Equivalent of sequence convergence in Hilbert space

Question

$H$ is a Hilbert space, $\{x_{n}\}$ is a orthogonal family in $H$(not need to be Orthonormal), prove that the following conditions are equivalent:

1. $\sum_{n=1}^{\infty} x_{n}$ convergence
2. $\forall y \in H ,\sum_{n=1}^{\infty}\langle x_{n},y \rangle$ convergence
3. $\sum_{n=1}^{\infty} \|x_{n}\|^2$ convergence

I have known that how to show (1)$\iff $ (3):

Calculate $\|\sum_{k=1}^{m} x_{k}-\sum_{k=1}^{n} x_{k}\|^2=\sum_{k=1}^{m} \|x_{k}\|^2-\sum_{k=1}^{n} \|x_{k}\|^2$ to show that $\{\sum_{k=1}^{n} x_{k}\}$ is Cauchy sequence $\iff$ $\sum_{k=1}^{n} \|x_{k}\|^2$ is Cauchy sequence, then use the condition that $H$ is Hilbert space.

And (1)$\Rightarrow $(2):

Take $x=\sum_{k=1}^{\infty}x_n$ then $\langle x,y \rangle=\langle \lim_{n \to \infty}\sum_{k=1}^{n}x_k,y \rangle=\sum_{n=1}^{\infty}\langle x_{n},y \rangle$.

But I still have no idea know how to show (2)$\Rightarrow$(1) or (2)$\Rightarrow$(3).

Thanks if you can give me any advice of any form.

Answer 1

Define $T_{n}(y)=\langle\sum_{k=1}^{n}x_{k},y\rangle$ . Then $||T_{n}||=\sqrt{\sum_{k=1}^{n}||x_{k}||^{2}}$

and $T_{n}(y)$ converges for each $y$ implies $\sup_{n}|T_{n}(y)|\leq C_{y}$

Thus by Banach Steinhauss/Uniform Boundedness Principle, $$\sup_{n}||T_{n}||<\infty$$

which also means that $\sup_{n}\sum_{k=1}^{n}||x_{k}||^{2}<\infty$ which is what is required.

Answer 2

Consider the sequence of linear functionals $\varphi_N$ on $\mathcal{H}$ defined by $$\varphi_N(y)=\left \langle \sum_{n=1}^Nx_n,y\right \rangle =\sum_{n=1}^N\langle x_n,y\rangle$$ Then $$\|\varphi_N\|^2=\left \|\sum_{n=1}^Nx_n\right \|^2=\sum_{n=1}^N\|x_n\|^2$$ By (2) the sequence $\varphi_N(y)$ is convergent for every $y,$ hence it is bounded. By the Banach-Steinhaus theorem the norms $\|\varphi_N\|$ are uniformly bounded. Hence the series $\sum_{n=1}^\infty \|x_n\|^2 $ is convergent.

Remark The completness of the space is essential. Indeed if $V={\rm span}\{x_n\}_{n=1}^\infty$ then the condition (2) holds for any orthonormal sequence $\{x_n\},$ but the condition (3) is not satisfied.