Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $H_1, H_2, \ldots, H_n$ be a countable family of Hilbert spaces. Let H be the set of tuples $x = (x_1, \ldots, x_n,\ldots)\in \prod_n H_n$ with the property that $$\|x \| ^2 =\sum_n \| x_n \| _{H_n}^2 <\infty.$$ Show that H is a Hilbert space.

I have no problem showing that that it is a normed, but I have some problem with the completeness. If we define $x^*$ as the coordinatewise limit points. How do we then show that $$\| x^* \| <\infty$$ and $$\lim_{k\rightarrow \infty} \| x^* - x_k \| \rightarrow 0 $$

share|improve this question
1  
You use $x_n$ here for both an element of $H_n$ and for a sequence of element in your "new" Hilbert space, which is confusing. –  Thomas Andrews Nov 20 '12 at 17:06
    
Yes, sorry about that, it is changed now. –  Johan Nov 20 '12 at 21:02
    
would this work for showing that $$\|x^*\|$$ is in the set? $$\| x^*\| \leq liminf_{k \rightarrow \infty} \| x^k\| = liminf \| x^k + x^m - x^m\| \leq liminf \| x^k - x^m\| + \|x^m\| \leq \epsilon + b \leq \infty $$. The first inequality is due to Fatou. –  Johan Nov 21 '12 at 14:08
    
"\parallel" produces a rather obscure symbol used for indicating parallel lines. It gives far too much spacing to use as a norm symbol: $\parallel x \parallel$. For norms, use "\|": $\| x \|$. –  Chris Eagle Nov 21 '12 at 16:26

1 Answer 1

up vote 2 down vote accepted

Let $(x_n)_{n\geq 1}$ a Cauchy sequence in $H$, then given $\epsilon>0$, there exists $N_o$ positive such that

\begin{equation} \|x_n-x_m\|_H<\epsilon, \text{ for }n, m\geq N_o\qquad(*). \end{equation} from this equation we have, for each $k\geq 1$

$$\|x_n^k-x_m^k\|_{H_k}^2\leq\sum_{k}\|x_n^k-x_m^k\|_{H_k}^2=\|x_n-x_m\|^2<\epsilon^2;\text{ for } n, m\geq N_o.$$ This means that $(x_n^k)_{n\geq 1}$ is a cauchy secuence in $H_k$ for each $k\geq 1,$ so there exists $x_k\in H_k$ such that $x_n^k\to x_k$ as $n\to \infty.$ Now we will consider the element $x=(x_1,x_2,x_3,x_4,\ldots,x_n, \ldots )$ obtained in this way. It's not too much dificult to show that $x\in H$ and $x_n\to x$ as $n\to \infty.$ In order to show that $x\in H$, we will show that $x_{n_o}-x\in H$ for some $n_o,$ since $H$ is closed under subtraction it will follow that $x=x_{n_o}-(x_{n_o}-x)\in H.$ Pick $N\in \mathbb{N}$, by $(*)$ $$\sum_{k=1}^{N}\Arrowvert x_n^k-x_m^k\Arrowvert_{H_k}^2\leq \sum_{k}\Arrowvert x_n^k-x_m^k\Arrowvert_{H_k}^2=\Arrowvert x_n-x_m\Arrowvert_H^2<\epsilon^2, \text{ for } n,m\geq N_o.$$ Let $m\to \infty$ in the finite sum on the left hand side: we obtain $$\sum_{k=1}^{N}\Arrowvert x_n^k-x_k\Arrowvert_{H_k}^2<\epsilon^2, \text{ for } n\geq N_o,$$ Since this equation holds for all $N\in \mathbb{N}$ we have, on letting $N\to \infty$, $$\sum_{k=1}^{\infty}\Arrowvert x_n^k-x_k\Arrowvert_{H_k}^2<\epsilon^2, \text{ for } n\geq N_o,$$ that is $$\Arrowvert x_n-x\Arrowvert_H<\epsilon, \text{ for } n\geq N_o, \qquad(**)$$ In particular we have $$\Arrowvert x_{N_o}-x\Arrowvert_H<\epsilon,$$ and so $x_{N_o}-x\in H, $ this proves that $x\in H,$ at the same time from $(**)$ we get $x_n\to x$ as $n\to\infty$ with recpect the norm $\Arrowvert\cdot \Arrowvert_H$.$\clubsuit$

share|improve this answer
    
Thanks, can you please show the rest? This was the part of the problem that I figured out by myself, along with the fact that it is a norm. –  Johan Nov 21 '12 at 7:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.