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Let $M$ be the closed span of these vectors in the real l2 sequence space. A vector $\beta$ is orthogonal to $M$ exactly when ${\beta }_{k+2}-2r{\beta }_{k+1}+{\beta }_k=0$. The characteristic roots are $r\pm \sqrt{r^2-1}$; one grows geometrically and the other, $\rho$, lies in $(0,1)$. Square summability eliminates the growing solution, so $M^{\perp }=\mathrm{span}\{({\rho }^{j-1}{)}_{j\ge 1}\}$. Hence $M$ is a closed hyperplane, namely the kernel of the bounded functional $\alpha \mapsto {\sum }_j{\alpha }_j{\rho }^{j-1}$. This combines a recurrence with density by annihilators.

The closed linear span of these vectors is all of $c_0$. Otherwise the Hahn-Banach theorem gives a nonzero continuous functional annihilating them, represented by $a\in {\ell }_1$. At the first index $m$ where $a_m\ne 0$, divide ${\sum }_j{a}_j{t}_n^j=0$ by $t_n^m$. The remaining tail has magnitude at most $t_n{\sum }_{j>m}|a_j|\to 0$, forcing $a_m=0$, a contradiction. This is an elementary coefficient-isolation argument for density by annihilators.

Suppose the distinct complex numbers ${\beta }_k$ satisfy $|{\beta }_k|\le \delta <1$. The geometric sequences in the displayed formula belong to the absolutely summable sequence space and have dense linear span. Otherwise the Hahn-Banach theorem gives a nonzero annihilating functional, represented by a bounded sequence $b_j$. Its power series $F(z)={\sum }_{j\ge 1}{b}_j{z}^{j-1}$ is holomorphic in the unit disk and vanishes at every ${\beta }_k$. Compactness supplies an accumulation point inside that disk. The identity theorem forces $F=0$ and hence all its coefficients zero, a contradiction. This is a useful form of density by annihilators that converts a Banach-space problem into uniqueness of an analytic function.

The nonzero bounded functionalis continuous on $L^1$. For every translate, $\Lambda (f(\cdot -t))=e^{-2\pi it}\hat{f}(1)=0$. Thus it annihilates $Y$, and density by annihilators shows that $Y$ is not dense in $L^1$.