- #1
e(ho0n3
- 1,357
- 0
[SOLVED] Consequence of Schwarz Inequality
Use the Schwarz inequality to demonstrate the following inequality:
[tex]\left(\sum_{j=1}^n |a_j + b_j|^2\right)^{1/2} \le \left(\sum_{j=1}^n |a_j|^2\right)^{1/2} + \left(\sum_{j=1}^n |b_j|^2\right)^{1/2}[/tex]
The Schwarz inequality:
[tex]\left|\sum_{j=1}^n a_j \bar{b}_j\right|^2 \le \left(\sum_{j=1}^n |a_j|^2\right) \left(\sum_{j=1}^n |b_j|^2\right)[/tex]
Expanding the term on the LHS, squaring both sides and simplifying the inequality of the problem produces:
[tex]\sum_{j=1}^n \Re(a_j\bar{b}_j) \le \left(\sum_{j=1}^n |a_j|^2\right) \left(\sum_{j=1}^n |b_j|^2\right)[/tex]
At this point, I use the Schwarz inequality: If I can show that
[tex]\sum_{j=1}^n \Re(a_j\bar{b}_j) \le \left|\sum_{j=1}^n a_j \bar{b}_j\right|^2[/tex]
then the problem is solved. [itex]a_j\bar{b}_j = \Re(a_j\bar{b}_j) + i \Im(a_j\bar{b}_j)[/itex]. Let [itex]\alpha[/itex] be
[tex]\sum_{j=1}^n \Re(a_j\bar{b}_j)[/tex]
and let [itex]\beta[/itex] be
[tex]\sum_{j=1}^n \Im(a_j\bar{b}_j)[/tex]
Then [itex]\alpha^2 \le |\alpha + i\beta|^2[/itex]. I would like to think that [itex]\alpha \le \alpha^2[/itex], but this only works if [itex]\alpha \ge 1[/itex] which is not necessarily true. What can I do?
Homework Statement
Use the Schwarz inequality to demonstrate the following inequality:
[tex]\left(\sum_{j=1}^n |a_j + b_j|^2\right)^{1/2} \le \left(\sum_{j=1}^n |a_j|^2\right)^{1/2} + \left(\sum_{j=1}^n |b_j|^2\right)^{1/2}[/tex]
Homework Equations
The Schwarz inequality:
[tex]\left|\sum_{j=1}^n a_j \bar{b}_j\right|^2 \le \left(\sum_{j=1}^n |a_j|^2\right) \left(\sum_{j=1}^n |b_j|^2\right)[/tex]
The Attempt at a Solution
Expanding the term on the LHS, squaring both sides and simplifying the inequality of the problem produces:
[tex]\sum_{j=1}^n \Re(a_j\bar{b}_j) \le \left(\sum_{j=1}^n |a_j|^2\right) \left(\sum_{j=1}^n |b_j|^2\right)[/tex]
At this point, I use the Schwarz inequality: If I can show that
[tex]\sum_{j=1}^n \Re(a_j\bar{b}_j) \le \left|\sum_{j=1}^n a_j \bar{b}_j\right|^2[/tex]
then the problem is solved. [itex]a_j\bar{b}_j = \Re(a_j\bar{b}_j) + i \Im(a_j\bar{b}_j)[/itex]. Let [itex]\alpha[/itex] be
[tex]\sum_{j=1}^n \Re(a_j\bar{b}_j)[/tex]
and let [itex]\beta[/itex] be
[tex]\sum_{j=1}^n \Im(a_j\bar{b}_j)[/tex]
Then [itex]\alpha^2 \le |\alpha + i\beta|^2[/itex]. I would like to think that [itex]\alpha \le \alpha^2[/itex], but this only works if [itex]\alpha \ge 1[/itex] which is not necessarily true. What can I do?