- #1

#### rjw5002

## Homework Statement

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,...) | [tex]\Sigma[/tex]xi < [tex]\infty[/tex]}

we have d(x,y) = [tex]\sqrt{\Sigma (xi-yi)^2}[/tex]

I need to show that this is a metric, starting by showing that if [tex]xi,yi\in[/tex] then [tex]xi-yi\in[/tex]

## Homework Equations

definition of a metric

## 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 [tex]xi,yi\in[/tex], we must show that [tex]\Sigma (xi -yi) < \infty[/tex].

First, I broke up the squared quantity to get [tex]\Sigma (xi^2- 2xi*yi + yi^2) [/tex].

Carrying the sumation through, [tex]\Sigma xi^2 < \infty [/tex] (as for yi^2).

But where I get stuck is how to show or determine that [tex]\Sigma xi*yi < \infty [/tex].

So then the first three parts of the definition of a metric ([tex]d(p,p) = 0, p\rightarrow d(p,q) > 0, [/tex] and [tex] d(p,q) = d(q,p)[/tex]) are easy enough to prove assuming [tex]xi-yi \in L2[/tex].

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.