Infinite sum of squares converges

1. Feb 25, 2008

rjw5002

1. The problem statement, all variables and given/known data
So, L2 is defined to be the set of all infinite sequences of real numbers, s.t. the sum of their squares converges:
L2 = {x=(x1,...,xn,...) | $$\Sigma$$xi < $$\infty$$}
we have d(x,y) = $$\sqrt{\Sigma (xi-yi)^2}$$

I need to show that this is a metric, starting by showing that if $$xi,yi\in$$ then $$xi-yi\in$$
2. Relevant equations

definition of a metric

3. The attempt at a solution
so my initial problem was with the first step:
x = (x1,..., xn,....), y = (y1,...,yn,...), and then x - y = (x1 - y1, ...,xn-yn,...)
to show $$xi,yi\in$$, we must show that $$\Sigma (xi -yi) < \infty$$.
First, I broke up the squared quantity to get $$\Sigma (xi^2- 2xi*yi + yi^2)$$.
Carrying the sumation through, $$\Sigma xi^2 < \infty$$ (as for yi^2).
But where I get stuck is how to show or determine that $$\Sigma xi*yi < \infty$$.

So then the first three parts of the definition of a metric ($$d(p,p) = 0, p\rightarrow d(p,q) > 0,$$ and $$d(p,q) = d(q,p)$$) are easy enough to prove assuming $$xi-yi \in L2$$.
I get caught up again in proving that d(p,q) < d(p,r) + d(r,q).
I tried squaring both sides and distributing the summation, but found the resulting right hand side of the equation did not really simplify in a useful way. I'm not really sure where to go from there... any ideas for either of these roadblocks would be greatly appreciated. Thanks.

2. Feb 25, 2008

Dick

Your missing ingredients are various facets of the 'Cauchy-Schwarz inequality'. Look that up.