Existence of normalized dual basis for a linearly independent set in normed spaces

Question

Let $X$ be a normed space and $\{x_n\}_{n\in\mathbb{N}}$ a countable linearly independent subset. I'm trying to determine whether there always exists a dual set $\{f_n\}_{n\in\mathbb{N}} \subset X^*$ such that:

1. $f_i(x_j) = \delta_{ij}$ (Kronecker delta), and
2. $\|f_n\| = 1$ for all $n$.

I considered using the Hahn-Banach theorem to construct $f_i$ that equals 1 at $x_i$ and vanishes on $\mathcal{M}:=\text{span}\{x_j\}_{j\neq i}$. However, this doesn't guarantee $\|f_i\|=1$.

If I consider $\delta=\inf_{y\in \mathcal{M}}\Vert x_i-y\Vert$, $f_i$ can be taken to satisfy $f_i|_{\mathcal{M}}=0$,$\Vert f_i\Vert =1$,and $f_i(x_i)=\delta$. This doesn’t guarantee $f_i(x_i)=1$.

Is there an analog of Gram-Schmidt orthogonalization for this setting? Any references or insights would be appreciated!

Answer 1

Define $\mathcal{M}_n := \text{span}\{e_i : i \ne n\}$. Since the problem doesn't provide conditions like $\mathrm{dist}(x_n, \mathcal{M}_n) = 1$, it's possible that some $x_n$ may belong to the closure of $\mathcal{M}_n$.

For example, in $\ell^2$, take an element $x_1$ which is naturally a countable linear combination of the standard orthonormal basis. If we choose the other $x_n$ to be the standard basis vectors, then $x_1 \in \overline{\mathcal{M}_1}$, yet $x_1$ cannot be expressed as a finite linear combination of the other $x_n$, meaning $(x_n)_{n \ge 1}$ is a linearly independent set.

However, if we assume the existence of $f_1$ satisfying the given conditions (linear and continuous), we'd arrive at the contradiction $1 = f_1(x_1) = 0$, since $x_1 \in \overline{\mathcal{M}_1}$ implies $f_1(x_1) = 0$.

Answer 2

Let $X$ be an infinite-dimensional normed space and $(x_{n})_{n\in\mathbb{N}}$ a linearly independent sequence in $X$, that is the set $\{x_{n} : n\in\mathbb{N}\}$ is linearly independent with $x_{n}\neq x_{m}$ whenever $m,n\in\mathbb{N}$ and $m\neq n$. The sequence $(f_{n})_{n\in\mathbb{N}}$ in $X^{*}$ is called a biorthogonal sequence associated with $(x_{n})_{n\in\mathbb{N}}$ if $f_{i}(x_{j}) = \delta_{ij}$ for all $i,j\in\mathbb{N}$. If in addition $\|f_{n}\| = 1$ for every $n\in\mathbb{N}$, then $(f_{n})_{n\in\mathbb{N}}$ is called a normalised biorthogonal sequence associated with $(x_{n})_{n\in\mathbb{N}}$. [Please mention if this does not match up with the standard terminology for this concept.] We have the following general result.

Let $X$ be an infinite-dimensional normed space. There exists a linearly independent sequence $(x_{n})_{n\in\mathbb{N}}$ in $X$ that has no biorthogonal sequence associated with $(x_{n})_{n\in\mathbb{N}}$.

With the argument in the answer by PauseAndPonder in mind, it is sufficient to show the following.

Let $X$ be an infinite-dimensional normed space and $x_{0}\in X$. Then there is a linearly independent sequence $(x_{n})_{n\in\mathbb{N}}$ in $X$ where $x_{1} = x_{0}$ and $\lim_{n\to\infty} x_{n} = x_{0}$.

We construct the sequence recursively in a similar manner to this construction. Let $n\in\mathbb{N}$ and suppose $x_{1}, \ldots , x_{n}\in X$ are given such that $x_{1} = x_{0}$, $x_{j}\not\in {\rm span}\{x_{1}, \ldots , x_{j-1}\}$ for each $j\in \{2, \ldots , n\}$ and $\|x_{j} - x_{0}\| \leq \tfrac{1}{j-1}$ for each $j\in \{2, \ldots , n\}$. Define $E := {\rm span}\{x_{1}, \ldots , x_{n}\}$. Since $X$ is infinite-dimensional, the proper finite-dimensional subspace $E$ has empty interior in $X$. Hence there is $x_{n+1}\in \{x\in X : \|x - x_{0}\| \leq \tfrac{1}{n}\} \setminus E$. Then $x_{n+1}\not\in {\rm span}\{x_{1}, \ldots , x_{n}\}$ and $\|x_{n+1} - x_{0}\| \leq \tfrac{1}{n}$. By recursion we obtain a sequence with the desired property.

For another interesting counterexample, consider the Banach space $(C[0,1], \|\cdot\|_{\infty})$ and define $x_{n}\in C[0,1]$ by $x_{n}(t) := t^{n}$ for every $n\in\mathbb{N}_{0}$. Then $(x_{n})_{n\in\mathbb{N}_{0}}$ is a linearly independent sequence in $C[0,1]$. [See here.] Now assume for a contradiction there is a biorthogonal sequence $(f_{n})_{n\in\mathbb{N}_{0}}$ associated with $(x_{n})_{n\in\mathbb{N}_{0}}$. Then for each $k\in\mathbb{N}$ we have $\{x_{j} : j\in\mathbb{N}_{0}\setminus\{k\} \} \subseteq \ker f_{k}$ and hence $x_{k}\not\in \overline{{\rm span}}\{x_{j} : j\in\mathbb{N}_{0}\setminus\{k\} \}$. However, if $F\subseteq \mathbb{N}$ is a nonempty finite set and $n = \max F$, then by applying the Stone-Weierstrass theorem to the algebra generated by the set $\{x_{k(n+1)} : k\in\mathbb{N}_{0}\}$, we have $\overline{{\rm span}}\{x_{k(n+1)} : k\in\mathbb{N}_{0}\} = C[0,1]$ and hence $\overline{{\rm span}}\{x_{k} : k\in\mathbb{N}_{0}\setminus F\} = C[0,1]$ which is a contradiction. As a consequence, there is no biorthogonal sequence associated with $(x_{n})_{n\in\mathbb{N}_{0}}$.