Show that $\{e_n\}_{n≥2}$ is an orthonormal system in which is not complete. However, if $f \in E$ and $f⊥e_n$, for all n≥2, then =0.

Question

Consider $x_0 = (1/j)_{j≥1} \in l^2(\mathbb{N})$ and $\{_\}$ the usual canonical vectors of $ l^2(\mathbb{N})$. Then $E = span(x_0,e_n)_{n≥2}$ is a pre-Hilbert space. Show that $\{e_n\}_{n≥2}$ is an orthonormal system in which is not complete. However, if $f \in E$ and $f⊥e_n$, for all n≥2, then =0.

My specific question is in the part "if $f \in E$ and $f⊥e_n$, for all n≥2, then =0." I have to prove that $\{_\}⊥$ and, clearly $||\{_\}||=1$ but also $f=x_0-\sum_{n=2}^\infty \langle x_0,e_n\rangle e_n$ nd ⊥,for all ≥2, $f⊥e_n$, for all n≥2, while ≠0

Answer 1

For $f \in l^2(\mathbb{N})$ it is obviously false (take $f = e_1$). So let $f \in E$, i.e. $f = \lambda_1 x_0 + \sum_{n=2}^\infty \lambda_n e_n \in E$ where all but finitely many $\lambda_n = 0$. We have $$f = \lambda_1 e_1 +\sum_{n=2}^\infty (\lambda_1 /n + \lambda_n) e_n .$$ Since $f⊥e_n$ for $n \ge 2$, we get $\langle f, e_n \rangle = \lambda_1 /n + \lambda_n = 0$ for $n \ge 2$. But $\lambda_n = 0$ for $n \ge n_0$ which shows that $\lambda_1 /n = 0$ for $n \ge n_0$. Thus $\lambda_1 = 0$. This implies that $\lambda_n = 0$ for all $n \ge 2$. Therefore $f = 0$.