I want to prove that $$c = \{ (x_n)_{n\geq 0} \mid x_n \in \Bbb C \text{ and the sequence converges} \}$$ with the norm $$ \left\|(x_n)_{n\geq 0}\right\|_{\infty} =\sup_{n\geq 0} |x_n|$$ is a Banach space. I have done some work, but I am having trouble concluding.

So far: let $(\xi_n)_{n\geq 0} = \left( (x_k^{(n)})_{k\geq 0} \right)_{n\geq 0}$ be a $\|\cdot\|_{\infty}$-Cauchy sequence. Let $\epsilon > 0$, there exists $n_0 \in \Bbb N$ such that: $$\begin{align} \| \xi_n - \xi_m \|_{\infty} &< \epsilon, \quad \forall\, n,m > n_0 \\ \sup_{k \geq 0} |x_k^{(n)}-x_k^{(m)}| &< \epsilon, \quad \forall\,n,m > n_0 \\ |x_k^{(n)} - x_k^{(m)}| &< \epsilon, \quad \forall\, k\geq 0, \quad \forall\,n,m > n_0 \end{align}$$ So fixed $k$, $(x_k^{(n)})_{n \geq 0}$ is a $|\cdot |$-Cauchy sequence, and since $\Bbb C$ is Banach, there exists a limit $\lim_{n\to \infty} x_k^{(n)} =: x_k$. Then define $\xi = (x_k)_{k \geq 0}$.

Now I understand I have two things to do:

  • prove that $\xi_n \stackrel{\|\cdot \|_{\infty}}{\longrightarrow} \xi $: we proceed as before. Let $\epsilon > 0$. Now, there is $n_0 \in \Bbb N$ such that: $$\begin{align} \| \xi_n - \xi_m \|_{\infty} &< \epsilon, \quad \forall\, n,m > n_0 \\ \sup_{k \geq 0} |x_k^{(n)}-x_k^{(m)}| &< \epsilon, \quad \forall\,n,m > n_0 \\ |x_k^{(n)} - x_k^{(m)}| &< \epsilon, \quad \forall\, k\geq 0, \quad \forall\,n,m > n_0 \\ \lim_{m \to \infty} |x_k^{(n)} - x_k^{(m)}| &\leq \epsilon, \quad \forall\,k\geq 0, \quad \forall\,n >n_0 \\ |x_k^{(n)} - x_k| &\leq \epsilon, \quad \forall\,k\geq 0 \quad \forall\, n>n_0 \\ \sup_{k\geq 0} |x_k^{(n)} - x_k| &\leq \epsilon,\quad \forall\, n > n_0 \\ \| \xi_n - \xi\|_{\infty} &\leq\epsilon, \quad \forall\,n>n_0, \end{align}$$ so ok.

  • prove that $\xi \in c$. I'm not sure of how to do this. I think I must use that every $(x_k^{(n)})_{k \geq 0}$ converge, because I don't seem to have used this yet.

How can I prove that $\xi \in c $?



Here is a sketch. I will use all of the same notations you are using (You should definitely check it for mistakes)

For each $n$, then sequence $\xi_n=(x^{(n)}_k)_{k\geq 0}$ is a convergent sequence (because it is in $c$), so let us say that $\xi_n=(x^{(n)}_k)_{k\geq 0}$ converges to $u_n \in \mathbb{C}$.

You can check that $|u_m-u_n|\leq ||\xi_m-\xi_n||_{\infty}$ for all $m,n \in \mathbb{N}$, so since $(\xi_n)_{n\geq 0}$ is a Cauchy sequence in $c$, it follows that $(u_n)_{n \geq 0}$ is a Cauchy sequence in $\mathbb{C}$, so there exists $u \in \mathbb{C}$ such that $\lim_{n\to \infty}u_n=u$.

