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PostPosted: Sat Jan 30, 2016 8:43 am 
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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.


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PostPosted: Sat Jan 30, 2016 8:44 am 

Joined: Mon Jan 18, 2016 4:33 pm
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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.


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PostPosted: Sat Jan 30, 2016 8:45 am 
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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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