- #1
Tsunoyukami
- 215
- 11
Here's yet another assigned problem that I'm having difficulty with. I think I'm close to the end but am "nervous" (for lack of a better word) about whether or not I have used summation notation properly throughout the problem. Here it is:
"Use the Schwarz inequality to establish that
[itex]( \sum |a_{j} + b_{j}|^{2})^{1/2} \leq (\sum |a_{j}|^{2})^{1/2} + (\sum |b_{j}|^{2})^{1/2}[/itex]" (From Complex Variables, 2nd Edition by Fisher, pg. 10; #24)
The Schwarz Inequality is:
[itex]| \sum a_{j}\overline{b_{j}}|^{2} \leq (\sum |a_{j}|^{2})(\sum |b_{j}|^{2})[/itex]
I will now describe my approach without writing out each step explicitly (I'm finding summation notation tedious to write for LaTeX and this is for a graded assignment so I'd rather not have the entire answer floating on the web). If anything I say is unclear or you require more details let me know and I can fill it out.
Also, please note that the sums in both the above inequalities are from j=1 to j=n. I was unsure how to code that properly in LaTeX.
I squared both side of the given inequality and expanded the left-hand side. After some algebra I obtained the following expression:
[itex](\sum Re(a_{j}\overline{b_{j}}))^{2} \leq (\sum |a_{j}|^{2})(\sum |b_{j}|^{2})[/itex]
The right-hand sides of this expression and the Schwarz Inequality are exactly equal. I argue that this inequality is true if it's left-hand side is less than or equal to the left-hand side of the Schwarz Inequality, that is:
[itex](\sum Re(a_{j}\overline{b_{j}}))^{2} \leq | \sum a_{j}\overline{b_{j}}|^{2} [/itex]
To me this is obvious when j = 1. Let [itex]w_{j} = a_{j}\overline{b_{j}}[/itex], and can be written in the form [itex]w = x + iy[/itex], then:
[itex]\sum Re(w_{j}) \leq | \sum w_{j}| [/itex]
[itex]Re(w_{j}) \leq | w_{j}| [/itex]
[itex]x \leq (x^{2} + y^{2})^{1/2} [/itex]
[itex]x^{2} \leq (x^{2} + y^{2}) [/itex]
[itex]0 \leq y^{2}[/itex]
So I am done. But how can I extend this to the case when [itex]j \neq 1[/itex]?
If anything I've described above seems incorrect please let me know and I can write out the steps I took explicitly to see where I went wrong. Thanks a lot in advance for any assistance! :)
"Use the Schwarz inequality to establish that
[itex]( \sum |a_{j} + b_{j}|^{2})^{1/2} \leq (\sum |a_{j}|^{2})^{1/2} + (\sum |b_{j}|^{2})^{1/2}[/itex]" (From Complex Variables, 2nd Edition by Fisher, pg. 10; #24)
The Schwarz Inequality is:
[itex]| \sum a_{j}\overline{b_{j}}|^{2} \leq (\sum |a_{j}|^{2})(\sum |b_{j}|^{2})[/itex]
I will now describe my approach without writing out each step explicitly (I'm finding summation notation tedious to write for LaTeX and this is for a graded assignment so I'd rather not have the entire answer floating on the web). If anything I say is unclear or you require more details let me know and I can fill it out.
Also, please note that the sums in both the above inequalities are from j=1 to j=n. I was unsure how to code that properly in LaTeX.
I squared both side of the given inequality and expanded the left-hand side. After some algebra I obtained the following expression:
[itex](\sum Re(a_{j}\overline{b_{j}}))^{2} \leq (\sum |a_{j}|^{2})(\sum |b_{j}|^{2})[/itex]
The right-hand sides of this expression and the Schwarz Inequality are exactly equal. I argue that this inequality is true if it's left-hand side is less than or equal to the left-hand side of the Schwarz Inequality, that is:
[itex](\sum Re(a_{j}\overline{b_{j}}))^{2} \leq | \sum a_{j}\overline{b_{j}}|^{2} [/itex]
To me this is obvious when j = 1. Let [itex]w_{j} = a_{j}\overline{b_{j}}[/itex], and can be written in the form [itex]w = x + iy[/itex], then:
[itex]\sum Re(w_{j}) \leq | \sum w_{j}| [/itex]
[itex]Re(w_{j}) \leq | w_{j}| [/itex]
[itex]x \leq (x^{2} + y^{2})^{1/2} [/itex]
[itex]x^{2} \leq (x^{2} + y^{2}) [/itex]
[itex]0 \leq y^{2}[/itex]
So I am done. But how can I extend this to the case when [itex]j \neq 1[/itex]?
If anything I've described above seems incorrect please let me know and I can write out the steps I took explicitly to see where I went wrong. Thanks a lot in advance for any assistance! :)