# Completeness of Vector Space

• May 20th 2010, 08:22 AM
ejgmath
Completeness of Vector Space
Let $H$ be the space of sequences $(x(1),x(2),x(3),...)$ such that $\sum^{\infty}_{j=1}\frac{|x(j)|^2}{j^2}<\infty$, with inner product $=\sum^{\infty}_{j=1}\frac{x(j)\overline{y(j)} }{j^2}$ and norm $\|x\|=^{1/2}$.

I need to show that $H$ is complete with respect to the inner product i.e. a Hilbert Space. I am assuming that $\mathbb{C}$ is complete.

I know that I need to show that every Cauchy sequence in H is convergent. I can see that this follows relatively easily from the definition of H.

So I have taken a sequence $x_{n}(j)$ but for this to be Cauchy $|x_{n}(j)-x_{m}(j)|<\epsilon$. But I can't see how to show that, given that the inner product on $H$ has a $j^2$ below it. Any help would be great. Thanks
• May 20th 2010, 12:46 PM
Focus
Quote:

Originally Posted by ejgmath
Let $H$ be the space of sequences $(x(1),x(2),x(3),...)$ such that $\sum^{\infty}_{j=1}\frac{|x(j)|^2}{j^2}<\infty$, with inner product $=\sum^{\infty}_{j=1}\frac{x(j)\overline{y(j)} }{j^2}$ and norm $\|x\|=^{1/2}$.

I need to show that $H$ is complete with respect to the inner product i.e. a Hilbert Space. I am assuming that $\mathbb{C}$ is complete.

I know that I need to show that every Cauchy sequence in H is convergent. I can see that this follows relatively easily from the definition of H.

So I have taken a sequence $x_{n}(j)$ but for this to be Cauchy $|x_{n}(j)-x_{m}(j)|<\epsilon$. But I can't see how to show that, given that the inner product on $H$ has a $j^2$ below it. Any help would be great. Thanks

You can just get rid of the j. Suppose that $x_n$ is Cauchy, then
$
\sum\frac{|x_n(j)-x_m(j)|^2}{j^2} \rightarrow 0$

In particular for each j
$
\frac{|x_n(j)-x_m(j)|^2}{j^2} \leq ||x_n-x_m|| \rightarrow 0$

which can only happen if the top part of the LHS goes to zero (I am taking m,n to infinity).
• May 20th 2010, 06:15 PM
ejgmath
Thanks for the reply, okay so I was already assuming that $x_n$ was Cauchy, the bit I was confused about was the inequality

$\frac{|x_n(j)-x_m(j)|}{j}\leq\|x_n-x_m\|<\epsilon$

Which is need to show that $x_n(j)$ is also Cauchy, for each j.

My question is: If you have the $\frac{1}{j}$ in the inequality, does that still satisfy the definition of Cauchy?
• May 20th 2010, 06:27 PM
Focus
Quote:

Originally Posted by ejgmath
Thanks for the reply, okay so I was already assuming that $x_n$ was Cauchy, the bit I was confused about was the inequality

$\frac{|x_n(j)-x_m(j)|}{j}\leq\|x_n-x_m\|<\epsilon$

Which is need to show that $x_n(j)$ is also Cauchy, for each j.

My question is: If you have the $\frac{1}{j}$ in the inequality, does that still satisfy the definition of Cauchy?

Why would it? You are considering j fixed, so if you really are worried about it just pick $\frac{\epsilon}{j}$.

A formal way would be to say, fix j, and let epsilon be greater than zero, then there exists an N such that for all n,m>N;
$
\frac{|x_n(j)-x_m(j)|^2}{j^2} \leq ||x_n-x_m||< \frac{\epsilon^2}{j^2}$

Thus for n,m>N
$
|x_n(j)-x_m(j)|< \epsilon$

i.e. x_n(j) is Cauchy.