 Solution

Put $\rho =r-\sqrt{r^2-1}$. Since $r>1$, $0<\rho <1$ and ${\rho }^2-2r\rho +1=0$. The nonzero square-summable sequence ${\beta }_j={\rho }^{j-1}$ defines a continuous linear functional on ${\ell }_2$ byThe Cauchy-Schwarz inequality proves continuity. For every $k$,Hence $f$ vanishes on the closed linear span $M$, but $f(e_1)=1$. ThereforeMore precisely, any vector in the orthogonal complement of $M$ satisfies the recurrence ${\beta }_{k+2}-2r{\beta }_{k+1}+{\beta }_k=0$. Its roots are $r\pm \sqrt{r^2-1}$, one less than one and the other greater than one. Square summability eliminates the growing solution, so $M^{\perp }$ is exactly the span of $({\rho }^{j-1})$. Thus $M=\ker f$ is a closed hyperplane of codimension one. This is annihilator of a second-difference span in l2.