Show that $\{e_n\}_{n≥2}$ is an orthonormal system in $span(x_0,e_n)_{n≥2}$ which is not complete.

Question

Consider $x_0 = (1/j)_{j≥1} \in l^2(\mathbb{N})$ and $e_n$ 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 $E$ which is not complete. However, if $f \in E$ and $f⊥e_n$, for all $n≥2$, then $f= 0.$

Can you help me with this proof?

My specific question is in the part "Show that $\{e_n\}_{n≥2}$ is an orthonormal system in $E$ which is not complete." I have to prove that $\{e_n\} ⊥ E$ and, clearly $||\{e_n\}||=1$ but also, how can I define a Cauchy sequence in this to prove the incompleteness?

Answer 1

The orthonormal set $\{e_n\}_{n\ge 2}$ is not a basis of $E$ since $$ x_0-\sum_{n=2}^\infty \langle x_0,e_n\rangle e_n=e_1\ne 0, $$ where $$ x_0=\left(1,\frac{1}{2},\ldots,\frac{1}{n},\ldots\right). $$ Note that $\sum_{n=2}^\infty \langle x_0,e_n\rangle e_n\in E$, since $$ \sum_{n=2}^\infty |\langle x_0,e_n\rangle|^2= \sum_{n=2}^\infty\frac{1}{n^2}<\infty. $$ But so does $$ w=x_0-\sum_{n=2}^\infty \langle x_0,e_n\rangle e_n $$ and $$ w\perp e_n,\quad\text{for all $n\ge 2$}, $$ while $w\ne 0$.

Answer 2

The completeness in that case means that $\overline{\text{span}(e_{n})_{n\geq 2}}\ne E$. Note that every finite linear combination of $(e_{n})_{n\geq 2}$ has the first entry $0$, so does their limit, so $x_{0}$ does not belong to $\overline{\text{span}(e_{n})_{n\geq 2}}$.