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.

Have I done this correct? Thanks in advance for your comments.

  • $\begingroup$ A Banach space is a space such that every Cauchy sequence converges. So a hint: what is your limit of your Cauchy sequence? $\endgroup$ – ehochix Oct 22 '17 at 20:57

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 \begin{equation} |f_z(a)-f_z(b)|<\varepsilon \end{equation} 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 \begin{equation} |f(a)-f_{m_a}(a)|<\varepsilon. \end{equation} 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 $n\geq N$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.