Convergence of xy Product in $\ell^2$

Click For Summary
SUMMARY

The discussion centers on proving that if x and y are elements of the space $\ell^2$, then the series $\sum_{i=1}^{\infty} |x_i y_i|$ converges. Participants confirm that since both $\sum_{i=1}^{\infty} (x_i)^2$ and $\sum_{i=1}^{\infty} (y_i)^2$ converge, it follows that the product series also converges. The Cauchy-Schwarz inequality is suggested as a more rigorous approach to solidify the proof.

PREREQUISITES
  • Understanding of $\ell^2$ space and its properties
  • Familiarity with series convergence criteria
  • Knowledge of the Cauchy-Schwarz inequality
  • Basic concepts of mathematical proofs in analysis
NEXT STEPS
  • Study the Cauchy-Schwarz inequality in detail
  • Explore convergence tests for series in functional analysis
  • Review properties of $\ell^p$ spaces
  • Practice proving convergence of product series using various methods
USEFUL FOR

Mathematics students, particularly those studying functional analysis, and anyone interested in series convergence and properties of $\ell^2$ spaces.

tylerc1991
Messages
158
Reaction score
0

Homework Statement



Show that if x=(x1, x2,...) and y=(y1, y2,...) \in \ell2 then The sum from i=1 to infinity of |xiyi| converges

Homework Equations





The Attempt at a Solution



since x and y are elements of l^2 then The sum from i=1 to infinity of (xi)2 and (yi)2 both converge, which implies that the sum of the product xy converges

Is this right? and more importantly, is it enough? Thank you for your help.
 
Physics news on Phys.org
tylerc1991 said:

Homework Statement



Show that if x=(x1, x2,...) and y=(y1, y2,...) \in \ell2 then The sum from i=1 to infinity of |xiyi| converges

Homework Equations




The Attempt at a Solution



since x and y are elements of l^2 then The sum from i=1 to infinity of (xi)2 and (yi)2 both converge, which implies that the sum of the product xy converges

Is this right? and more importantly, is it enough? Thank you for your help.

Restating the problem is hardly a proof. Think about using Cauchy-Schwarz.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K