Convergent sequence in Hilbert space

Question

Let $H$ be a Hilbert space with orthonormal basis $(e_n)_n$ and $(a_n)_n \in \ell^2(\mathbb{N})$, so $$\sum_{n=0}^{\infty} |a_n|^2 < \infty$$ Now, define the sequence $(\xi_n)_n$ in $H$ as $$\xi_n= \sum_{k=0}^{n} a_ke_k $$

I now have to proof that this sequence is convergent in $H$ or, if not, give a counterexample. I think that it's true because $a_n \to 0$ when $n \to \infty$ but I don't know how I can proof this.

Answer 1

I claim the sequence

$\xi_n = \displaystyle \sum_{k = 0}^n a_k e_k \tag 1$

is Cauchy in $H$; for, letting $\langle \cdot, \cdot \rangle$ be the inner product on $H$, and $\Vert \cdot \Vert$ the corresponding norm, so that

$\Vert x \Vert^2 = \langle x, x \rangle, \; \forall x \in H, \tag 2$

we compute, for $m > l$,

$\Vert \xi_m - \xi_l \Vert^2 = \langle \xi_m - \xi_l, \xi_m - \xi_l \rangle = \left \langle \displaystyle \sum_{l + 1}^m a_k e_k, \sum_{l + 1}^m a_k e_k \right \rangle = \displaystyle \sum_{l + 1}^m \vert a_k \vert^2 \tag 3$

by the orthonormality of the $e_i$; we have

$\displaystyle \sum_{n = 0}^\infty \vert a_i\vert^2 < \infty, \tag 4$

from which it follows that, given $\epsilon > 0$, there is a positive integer $N$ such that, for $l > N$,

$\displaystyle \sum_{l + 1}^\infty \vert a_i \vert^2 < \epsilon; \tag 5$

then for $m > l > N$,

$\Vert \xi_m - \xi_l \Vert^2 = \displaystyle \sum_{l + 1}^m \vert a_k \vert^2 \le \sum_{l + 1}^\infty \vert a_i \vert^2 < \epsilon; \tag 6$

this indicates that $\xi_n$ is Cauchy in $H$; since $H$ is by definition Cauchy complete, $\xi_n$ converges to a limit $\xi$ in $H$:

$\xi_n \to \xi. \tag 7$

Answer 2

I don't think the observation of $a_k\to 0$ as $k\to\infty$ is going to help you much.

It suffices to show that $(\xi_n)$ is a Cauchy sequence. Essentially you need to estimate $$ \|\xi_p-\xi_q\|=\left\|\sum_{k=q+1}^p a_ke_k\right\|.\tag{1} $$ But the orthonormal property tells you that $$ \left\|\sum_{k=q+1}^p a_ke_k\right\|^2=\sum_{k=q+1}^p \|a_ke_k\|^2=\sum_{k=q+1}^p |a_k|^2.\tag{2} $$ On the other hand, you know that $$ \sum |a_k|^2<\infty $$ which gives you an estimate for (2).