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 Post subject: Hilbert space and sequence Posted: Sat Jan 30, 2016 8:43 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Let $\displaystyle{\left(H,<,>\right)}$ be a $\displaystyle{\rm{Hilbert}}$ space and $\displaystyle{\left(x_{n}\right)_{n\in\mathbb{N}}}$

be a sequence of $\displaystyle{H}$ such that $\displaystyle{<x_{n},x_{m}>=0\,,n\,,m\in\mathbb{N}\,,n\neq m}$ .

Prove that the series $\displaystyle{\sum_{n=1}^{\infty}x_{n}}$ converges if, and only if, the real series $\displaystyle{\sum_{n=1}^{\infty}||x_{n}||^2}$

converges.

Top   Post subject: Re: Hilbert space and sequence Posted: Sat Jan 30, 2016 8:44 am

Joined: Mon Jan 18, 2016 4:33 pm
Posts: 4
Hello!Let $a_{k}=\sum_{n=0}^{k} x_{n}$ and $l\geq m\in\mathbb N$.Since $x_{n}$ are mutually orthogonal,applying Pythagoras' theorem gives us that:
$$\|a_{l}-a_{m}\|^{2}=\|\sum_{n=m+1}^{l} x_{n}\|^{2}=\sum_{n=m+1}^{l}\|x_{n}\|^2=\sum_{n=1}^{l}\|x_{n}\|^2-\sum_{n=1}^{m}\|x_{n}\|^2$$
It follows that $(a_{n})$ is a Cauchy sequence (i.e. $\|a_{l}-a_{m}\|\rightarrow 0$, as $n,m\rightarrow\infty$) if and only if $(\sum_{n=1}^{l}\|x_{n}\|^2)_{l}$ is a Cauchy sequence.
Since $( H,<,> )$ is a Hilbert space ,every Cauchy sequence in $H$ converges in $H$(also $\mathbb R$ is complete), so we can equivalently deduce that:
$\sum_{n=1}^{\infty}x_{n}$ converges in $H$ if and only if $\sum_{n=1}^{\infty}\|x_{n}\|^2$ converges.

Top   Post subject: Re: Hilbert space and sequence Posted: Sat Jan 30, 2016 8:45 am
 Team Member

Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Thank you Gigaster for your solution. Here is another solution (similar to Gigaster's solution.)

Obviously, $\displaystyle{||\sum_{k=1}^{1}x_{k}||^2=||x_{1}||^2}$ .

$\displaystyle{||x_1+x_2||^2=||x_1||^2+||x_2||^2+2\,<x_1,x_2>=||x_1||^2+||x_2||^2}$ .

Let $\displaystyle{||\sum_{k=1}^{n}x_{k}||^2=\sum_{k=1}^{n}||x_{k}||^2\,\,(I)}$ for some $\displaystyle{n\in\mathbb{N}}$ .

Now, by using the fact that the inner product is linear, we get:

\displaystyle{\begin{aligned} ||\sum_{k=1}^{n+1}\,x_{k}||^2&=||\sum_{k=1}^{n}x_{k}+x_{n+1}||^2\\&=||\sum_{k=1}^{n}x_{k}||^2+2\,<\sum_{k=1}^{n}x_{k}\,x_{n+1}>+||x_{n+1}||^2\\&\stackrel{(I)}{=}\sum_{k=1}^{n}||x_{k}||^2+||x_{n+1}||^2\\&=\sum_{k=1}^{n+1}||x_{k}||^2\end{aligned}}

So, $\displaystyle{||\sum_{k=1}^{n}x_{k}||^2=\sum_{k=1}^{n}||x_{k}||^2\,,\forall\,n\in\mathbb{N}}$ .

If the series $\displaystyle{\sum_{n=1}^{\infty}x_{n}}$ converges in $\displaystyle{\left(H,||\cdot||\right)}$, then, since the $\displaystyle{||\cdot||}$

is continuous, we have that :

$\displaystyle{\lim_{n\to \infty}\sum_{k=1}^{n}||x_{k}||^2=\lim_{n\to \infty}||\sum_{k=1}^{n}x_{k}||^2=||\sum_{n=1}^{\infty}x_{n}||^2}$ and thus

$\displaystyle{\sum_{n=1}^{\infty}||x_{n}||^2=||\sum_{n=1}^{\infty}x_{n}||^2}$ .

On the other hand, suppose that the series $\displaystyle{\sum_{n=1}^{\infty}||x_{n}||^2}$ converges. Since the normed-space $\displaystyle{\left(H,||\cdot||\right)}$

is a $\displaystyle{\rm{Banach}}$ space, it is sufficient to prove that the sequence $\displaystyle{\left(\sum_{k=1}^{n}x_{k}\right)_{n\in\mathbb{N}}}$

is a $\displaystyle{\rm{Cauchy}}$ sequence. This is true, because, for every $\displaystyle{n\,,m\in\mathbb{N}\,,n\neq m}$ we get :

$\displaystyle{||\sum_{k=1}^{n}x_{k}-\sum_{k=1}^{m}x_{k}||\leq ||\sum_{k=1}^{n}x_{k}||+||\sum_{k=1}^{m}x_{k}||=\sqrt{\sum_{k=1}^{n}||x_{k}||^2}+\sqrt{\sum_{k=1}^{m}||x_{k}||^2}}$

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