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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 $\ell_1$.

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    @keaton I think it should work, just as you say2012-04-01

1 Answers 1

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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 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 $\sup\limits_{n}|x_n-y_n|\leq \epsilon$.