Linear Independence and Biorthogonal sets

Question

Let $X$ be an Hilbert space we say that $f_n\in X$ and $g_n\in X$ are biorthogonal if for any $n,m \in \mathbb{N}$ we have $\langle f_n,g_m \rangle=\delta_{nm}$. Now I was able to show that a necessary condition for this to happen is that the $f_n's$ need to be linearly independent and I am wondering if this is also sufficient? I know that if we are working with a finite set this will be true because we can use the Hahn-Banach Theorem to find linear functions that do what we want and then we can use the Riesz-Representation theorem , but I am not sure this works for infinite dimensions and I wasn't able to come with a counterexample, so any help guiding me in the right direction is appreciated.

Answer 1

Such a biorthogonal family exists if and only if $$ f_n \notin V_n := \overline{\mathrm{span}} \{ f_m : m \neq n \} \qquad \forall \, n \in \Bbb{N} . \tag{$\ast$} $$ Indeed, if $(\ast)$ holds, then $g_n^{(0)} := \pi_{V_n^\perp} f_n \neq 0$ (here, $\pi_{V_n^\perp}$ is the orthogonal projection on the orthogonal complement of $V_n$), and $\langle f_n, g_n^{(0)} \rangle = \| g_n^{(0)} \|^2 > 0$, so that $\langle f_n ,g_n \rangle = 1$ for $g_n = \alpha_n \, g_n^{(0)}$ with $\alpha_n > 0$ suitable. Furthermore, if $n \neq m$, then $f_n \in V_m$ and $g_m \in V_m^\perp$, so that $\langle f_n, g_m \rangle = 0$.

Conversely, if a biorthogonal family $(g_m)_{m}$ exists, then it is easy to see $\langle f, g_m \rangle = 0$ for all $f \in V_m$, and hence $f_m \notin V_m$, since $\langle f_m, g_m \rangle = 1 \neq 0$.

Now for a specific counterexample where linear independence holds, but where $(\ast)$ is not satisfied, take $X = \ell^2(\Bbb{N})$ with $\Bbb{N} = \{ 1,2,\dots \}$, and take $f_1 = (1/k)_{k \in \Bbb{N}}$ and $f_{n+1} = \delta_n$ for $n \in \Bbb{N}$. I leave it to you to verify that $(f_n)_n$ is linearly independent, but that $f_1 \in V_1$.