Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This may be a silly question, but here goes. To ensure clarity, $\ell_1$ is the space of absolutely summable sequences, and $c_0$ the space of bounded sequences with limit 0. So we know that $\ell_1\subset c_0$ by basic principles. My question is: is $\ell_1$ when equipped with the sup-norm dense in $c_0$?

Here is my thought, and I would appreciate a comment on correctness or if something went wrong:

Let $\xi\in c_0$ and write $\xi=\{\xi_1,\xi_2,\xi_3,\dots\}$. Now define $P_n:\ell_1\to c_0$ by $$P_n(\eta)=\{\eta_1,\eta_2,\dots,\eta_n\}$$ So if $\xi\in c_0$, we can say $$\xi=\underset{n\to\infty}{\lim}P_n\xi$$

So does this get us all of $c_0$?

A typical example would be the harmonic sequence $\{1, 1/2, \dots, 1/n,\dots\}$. This is in $c_0$ but not $\ell_1$, but taking finite pieces of this sequence at a time guarantees us to remain in $\ell_1$, and we can approximate the sequence in $c_0$ as the limit of elements of $\ell_1$.

share|improve this question
1  
Your proof for the harmonic sequence generalizes. –  dls Mar 28 '12 at 23:41
3  
$c_{00}$ is dense in $c_0$, so any set containing $c_{00}$ is dense in $c_0$. –  user16299 Mar 29 '12 at 2:50
    
@Yemon Will the same reasoning always hold for sums of Banach spaces? It seems like the same arguments will work to show that the $\ell_1$ sum of Banach spaces is dense in the $c_0$ sum of Banach space, and so on. Or is there an example out there where this would break down? –  Keaton Mar 29 '12 at 16:04
    
@keaton I think it should work, just as you say –  user16299 Apr 1 '12 at 0:18
add comment

1 Answer

up vote 4 down vote accepted

Yes. Let $(x_n)\in c_0$ and take $\epsilon>0$. Since $x_n\to 0$, we have some $N$ such that $n\geq N\implies |x_n|<\epsilon$. Define the sequence $(y_n)$ by $y_n=x_n$ for $n<N$ and $y_n=0$ for $n\geq N$. Clearly $(y_n)\in \ell^1$, and for any $n$, $|x_n-y_n|\leq \epsilon$, hence $\sup\limits_{n}|x_n-y_n|\leq \epsilon$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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