# Closed subspace of $l^\infty$

I've got here this exercise that says: "Show that $c$ is a closed subspace of $l^{\infty}$" (with $c$ I mean the sequences of $l^{\infty}$ that converge in $l^{\infty}$, with respect to the norm of $l^{\infty}$). I've done it, but I cannot say if it is correct.

In order to show that $c$ is a closed subspace of $l^{\infty}$, I have to prove that any convergent sequence $\{c_n\}_n$ of elements of $c$ converges to $x\in c$. I know that, since $l^{\infty}$ is complete, $\{c_n\}$ converges to $x\in l^{\infty}$, so it is enough prove that $x\in c$. Since $\{c_n\}$ converges to $x\in l^{\infty}$, we have that $||c_n-x||_\infty\to 0$ for $n\to \infty$, i.e. $\sup_{j\in \mathbb{N}}|c_n-x|\to 0$ (here $j$ runs over the elements of the sequence $c_n-x$), i.e. for every $\epsilon>0$ there exists $N>0$ such that $\sup_{j\in \mathbb{N}}|c_n-x|<\epsilon$ for $n>N$.

Now, since $c_n\in c$, we have that $c_n\to \xi\in l^{\infty}$ for $n\to \infty$, and so for every $\epsilon>0$

\begin{equation*} \sup_{j\in \mathbb{N}}|\lim_{n\to \infty}(c_n-x)|=\sup_{j\in \mathbb{N}}|\xi-x|<\epsilon, \end{equation*}

which means that $x\in c$.

What do you think? Is there anybody that could suggest me a different argument?

• This inequality may help: $\|x\|_\infty\leq\|x-c_n\|_\infty+\|c_n\|_\infty$. Nov 8, 2013 at 16:25
• Can you define explicitly the set $c$, please? It's not at all clear to me what its elements are. Nov 8, 2013 at 16:29
• @user40615: Thanks for the hint, I was thinking about that too, but I only know that $||c_n||_\infty$ is bounded, how can I say that $||c_n||_\infty\rightarrow 0$ for $n\rightarrow \infty$? Nov 8, 2013 at 16:38
• As I remember $c$ is the subspace of $\ell_\infty$ with the property that all sequences in $c$ converges zero. Nov 8, 2013 at 16:41
• @user40615: For me, the subspace of $l^\infty$ whose elements converge to zero is $c_0$. My professor has defined $c$ as the subspace of the convergent sequence. Nov 8, 2013 at 16:46

I think your argument has the right ideas, but it still needs a little touch since your $\xi$ depends on $n$.

It is probably easier to just show that $x$ is Cauchy: for $\varepsilon>0$, choose $k$ with $\|x-c_k\|_\infty<\varepsilon$, and $n_0$ such that $\|c_k(m)-c_k(n)|<\varepsilon$ for all $m,n>n_0$. Then $$|x(m)-x(n)|\leq |x(m)-c_k(m)|+|c_k(m)-c_k(n)|+|c_k(n)-x(n)|\\\leq2\|x-c_k\|_\infty+|c_k(m)-c_k(n)| \leq3\varepsilon.$$

Careful book-keeping is needed for a proof like this. Denote by $c_n = (c_n^{(j)})_{j\in\mathbb{N}}$ a generic term of a sequence of elements of $l^\infty$. Define $c:=\{(x^{(j)})_{j\in\mathbb{N}}\in l^\infty ~|~ \exists \lim\limits_{j\rightarrow\infty} x^{(j)} \in \mathbb{C} \}$, a subspace of $l^\infty$. Note that the elements of $c$ are precisely the convergent sequences in $\mathbb{C}$ (since these are automatically bounded). To avoid confusion, I will refer only to convergence in $\mathbb{C}$ using the notation $\lim\limits_{j\rightarrow\infty}$.

Now suppose the sequence $(x_n)_{n\in\mathbb{N}}$ of elements of $c$ has a sup-norm limit $x$. By completeness, as you said, $x = (x^{(j)})_{j\in\mathbb{N}} \in l^\infty$, i.e., it's a bounded sequence of complex numbers, and by the definition of $l^\infty$-convergence we have that $\|x_n-x\|_\infty\rightarrow 0$ as $n\rightarrow \infty$.

To prove that $x \in c$ we need to show that there is $\xi \in \mathbb{C}$ which is the limit of $(x^{(j)})_{j\in\mathbb{N}}$, i.e., $\lim\limits_{j\rightarrow\infty} x^{(j)} = \xi \in \mathbb{C}$. The obvious candidate is the limit (if it exists!) of the sequence $(\xi_n)_{n\in\mathbb{N}}$, where $\xi_n:=\lim\limits_{j\rightarrow\infty} x_n^{(j)}$ exists in $\mathbb{C}$ for each $n\in\mathbb{N}$ since each $x_n$ is in $c$.

Now, $|\xi_n - \xi_m| \leq |\xi_n - x_n^{(j)}| + \|x_n- x_m\|_\infty + |x_m^{(j)} - \xi_m| \rightarrow 0$, as $m,n\rightarrow \infty$ (since the LHS is independent of the $j$ in the RHS, so we can take $j\rightarrow \infty$ as well). Using the completeness of $\mathbb{C}$, there is a limit $\xi = \lim\limits_{n\rightarrow\infty} \xi_n \in \mathbb{C}$.

Then, $|x^{(j)}-\xi| \leq \|x - x_n\|_\infty + |x_n^{(j)} - \xi_n| + |\xi_n - \xi|\rightarrow 0$ as $j\rightarrow \infty$ since, similarly to earlier, the LHS is independent of $n$ in the RHS. The proof is complete.

• Thank you for the answer. But I still have a (stupid) doubt on what you've said. It's not clear when you've used the triangle inequality to show that the sequence $(\xi_n)$ is Cauchy. Could you explain it more specifically? Nov 9, 2013 at 9:24
• No no ok!! Now I understand what you mean. Perfect, thanks for the explanation! Nov 9, 2013 at 9:28
• The trick in many similar cases (e.g., proving completeness of a sequence space) is to find a plausible candidate for the limit of a sequence, and then prove using the triangle inequality that the sequence does indeed converge to this limit. Also, just adding zero is another favourite little trick of most analysts. Nov 9, 2013 at 9:39
• Martin Argerami's answer is also sound, and more concise than mine; mine is a construction of the limit, whereas his is simply a non-constructive proof of its existence, which is still enough to establish the claim. Nov 9, 2013 at 9:49

For any $\epsilon$ it is true that $\|x-c_n\|\leq \epsilon$ and $\|c_n\|\leq M$ for some $M>0$(since $c_n\in\ell_{\infty}$). Thus you have $\|x\|_\infty\leq \epsilon+M$ (Choose $\epsilon=1$).