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?

  • $\begingroup$ This inequality may help: $\|x\|_\infty\leq\|x-c_n\|_\infty+\|c_n\|_\infty$. $\endgroup$
    – daulomb
    Nov 8, 2013 at 16:25
  • $\begingroup$ Can you define explicitly the set $c$, please? It's not at all clear to me what its elements are. $\endgroup$
    – Stromael
    Nov 8, 2013 at 16:29
  • $\begingroup$ @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 $n\rightarrow \infty$? $\endgroup$
    – batman
    Nov 8, 2013 at 16:38
  • $\begingroup$ As I remember $c$ is the subspace of $\ell_\infty$ with the property that all sequences in $c$ converges zero. $\endgroup$
    – daulomb
    Nov 8, 2013 at 16:41
  • $\begingroup$ @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. $\endgroup$
    – batman
    Nov 8, 2013 at 16:46

3 Answers 3


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.

  • $\begingroup$ 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? $\endgroup$
    – batman
    Nov 9, 2013 at 9:24
  • $\begingroup$ No no ok!! Now I understand what you mean. Perfect, thanks for the explanation! $\endgroup$
    – batman
    Nov 9, 2013 at 9:28
  • $\begingroup$ 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. $\endgroup$
    – Stromael
    Nov 9, 2013 at 9:39
  • $\begingroup$ 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. $\endgroup$
    – Stromael
    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 $\epsilon=1$).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .