is the following set complete in $l_p$?

Question

I'm not sure what is the correct term in English, but by 'complete set' I mean that the span (of finite combinations) of the set is dense in the space.

I have to show for which p ($ 1\leq p <\infty $) the following is a complete set in the sequence space $ l_p $ : $$\{ x_1=(1,-1,0,0,...) \\ x_2=(0,1,-1,0,0,...) \\ ...\\ x_n=(0,...,0,1,-1,0,...)\\ ...\}$$ (for $x_n$ the nth index is 1, followed by -1).

It is easy to show that it is true for p=2 by showing that the only sequence normal to each of these vectors is the zero sequence.

I have tried proving that it is not true for p other than 2 by showing that $\|e_1-\Sigma c_kx_k\|_p$ is always greater than 1, for the unit vector $e_1=(1,0,0,...)$ and any constants $c_k$. i didn't manage to prove it, and I can't see why it would be different for p=2, which might say that my assumption is wrong.

I have seen this question with slight variations, in which the set becomes for example $x_n=(0,...,0,1,\sqrt{1+1/n},0,...)$.

Answer 1

A different way for $p>1$:

It's not to hard to show the linear span of the $x_i$ contains the vectors $$y_n=(1,\underbrace{-1/n,-1/n,\ldots,-1/n}_{n\text{-terms}},0,\ldots).$$

Since, for $p>1$, we have $\Vert e_1-y_n\Vert_p =1/n^{1-1/p}\rightarrow0$ as $n\rightarrow\infty$, $e_1$ is in the closed linear span of the $x_i$. But then every $e_i$ is in the closed linear span of the $x_i$.

It follows that $\{x_i \mid i\in\Bbb N\}$ is dense in $\ell_p$, for $p>1$.

For $p=1$, see my comment above.

Answer 2

Hint: try the similar idea as for $p=2.$ I.e. use the mapping $\langle\cdot,\cdot\rangle:\ell_p\times\ell_q\to\mathbb C$ instead of the scalar product.

Statement: a vector subspace $F$ of a Banach space $E$ is dense in $E$ if and only if $\text{for every functional}\ f\in E'$ holds:

$$\text{the restriction}\ f|_F=0\ \text{if and only if}\ f=0\ \text{on}\ E$$