- #1
joecharland
- 3
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I have worked my way though the proof of the Cauchy Schwarz inequality in Rudin but I am struggling to understand how one could have arrived at that proof in the first place. The essence of the proof is that this sum:
##\sum |B a_j - C b_j|^2##
is shown to be equivalent to the following expression:
##B(AB - |C|^2)##
Now since each term of the first sum is positive, it is clearly greater than or equal to zero, so that the expression $$B(AB - |C|^2)$$ is also greater than or equal to zero. Now if $$B = 0$$ the theorem is trivial, so assume that $$B \geq 0$$ and then the inequality $$B(AB - |C|^2) \geq 0$$ implies that $$AB - |C|^2 \geq 0$$ which is the theorem.
Now naturally what I want to understand is how to arrive at this proof in the first place. Some intuition to start with is that if $$AB - |C|^2$$ can be made equivalent to a single sum, each term of which is nonnegative, this would give the desired result. But Rudin added a step to this, by showing that $$B(AB - |C|^2)$$ can be made equivalent to a single sum and then the B can be canceled out. What train of thought would have led Rudin to this proof?
There is an explanation offered here:
http://math.berkeley.edu/~gbergman/ug.hndts/06x2+03F_104_q+a.txt
But I am still struggling to figure out that explanation too. Can anyone either help or direct me to a useful resource?
Thanks!
##\sum |B a_j - C b_j|^2##
is shown to be equivalent to the following expression:
##B(AB - |C|^2)##
Now since each term of the first sum is positive, it is clearly greater than or equal to zero, so that the expression $$B(AB - |C|^2)$$ is also greater than or equal to zero. Now if $$B = 0$$ the theorem is trivial, so assume that $$B \geq 0$$ and then the inequality $$B(AB - |C|^2) \geq 0$$ implies that $$AB - |C|^2 \geq 0$$ which is the theorem.
Now naturally what I want to understand is how to arrive at this proof in the first place. Some intuition to start with is that if $$AB - |C|^2$$ can be made equivalent to a single sum, each term of which is nonnegative, this would give the desired result. But Rudin added a step to this, by showing that $$B(AB - |C|^2)$$ can be made equivalent to a single sum and then the B can be canceled out. What train of thought would have led Rudin to this proof?
There is an explanation offered here:
http://math.berkeley.edu/~gbergman/ug.hndts/06x2+03F_104_q+a.txt
But I am still struggling to figure out that explanation too. Can anyone either help or direct me to a useful resource?
Thanks!