Completeness of Vector Space

Jan 2010
32
1
Let \(\displaystyle H\) be the space of sequences \(\displaystyle (x(1),x(2),x(3),...)\) such that \(\displaystyle \sum^{\infty}_{j=1}\frac{|x(j)|^2}{j^2}<\infty\), with inner product \(\displaystyle <x,y>=\sum^{\infty}_{j=1}\frac{x(j)\overline{y(j)}}{j^2}\) and norm \(\displaystyle \|x\|=<x,x>^{1/2}\).

I need to show that \(\displaystyle H\) is complete with respect to the inner product i.e. a Hilbert Space. I am assuming that \(\displaystyle \mathbb{C}\) 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 \(\displaystyle x_{n}(j)\) but for this to be Cauchy \(\displaystyle |x_{n}(j)-x_{m}(j)|<\epsilon\). But I can't see how to show that, given that the inner product on \(\displaystyle H\) has a \(\displaystyle j^2\) below it. Any help would be great. Thanks
 
Aug 2009
228
80
Let \(\displaystyle H\) be the space of sequences \(\displaystyle (x(1),x(2),x(3),...)\) such that \(\displaystyle \sum^{\infty}_{j=1}\frac{|x(j)|^2}{j^2}<\infty\), with inner product \(\displaystyle <x,y>=\sum^{\infty}_{j=1}\frac{x(j)\overline{y(j)}}{j^2}\) and norm \(\displaystyle \|x\|=<x,x>^{1/2}\).

I need to show that \(\displaystyle H\) is complete with respect to the inner product i.e. a Hilbert Space. I am assuming that \(\displaystyle \mathbb{C}\) 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 \(\displaystyle x_{n}(j)\) but for this to be Cauchy \(\displaystyle |x_{n}(j)-x_{m}(j)|<\epsilon\). But I can't see how to show that, given that the inner product on \(\displaystyle H\) has a \(\displaystyle j^2\) below it. Any help would be great. Thanks
You can just get rid of the j. Suppose that \(\displaystyle x_n\) is Cauchy, then
\(\displaystyle
\sum\frac{|x_n(j)-x_m(j)|^2}{j^2} \rightarrow 0\)

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

which can only happen if the top part of the LHS goes to zero (I am taking m,n to infinity).
 
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Jan 2010
32
1
Thanks for the reply, okay so I was already assuming that \(\displaystyle x_n\) was Cauchy, the bit I was confused about was the inequality

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

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

My question is: If you have the \(\displaystyle \frac{1}{j}\) in the inequality, does that still satisfy the definition of Cauchy?
 
Aug 2009
228
80
Thanks for the reply, okay so I was already assuming that \(\displaystyle x_n\) was Cauchy, the bit I was confused about was the inequality

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

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

My question is: If you have the \(\displaystyle \frac{1}{j}\) 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 \(\displaystyle \frac{\epsilon}{j}\).

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;
\(\displaystyle
\frac{|x_n(j)-x_m(j)|^2}{j^2} \leq ||x_n-x_m||< \frac{\epsilon^2}{j^2}\)

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

i.e. x_n(j) is Cauchy.
 
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