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Completeness

Impo

New member
Apr 2, 2013
17
Hi,

Let [tex] H = \{(x_n)_n \subseteq \mathbb{R} | \sum_{n=1}^{\infty} x_n < \infty \}[/tex] and for $(x_n)_n \in H$ define
$$\|(x_n)_n\|_H = \sup_{n} \left|\sum_{k=0}^{n} x_k \right|$$

Prove that $H$ is complete. Is $H$ a Hilbert space?

What is the best way to prove $H$ is complete?
To prove it's a Hilbert space, is it enough to prove that $\|.\|_H$ satisfies the parallellogram law?

Thanks in advance!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I have moved this topic from the Calculus sub-forum to our Analysis sub-forum as I feel you are more likely to get help here.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,701
Let [tex] H = \{(x_n)_n \subseteq \mathbb{R} | \sum_{n=1}^{\infty} x_n < \infty \}[/tex] and for $(x_n)_n \in H$ define
$$\|(x_n)_n\|_H = \sup_{n} \left|\sum_{k=0}^{n} x_k \right|$$

Prove that $H$ is complete. Is $H$ a Hilbert space?

What is the best way to prove $H$ is complete?
I suppose you should start by checking that $\|\,.\,\|_H$ is indeed a norm. Assuming that has been done, notice next that if $\|(x_n)\|_H < \varepsilon$ then (for each $n$) \(\displaystyle \biggl|\sum_{k=0}^{n} x_k \biggr| < \varepsilon\) and also \(\displaystyle \biggl|\sum_{k=0}^{n-1} x_k \biggr| < \varepsilon\). Take the difference and use the triangle inequality to see that $|x_n|<2\varepsilon.$

It follows that if $\{(x^{(m)}_n)\, \mid\, (m=1,2,\ldots\}$ is a Cauchy sequence for the $\|\,.\,\|_H$-norm, then for each fixed $n$ the sequence $(x^{(m)}_n)$ is Cauchy in $\mathbb{R}$ and hence converges to a limit $y_n$ say. In other words, $(x^{(m)}_n)$ converges coordinatewise to a sequence $(y_n).$ That gives you a candidate for the limit of the sequence. You then have to prove (i) that $(y_n)\in H$, and (ii) that $\|(x^{(m)}_n)-(y_n)\|_H \to0.$ (In other words, $(x^{(m)}_n) \to (y_n)$ in the $\|\,.\,\|_H$-norm and not just coordinatewise.)

You should probably model your proof on similar arguments that you may have seen for showing that spaces like $\ell^1(\mathbb{R})$ are complete.

To prove it's a Hilbert space, is it enough to prove that $\|.\|_H$ satisfies the parallellogram law?
To see whether or not $H$ satisfies the parallelogram law, try it out on some simple sequences, for example $(x_n) = (1,0,0,0,\ldots)$ and $(y_n) = (0,1,0,0,\ldots)$.
 

Impo

New member
Apr 2, 2013
17
Thanks for the help Opalg!


1. Let's check [tex](y_n)_n \in H[/tex], that is [tex]\sum_{n=1}^{\infty} y_n < \infty[/tex]. We have [tex]\sum_{n=1}^{\infty} y_n = \sum_{n=1}^{\infty} \lim_{m \to \infty} x_n^{(m)} = \lim_{m \to \infty} \sum_{n=1}^{\infty} x_n^{(m)} < \infty[/tex]. The last inequality follows from the fact that [tex]x_n^{(m)} \in H[/tex] ($\forall m$)

2. I'm not quite sure how I can work this out ...
 

TheBigBadBen

Active member
May 12, 2013
84
It is indeed sufficient to show that $\left\|\cdot\right\|_H$ satisfies the parallelogram law since we've already shown H to be a Banach (i.e. normed vector-) space. From there, we use the polarization identity to say that
$$\langle x, \ y \rangle = \frac{1}{4} \left(\|x + y \|^2 - \|x-y\|^2 \right)\ \forall \ x,y \in H$$
Gives us a valid inner product.

Try to approach this by extending the finite case. In other words, how would this work for the $\left\|\cdot\right\|_1$ norm over $\mathbb{R}^n$?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,701
1. Let's check [tex](y_n)_n \in H[/tex], that is [tex]\sum_{n=1}^{\infty} y_n < \infty[/tex]. We have [tex]\sum_{n=1}^{\infty} y_n = \color{red}{\sum_{n=1}^{\infty} \lim_{m \to \infty}} x_n^{(m)} = \color{red}{\lim_{m \to \infty} \sum_{n=1}^{\infty}} x_n^{(m)} < \infty[/tex]. The last inequality follows from the fact that [tex]x_n^{(m)} \in H[/tex] ($\forall m$)
You're playing with fire when you interchange limits like that! In general it is not legitimate to do so. What you need to do here is to replace the infinite sum by a finite sum so as avoid the interchange of limits. You are given a $\|\,.\,\|_H$-Cauchy sequence $(x^{(m)})$ in $H$, and a Cauchy sequence is necessarily bounded. Therefore there exists $K>0$ such that $\|x^{(m)}\|_H \leqslant K$ for all $m$, which means that \(\displaystyle \biggl|\sum_{k=1}^nx^{(m)}_k\biggr|\leqslant K\) for all $m$ and all $n$. Now let $m\to\infty$ in that inequality to see that \(\displaystyle \biggl|\sum_{k=1}^ny_k\biggr|\leqslant K\) (for all $n$). Hence $y\in H$, with $\|y\|_H\leqslant K$.

2. I'm not quite sure how I can work this out ...
Since the sequence $(x^{(m)})$ is Cauchy, given $\varepsilon>0$ there exists $M$ such that $\|x^{(m)} - x^{(r)}\|_H < \varepsilon$ whenever $m,r>M$. Thus \(\displaystyle \biggl|\sum_{k=1}^nx^{(m)}_k - x^{(r)}_k\biggr| < \varepsilon\) for all $n$. This is a finite sum, so we can safely let $r\to\infty$ to see that \(\displaystyle \biggl|\sum_{k=1}^nx^{(m)}_k - y_k\biggr| \leqslant \varepsilon\) for all $n$ (whenever $m>M$). Thus $\|x^{(m)} - y\|_H \leqslant \varepsilon$ for all $m>M$. That's enough to show that $x^{(m)}\to y$.

[In case it's not obvious, I have been using the abbreviation $x^{(m)}$ to denote the sequence $(x^{(m)}_n)_n$, and similarly $y$ for $(y_n)_n$.]
 
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