- #1
1MileCrash
- 1,342
- 41
Homework Statement
Show that if x = (x1, x2,...) and y = (y1, y2,...) are members of l^2, then
[itex]\sum^{\infty}_{i=1} |x_{i}y_{i}|[/itex]
Converges
Homework Equations
My book defines l^2 to be:
[itex]{ x=(x_{1}, x_{2}, ... ) \in ℝ^{\omega} : \sum^{\infty}_{i=1} (x_{i})^{2} converges }[/itex]
(should be set brackets around that, don't know why they don't show up.)
The Attempt at a Solution
Concerns:
-I have no idea if Cauchy-Schwarts inequality actually holds in an infinite dimensional space, and can't find any information regarding whether it does or not.
-I can't think of any way to relate my last line to the expression I am showing convergence for.
-I have no idea if just because two series are convergent, then their product is convergent. I assumed it was anyway.
______________
Proof:
Since x and y are both members of l^2,
[itex] \sum^{\infty}_{i=1} (x_{i})^{2}[/itex]
and
[itex]\sum^{\infty}_{i=1} (y_{i})^{2}[/itex]
converge.
Then, the root of these series must converge too. The roots of these series are the norms of x and y.
So, ||x|| and ||y|| converge.
The Cauchy-Schwarts inequality states that
[itex]|x \cdot y| \leq ||x|| ||y||[/itex]
Since ||x|| ||y|| converges, so does |x (dot) y|
SCRATCH that!
I thought that was so because of the comparison test for convergent series, but the comparison test requires a comparison to be made between the terms of the series, not the series themselves...
So, I'm at square one.
Any help with this one?