Show that $(e_1,e_2,...)$ is not a Schauder basis of $\ell^\infty$ where $e_i$ is the vector in $\mathbb R^\infty$ with 1 in the ith coordinate and 0 elsewhere and $\ell^\infty=\{(x_1,x_2,...)|x_i\in \mathbb R ~~and ~~ \|x\|_\infty<\infty\} ,\|x\|_\infty=\sup\{|x_1|,|x_2|,...\}$

I am wondering how to prove it and why this statement is true. Actually I think it is a basis since for any given vector $(x_1,x_2,...)=x_1e_1+x_2e_2+...$.

I already proved that $(e_1,e_2,...)$ is a Schauder basis of $\ell^p$ where p>1. I just can't see what changes when it comes to the case of $\ell^\infty$

  • 1
    $\begingroup$ Spaces with Schauder bases are separable. $\endgroup$ Sep 28, 2017 at 12:24
  • 1
    $\begingroup$ Or, directly, show that no sum $\sum_{i=1}^\infty \alpha_i e_i$ can converge to $(1,1,\ldots)$. $\endgroup$ Sep 28, 2017 at 12:25
  • $\begingroup$ @DavidMitra did you mean "... no sum $\sum_{i=1}^n \alpha_ie_i$ can converge..."? $\endgroup$ Sep 28, 2017 at 12:36
  • $\begingroup$ @TZakrevskiy No? The infinite sum converges, of course, iff the sequence of partial sums does. $\endgroup$ Sep 28, 2017 at 12:48
  • $\begingroup$ @DavidMitra If every $\alpha_i=1$? $\endgroup$ Sep 28, 2017 at 13:06

2 Answers 2


Proof by contradiction. Assume it is a Schauder basis. Then there are $a_1, \ldots, a_n \in \mathbb{R}$ such that $$ \| a_1 e_1 + \ldots + a_n e_n - (1, 1, 1, \ldots) \| < \frac{1}{2} $$

But $a_1 e_1 + \ldots + a_n e_n = (a_1, a_2, \ldots, a_n, 0, 0, \ldots)$. Thus $$ \| a_1 e_1 + \ldots + a_n e_n - (1, 1, 1, \ldots) \| \geq \lvert0-1\rvert = 1 $$ Contradiction.


Hint: By contradiction

1.Show that, $(\mathbb R^\infty,\|\cdot\|_{\infty})$ is a Banach space

  1. For $n\in\mathbb N$ set $$\mathbb R_n = \{x=(x_1,x_2,...)\in\mathbb R^\infty :x_i =0 ~~\text{if}~~i>n~i.e~x=(x_1,x_2,\cdots,x_n,0,0,\cdots)\}$$
    Show that $\mathbb R_n$ is closed subspace of $(\mathbb R^\infty,\|\cdot\|_{\infty})$
  2. Since $(e_1,e_2,..)$ is Schauder Basis, observe that, $$\mathbb R^\infty = \bigcup_{n\in\mathbb N}\mathbb R_n $$
  3. From Baire Theorem's there exists $n_0$ such that, $\overset{\circ}{\mathbb R}_{n_0} \neq \emptyset$ that is there are $x_0\in \mathbb R^\infty$ and $r>0$ such that, $B(x_0,r)\subset\mathbb R_{n_0}$

5.Deduce from 4. that, $\mathbb R_{n_0}=\mathbb R^\infty$ contradiction.


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