# Completeness of Vector Space

#### ejgmath

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

#### Focus

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).

• ejgmath

#### ejgmath

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?

#### Focus

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.

• ejgmath