I claim that the sequence $\xi=(x_k)_{k\geq 0}$ converges to $u$. The reason is that for any $n$ we have $$|x_k-u| \leq |x_k-x^{(n)}_k|+|x^{(n)}_k-u_n|+|u_n-u| $$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\leq ||\xi-\xi_n||_{\infty}+|x^{(n)}_k-u_n|+|u_n-u|$$Let $\epsilon>0$. Choose $n_0 \in \mathbb{N}$ such that $|u_{n_0}-u|<\frac{\epsilon}{3}$ and $||\xi-\xi_{n_0}||<\frac{\epsilon}{3}$. Choose $K \in \mathbb{N}$ such that $|x_k^{(n_0)}-u_{n_0}|<\frac{\epsilon}{3}$ whenever $k \geq K$. Then for $k \geq K$, all three terms on the right hand side of the above equation are less than $\frac{\epsilon}{3}$, and thus $|x_k-u|<\epsilon$.

Thus $x_k \to u$. Therefore the sequence $\xi=(x_k)_{k\geq 0}$ converges, so $\xi \in c$.

Edit: (Alternative solution) Actually, I just realized that we don't really care about the limit of $\xi$, we only care that it converges. So it suffices to prove that it is Cauchy. To do this, you can just say that for any $n$ $$|x_k-x_l| \leq |x_k-x^{(n)}_k|+|x^{(n)}_k-x^{(n)}_l|+|x^{(n)}_l-x_l|$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\leq ||\xi-\xi_n||_{\infty}+|x^{(n)}_k-x^{(n)}_l|+||\xi-\xi_n||_{\infty}$$ and then argue as before.

  • $\begingroup$ After your claim, you use the symbol $|x_k-x^{(n)}_k|$ - how does this work? Isn't $x_k$ a sequence of sequences, while $x^{(n)}_k$ is a n-th sequence in $x_k$? How can we take a modulus of it? $\endgroup$ – Никита Васильев Nov 22 '20 at 12:59

Here is an alternative proof, using as a starting point that $\ell^{\infty}$ is a Banach space. As $c$ is a subspace of $\ell^{\infty}$, it suffices to show that $c$ is closed in $\ell^{\infty}$.

To see this let $\mathbf{x}=(x_1,x_2,\dotsc)\in\ell^{\infty}$, and suppose that $\{\mathbf{x}^n\}_{n=1}^{\infty}$ is a sequence in $c$ converging to $\ell^{\infty}$ in the $\|\,.\|_{\infty}$ norm. For $n\in\mathbb{N}$ write $\mathbf{x}^n=(x_1^n,x_2^n,\dotsc)$ so that $x_i^n$ is the $i$-th term of the sequence $\mathbf{x}^n$ in $c$.

Let $\varepsilon>0$. As $\mathbf{x}^n\to\mathbf{x}$ as $n\to\infty$, there exists $N\in\mathbb{N}$ such that $\|\mathbf{x}^N-\mathbf{x}\|_{\infty}<\varepsilon/3$. As $\mathbf{x}^N$ is in $c$, it is Cauchy, so there exists $K\in\mathbb{N}$ such that $|x_i^N-x_j^N|<\varepsilon/3$ for all $i,j\geq K$. But then for such $i,j$ we have

\begin{align*} |x_i-x_j|&\leq |x_i-x_i^N|+|x_i^N-x_j^N|+|x_j^N-x_j|\\ &\leq \|\mathbf{x}-\mathbf{x}^N\|_{\infty}+|x_i^N-x_j^N|+\|\mathbf{x}^N-\mathbf{x}\|_{\infty}\\ &<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon, \end{align*}

which proves that $\mathbf{x}$ is cauchy and hence it is in $c$, so $c$ is a Banach space.

On a side note, using that $c$ is a Banach space gives a very quick proof that $c_0=\{(x_n)_{n\geq 0}\in\ell^{\infty}:\lim_{n\to\infty}x_n=0\}$ is a Banach space: one simply observes that we have a bounded linear functional $T:c\to\mathbb{C}$ given by $(x_n)\mapsto \lim_{n\to\infty}x_n$ (check). Then $c_0=\mathrm{ker}(T)$ is a closed linear subspace of $c$, and hence $c_0$ is a Banach space as well.


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