[-n-1,-n]$. Linear means the graph is a straight line in that interval. $\endgroup$ – Kavi Rama Murthy Dec 15 '17 at 10:19