True or False: "The span of a set $U$ that is dense in the closed unit ball $S$ of a Banach Space $X$ is $X$"

Question

I think this statement is highly unlikely to be true. But can anyone produce a counterexample for this?

Equivalently, the statement can be written as "The maximal proper subspace of a Banach Space $X$ is not dense in $X$".

Answer 1

My functional analysis is a bit rusty, but this might work as a counterexample...

Look at $C [0,1] $ (continuous real functions over $[0,1] $), with the $\infty $ (maximum) norm, and look only at the polynomial functions as a subspace. They are dense not just in the unit ball, but everywhere, because every continuous function can be (uniformly) approximated by a polynomial function to any desired accuracy. Still, polynomials don't span all of $C [0,1] $.

Note: I know, should've taken just the polynomials with norm $\le 1$, but this should be just a detail to fill in... After all, for every $f\in C [0,1] $, either its norm is $\lt 1$, in which case the norm of the approximating polynomials can be made $\lt 1$ by the triangle inequality, or the norm of $f $ is 1, in which case let's approximate something like $(1-\varepsilon)f $ and use the triangle inequality to conclude $f $ itself is well-approximated.

Answer 2

Consider the Banach space $c_0$ with the norm $\|\cdot\|_\infty$ and the canonical vectors $$e_n = (\underbrace{0, \ldots, 0}_{n-1}, 1, 0, 0, \ldots)$$ for all $n \in \mathbb{N}$.

It is known that every $x = (x_1, x_2, \ldots) \in c_0$ can be written as $x = \sum_{n=1}^\infty x_ne_n$.

In particular, if $\|x\|_\infty \le 1$, then $$x = \lim_{n\to\infty}\sum_{i=1}^n x_ie_i = \lim_{n\to\infty} (x_1, \ldots, x_n, 0, 0, \ldots)$$ where $|x_n| \le 1$ for all $n \in \mathbb{N}$.

If $c_{00}$ denotes the set of all finitely-supported sequences, this means that the set $$U = \big\{(x_n)_{n=1}^\infty \in c_{00} : |x_n| \le 1, \forall n \in \mathbb{N}\big\} = c_{00} \cap \overline{B}(0,1)$$ is dense in $\overline{B}(0,1)$.

However, $\operatorname{span} U = c_{00}$ is strictly less than $c_0$ since for example $\left(\frac1n\right)_{n=1}^\infty \in c_0 \setminus c_{00}$.