# Banach space, $(c,||\cdot||_\infty)$

Let $c:=\{f:\mathbb{N}\to\mathbb{C}\colon \lim_{n\to\infty} f(n)~\text{exists}\}$ and $||f||_\infty=\sup_{n\in\mathbb{N}} |f(n)|$

Show, that $(c,||\cdot||_\infty)$ is a Banach space.

I have to show, that $||\cdot||_\infty$ defines a norm and is complete vectorspace. That $c$ is a vectorspace over $\mathbb{R}$ and $\mathbb{C}$ is obvious.

I show, that $||\cdot||_\infty$ is a norm.

1) Let $f\in c$ and suppose $||f||_\infty=\sup_{n\in\mathbb{N}} |f(n)|=0$, then we get:

$||f||_\infty=\sup_{n\in\mathbb{N}} |f(n)|=0\Leftrightarrow |f(n)|=0~\forall n\in\mathbb{N}\Leftrightarrow f(n)=0~\forall n\in\mathbb{N}\Leftrightarrow f=0$

2) Let $\lambda\in\mathbb{K}$ [$\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$], then:

$||\lambda f||_\infty=\sup_{n\in\mathbb{N}} |\lambda f(n)|=|\lambda|\sup_{n\in\mathbb{N}} |f(n)|=|\lambda|\cdot||f||_\infty$

3) $f,g\in c$, then

$$||f+g||_\infty=\sup_{n\in\mathbb{N}} |f(n)+g(n)|\leq\sup_{n\in\mathbb{N}}(|f(n)|+|g(n)|)\leq\sup_{n\in\mathbb{N}}|f(n)|+\sup_{n\in\mathbb{N}}|g(n)|\\=||f||_\infty+||g||_\infty$$

Which is all pretty obvious. Or should I be more specific at some point?

Now I want to show, that every Cauchy sequence converges. So let $(f_n)_{n\in\mathbb{N}}$ be a Cauchy sequence. Hence for every $\epsilon>0$ exists $N\in\mathbb{N}$ such that for every $r,s\geq N$ holds $d(f_r, f_s)<\epsilon$

Since $f_r, f_s\in c$ we know that they converge. Therefore:

$\lim_{n\to\infty} f_r(n)=x$ and $\lim_{n\to\infty} f_s(n)=y$

So there is $R,S\in\mathbb{N}$ such that $\sup_{n\geq R} |f_r(n)-x|\leq\frac{\epsilon}{2}$ and $\sup_{n\geq S} |f_s(n)-y|\leq\frac{\epsilon}{2}$. We take $N=\max\{S,R\}$ and proceed:

$d(f_r-f_s,x-y)=||f_r-f_s-(x-y)||_\infty\leq ||f_r-x||_\infty+||f_s-y||_\infty\leq\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

Hence $(c,||\cdot||_\infty)$ is a Banach space.

You need to show that your Cauchy sequence converges a limit $f$ in $c$. In the following we construct such limit. Since $(f_n)_n$ is a Cauchy sequence, for each $a\in \mathbb{N}$, you see there exists some $N_a\in\mathbb{N}$ such that $|f_l(a)-f_m(a)|<\varepsilon$ for all $l,m<N_a$. Therefore $(f_n(a))_n$ is a Cauchy seuquence in $\mathbb{R}$ and you obtain a limit $f(a)$ for each $a\in \mathbb{N}$. Now we wanna show that $f\in c$. Namely $\lim_{n\rightarrow\infty} f(n)$ exists. It suffices to show that $(f(n))_n$ is a Cauchy sequence. Notice that $(f_l)_l$ is a Cauchy sequence in $c$, i.e., for each $\varepsilon>0$ you find $N_{\varepsilon}\in\mathbb{N}$ such that $\sup_{n\in\mathbb{N}}|f_l(n)-f_m(n)|<\varepsilon$ for all $l,m>N_{\varepsilon}$. In this case, let \begin{align} |f(a)-f_{l_a}(a)|<\varepsilon \end{align} such that $l_a>N_{\varepsilon}$ for all $a\in \mathbb{N}$. This is valid due to the construction of $f$. On the other hand, let $z>N_{\varepsilon}$ be fixed such that $$|f_z(a)-f_z(b)|<\varepsilon$$ for all $a,b>M_{z,\varepsilon}\in \mathbb{N}$. This is valid since $f_z\in c$ and therefore $\lim_{n\rightarrow\infty}f_{z}(n)$ exists. Then we obtain We obtain that \begin{align} |f(a)-f(b)&|\leq|f(a)-f_{l_a}(a)|+|f_{l_a}(a)-f_{z}(a)|+|f_{z}(a)-f_{z}(b)|\\ &+|f_{z}(b)-f_{l_b}(b)+|f_{l_b}(b)-f(b)|\leq 5\varepsilon \end{align} for all $a,b>M_{z,\varepsilon}$. Therefore $(f(n))_n$ is a Cauchy sequence and hence $f\in c$. Now we want to show that $f_n\rightarrow f$ in $c$. Now for each $a\in\mathbb{N}$ let $m_a\in \mathbb{N}>N_{\varepsilon}$ sufficiently large such that $$|f(a)-f_{m_a}(a)|<\varepsilon.$$ This is valid due to the construction of $f$. Then for each $n>N_{\varepsilon}$ we obtain that \begin{align} |f_n(a)-f(a)|\leq |f_n(a)-f_{m_a}(a)|+|f_{m_a}(a)-f(a)|\leq 2\varepsilon \end{align} for each $a\in\mathbb{N}$. Therefore we obtain that $f_n\rightarrow f$ in $c$ and this completes the proof.
Your proof that $\|\cdot\|_\infty$ is a norm is fine. Your proof that $c$ is complete, however, is not quite correct. You need to show show that the Cauchy sequence $(f_n)$ converges to some $f\in c$ (in the metric induced by the norm). That is, given the Cauchy sequence you need to find some $f:\mathbb N\to\mathbb C$ such that $\lim_nf(n)$ exists and for every $\varepsilon>0$ there is some $N\in\mathbb N$ such that $\|f_n-f\|_\infty<\varepsilon$ whenever $n\geq N$.
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