Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be continuous such that $f$ vanishes at infinity. i.e. for all $\epsilon >0$, there exists $N\in \mathbb{N}$ such that $n>N$ implies $|f(x)|<\epsilon$. Let $C_0(\mathbb{R})$ denotes the set of all such functions. For $f\in C_0(\mathbb{R})$, define $||f||= \sup \{|f(x)|:x\in \mathbb{R}\}$. $C_0(\mathbb{R})$ is a normed subspace of $B(\mathbb{R})$. Prove that $C_0(\mathbb{R})$ is complete.

I have learned elementary theory of normed vector space. Hence I know it sufficed to prove that every absolute convergent series $\sum_{i=1}^{\infty}f_i(x)$ converges in $C_0(\mathbb{R})$. I can prove that $f=\sum_{i=1}^{\infty}f_i(x)$ is bouneded, but i cant prove that $f$ vanishes at infinity, please helps.

  • $\begingroup$ Have you proved that $f$ is continuous? $\endgroup$ – user99914 Nov 16 '14 at 9:55
  • $\begingroup$ @John Yes, because $\sum ||f_i||<\infty $ implies uniform convergence, hence $f$ is continuous? $\endgroup$ – Jessie Nov 16 '14 at 9:57

Assume that $\sum_{i=1}^\infty f_i$ is a series of absolute summable series in $C_0(\mathbb R)$. That is,

$$(*) \sum_{i=1}^\infty ||f_i|| <\infty$$

You know already that $f = \sum_{i=1}^\infty f_i$ is continuous and bounded. To show that $f$ vanishes at infinity, let $\epsilon >0$. Then by (*) there is $N$ so that

$$ \sum_{i=N}^\infty ||f_i|| < \epsilon /2 $$

For fix this $N$. For each $f_1, \cdots, f_{N-1}$, let $M_1, \cdots M_{N-1} >0$ such that

$$|f_i(x)| < \epsilon/2N$$

whenever $|x| \geq M_i$. Let $M = \max\{M_1, \cdots, M_{N-1}\}$. Then if $|x| \geq M$,

$$|f(x)| \leq \sum_{i=1}^\infty |f_i(x)| \leq |f_1(x)| + \cdots +|f_{N-1} (x)| + \sum_{i=N}^\infty |f_i(x)| < \frac{(N-1)\epsilon}{2N} + \frac{\epsilon}{2} <\epsilon. $$

As $\epsilon>0$ is arbitrary, we conclude that $f$ vanish at infinity.


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