This is a question that is based off of the book $\textit{An Introduction to Nonharmonic Fourier Series}$ by Young.
Assume that $\{e_n\}_{n = 1}^\infty$ is the canonical basis for $\ell^2(\mathbb{N})$. Consider the sequence $\{e_n + e_{n+1}\}_{n = 1}^\infty$. I want to know when this sequence is minimal. Being minimal is equivalent to the existence of another sequence $\{f_n\}_{n = 1}^\infty$ such that $(e_n + e_{n+1}, f_m) = \delta_{nm}$ (biorthogonal system). Here $(\cdot, \cdot)$ denotes the inner product of $\ell^2(\mathbb{N})$ and $$\delta_{nm} = \cases{0, n \neq m \\ 1, n = m}.$$ We know that the first element of the sequence $f_1$ must satisfy the equations $$\cases{(e_2, f_1) = 1 - (e_1, f_1) \\ (e_k, f_1) = (-1)^{k-2}(1 - (e_1, f_1))},$$ for $k \geq 2$. How can I solve this system for $f_1$?
More generally, what happens when I instead consider $\{c_1e_n + c_2e_{n+1}\}_{n=1}^\infty$ for scalar $c_1,c_2 \neq 0$?
We have similar equations that $f_1$ must satisfy $$(e_2, f_1) = \frac{1}{c_2}(1 - c_1(e_1, f_1)) \\ (e_k,f_1) = (-\frac{c_1}{c_2})^{k-2} \frac{1}{c_2}(1-c_1(e_1, f_1)),$$ for $k \geq 2$. It is not so obvious as in the first case however.
$\|f_1\|^{2}=\sum |(e_k,f_1)|^{2} <\infty$. Since $|(e_k,f_1)|$ is independent of $k$ for $k>1$ this is possible only when $1-(e_1,f_1)=0$. Thus, $(e_1,f_1)=1$ and $(e_k,f_1)=0$ for $k \neq 1$. This means $f_1=e_1$.
In the general case just use the fact that $f_1=\sum (f_1,e_k)e_k$.