Subspace of a Banach Space that is neither dense nor closed

Question

Does anyone know any basic examples of subspaces of Banach spaces that are neither dense in that space, nor closed? The kernel of any continuous functional is closed, and the kernel of any discontinuous functional is dense, so it would have to be a space that is not the kernel of any functional (just throwing out some ideas here).

Answer 1

A simple way to generate such subspaces $V$ in $\mathcal{C}([0,1])$ endowed with the supremum norm, is to take families of polynomials with a certain form: $$V = \mbox{Span}(\{x^n, n \in S\})$$ with $S$ an infinite subset of $\mathbb{N}$ such that $\sum \limits_{n \in S} \frac{1}{n} < \infty$.

Indeed:

• $V$ isn't closed. If it were, then as a closed subspace of a Banach space, it would be a Banach space itself. But a basis of any infinite dimensional Banach space is uncountable (this comes from the Baire category theorem), and $S$ is countable, which is absurd.
• $V$ is not dense due to the Müntz–Szász theorem.

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Another even more basic example would be in $L^2([-\pi,\pi])$: just consider $$V = \mbox{Span}(\{x \mapsto \sin(nx), n \in \mathbb{N}\})$$

It is not closed, due to Baire category theorem (cf supra). It is not dense either because any even function is orthogonal to all elements of $V$ and thus (except for the zero function) does not belong to $\overline{V}$.

Answer 2

This is basically the same example, but it may be useful to see it written in terms of sequence spaces:

Let $\ell_2 = \{(x_n)_{n \in \mathbb N}:\sum_{n=0}^{\infty} x_n^2 <\infty\}$ be the Hilbert space with a countable dense subset, and $d\colon \ell_2 \to \ell_2$ be the map given by $d((x_n)) = (y_n)$, where for all $k \in \mathbb N$, $y_{2k} = x_k$ and $y_{2k+1}=0$. Then $d$ maps $\ell_2$ isomorphically onto its image $\mathcal E$, a closed proper subspace of $\ell_2$. Now if $V$ is any dense subspace of $\ell_2$, such as the subspace of sequences which have only finitely many non-zero terms

$$V=\left\{(x_n)_{n \geq 0}: \exists N>0 \text{ such that } \forall n \geq N, x_n=0\right\},$$

then $d(V)$ is neither closed nor dense in $\ell_2$.

The above can also be put more abstractly: given a Banach space $X$ consider $Y = X\oplus X$ (where, for $(x,y)\in Y$, we set $\|(x,y)\| = \|x\|+\|y\|$). Any non-closed dense subspace $V$ of $X$ yields a subspace $V_1 = \{(v,0): v \in V\}$ of $Y$ whose closure is easily seen to be $X_1 = \{(x,0): x\in X\} \subsetneq Y$.