Show, that $(c_0, \|\cdot \|_\infty)$ with $c_0:=\{f:\mathbb{N}\to\mathbb{C}\colon \lim_{n\to\infty} f(n)=0\}$ is a banach-space.

I already showed, that $(c_0, \|\cdot \|_\infty)$ is a normed vectorspace. I am struggeling to show, that every cauchy-sequence converges.

Let $(f_k)_k\in c_0$ be a cauchy-sequence. I have to find a limit $f\in c_0$. Hence $\|f_k-f\|<\varepsilon$ for every $\varepsilon>0$

$(f_k)\in c_0$ means, that $\lim_{n\to\infty} f_k(n)=0$ for every $k\in\mathbb{N}$. Therefore there is an $N_k(\varepsilon)\in N$ such that for every $n>N(\varepsilon)$ we have $|f_k(n)|<\varepsilon$, for every $k\in\mathbb{N}$

I want to show, that $0$ is a limit of $(f_k)$. Since $(f_k)$ is a cauchy sequence we have $\|f_k(m)-f_k(m')\|_\infty<\varepsilon$ for every $m,m'>N$ and $f_k(m), f_k(m')\to 0$. Hence $|f_k(m)|<\varepsilon/2$ for every $m>N_m(\epsilon)$ and $|f_k(m')|<\varepsilon/2$ for every $m'>N_m'(\epsilon)$. Now we choose $N(\varepsilon)=\max\{N_m(\varepsilon), N_m'(\varepsilon)\}$

We proceed:

$\|f_k(m)-f_k(m')\|_\infty\leq \|f_k(m)\|_\infty+\|f_k(m'\|_\infty\\=\sup_{n\geq m}|f_k(m)|+\sup_{n\geq m'}|f_k(m')|\leq \varepsilon/2+\varepsilon/2=\varepsilon$

I doubt that this is right. Can you help me out? Is this at least a try in the right direction? Thanks in advance.

  • $\begingroup$ My hint is that completeness in the classical spaces comes from completeness of $\mathbb{R}$ itself. $\endgroup$ – Project Book Oct 26 '17 at 4:55

I'm not really following your proof. Remember, you're showing that every Cauchy sequence converges. You've set up the problem by assuming you have a Cauchy sequence $(f_k)_{k \in \mathbb{N}} \in c_0$, which is good, but now you need to nominate a limit for this function, and prove it.

The first thing to notice is that the "orthogonal" sequences $(f_k(n))_{k\in \mathbb{N}}$ are all Cauchy for all $n \in \mathbb{N}$. This is because, $$|f_k(n) - f_l(n)| = |(f_k - f_l)(n)| \le \sup_{m \in \mathbb{N}} |(f_k - f_l)(m)| = \|f_k - f_l\|_\infty,$$ and $(f_k)$ is Cauchy. Since $\mathbb{R}$ is complete, there exist limits for the sequences, $f_\infty(n)$, for each $n \in \mathbb{N}$.

You now need to show two things: $f_\infty \in c_0$ and $f_k \rightarrow f_\infty$ as $k \rightarrow \infty$. Let me know if you want any more help.

EDIT: We wish to show that $f_\infty(n) \rightarrow 0$. Fix $\varepsilon > 0$, and choose $N_1$ such that $$k, l \ge N_1 \implies \|f_k - f_l\| < \frac{\varepsilon}{3}.$$ Fix $k \ge N_1$. Since $f_k(n) \rightarrow 0$ as $n \rightarrow \infty$, there exists some $N_2$ such that $$n \ge N_2 \implies |f_k(n)| < \frac{\varepsilon}{3}.$$ From the inequality before the edit, we get $$n \ge N_2 \text{ and } l \ge N_1 \implies |f_l(n)| < \frac{2\varepsilon}{3}.$$ Consider the above inequality as $l \rightarrow \infty$. We may choose $N_3$ such that $$l \ge N_3 \implies |f_l(n) - f_\infty(n)| < \frac{\varepsilon}{3},$$ so given $n \ge N_2$ and $l \ge \max \lbrace N_3, N_1 \rbrace$, we have, $$|f_\infty(n)| \le |f_\infty(n) - f_l(n)| + |f_l(n)| < \frac{\varepsilon}{3} + \frac{2\varepsilon}{3} = \varepsilon.$$ Note that $N_2$ depended only on $\varepsilon$ (via fixed $k$ and $N_1$). Therefore, $f_\infty(n) \rightarrow 0$, i.e. $f_\infty \in c_0$.

Next, we show $f_k \rightarrow f_\infty$, i.e. $\|f_k - f_\infty\|_\infty \rightarrow 0$ as $k \rightarrow \infty$. Fix $\varepsilon > 0$. Using Cauchiness of $(f_k)$ as before, fix $N_1$ such that, $$k, l \ge N_1 \implies \|f_k - f_l\|_\infty < \frac{\varepsilon}{2} \implies \forall n \in \mathbb{N}, |f_k(n) - f_l(n)| < \frac{\varepsilon}{2}.$$ Taking the limit as $l \rightarrow \infty$, we obtain, $$k \ge N_1 \implies \forall n \in \mathbb{N}, |f_k(n) - f_\infty(n)| \le \frac{\varepsilon}{2} \implies \|f_k - f_\infty\|_\infty \le \frac{\varepsilon}{2} < \varepsilon,$$ as required.

  • $\begingroup$ More help is appreciated. $\endgroup$ – Cornman Oct 26 '17 at 5:21
  • $\begingroup$ Which part do you need help with? $\endgroup$ – Theo Bendit Oct 26 '17 at 5:41
  • $\begingroup$ Both parts would be nice. $\endgroup$ – Cornman Oct 26 '17 at 5:44
  • $\begingroup$ @Cornman: Edited. $\endgroup$ – Theo Bendit Oct 26 '17 at 6:28

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