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Proof of Schwarz inequality in Baby Rudin (theorem 1.35)
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I'm having trouble understanding the following step in the proof of Theorem 1.35 (Schwarz inequality) in Baby Rudin. The general idea of the proof is clear, however on this line:
$sum{|Ba_j - Cb_j|^2} = sum{(Ba_j - Cb_j)(B overline{a_j} - overline{Cb_j})}$
From my understanding expanding the left-handed expression leads to:
$sum{ sqrt{(Ba_j - Cb_j)( overline{Ba_j - Cb_j})}^2} = sum{(Ba_j - Cb_j)(overline{Ba_j - Cb_j})} = sum{(Ba_j - Cb_j)(overline{Ba_j} - overline{Cb_j})}$
However, I don't understand why $overline{Ba_j} = Boverline{a_j}$, i.e, why the conjugate doesn't apply to $B$ in Rudin's proof.
In this proof $B = sum{|b_j|^2}$ and $C = sum{a_joverline{b_j}}$, where $a_1, ..., a_n$ and $b_1, ..., b_n$ are complex numbers and in the sums j runs over the values $1, ..., n$
real-analysis complex-analysis analysis complex-numbers
asked Jan 23 at 14:58
jlocksjlocks
1
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add a comment |
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I'm having trouble understanding the following step in the proof of Theorem 1.35 (Schwarz inequality) in Baby Rudin. The general idea of the proof is clear, however on this line:
$sum{|Ba_j - Cb_j|^2} = sum{(Ba_j - Cb_j)(B overline{a_j} - overline{Cb_j})}$
From my understanding expanding the left-handed expression leads to:
$sum{ sqrt{(Ba_j - Cb_j)( overline{Ba_j - Cb_j})}^2} = sum{(Ba_j - Cb_j)(overline{Ba_j - Cb_j})} = sum{(Ba_j - Cb_j)(overline{Ba_j} - overline{Cb_j})}$
However, I don't understand why $overline{Ba_j} = Boverline{a_j}$, i.e, why the conjugate doesn't apply to $B$ in Rudin's proof.
In this proof $B = sum{|b_j|^2}$ and $C = sum{a_joverline{b_j}}$, where $a_1, ..., a_n$ and $b_1, ..., b_n$ are complex numbers and in the sums j runs over the values $1, ..., n$
real-analysis complex-analysis analysis complex-numbers
asked Jan 23 at 14:58
jlocksjlocks
1
$endgroup$
1
$begingroup$
$B$ is a real number, so $overline B=B$.
$endgroup$
– Aweygan
Jan 23 at 15:03
add a comment |
$begingroup$
I'm having trouble understanding the following step in the proof of Theorem 1.35 (Schwarz inequality) in Baby Rudin. The general idea of the proof is clear, however on this line:
$sum{|Ba_j - Cb_j|^2} = sum{(Ba_j - Cb_j)(B overline{a_j} - overline{Cb_j})}$
From my understanding expanding the left-handed expression leads to:
$sum{ sqrt{(Ba_j - Cb_j)( overline{Ba_j - Cb_j})}^2} = sum{(Ba_j - Cb_j)(overline{Ba_j - Cb_j})} = sum{(Ba_j - Cb_j)(overline{Ba_j} - overline{Cb_j})}$
However, I don't understand why $overline{Ba_j} = Boverline{a_j}$, i.e, why the conjugate doesn't apply to $B$ in Rudin's proof.
In this proof $B = sum{|b_j|^2}$ and $C = sum{a_joverline{b_j}}$, where $a_1, ..., a_n$ and $b_1, ..., b_n$ are complex numbers and in the sums j runs over the values $1, ..., n$
real-analysis complex-analysis analysis complex-numbers
asked Jan 23 at 14:58
jlocksjlocks
1
$endgroup$
I'm having trouble understanding the following step in the proof of Theorem 1.35 (Schwarz inequality) in Baby Rudin. The general idea of the proof is clear, however on this line:
$sum{|Ba_j - Cb_j|^2} = sum{(Ba_j - Cb_j)(B overline{a_j} - overline{Cb_j})}$
From my understanding expanding the left-handed expression leads to:
$sum{ sqrt{(Ba_j - Cb_j)( overline{Ba_j - Cb_j})}^2} = sum{(Ba_j - Cb_j)(overline{Ba_j - Cb_j})} = sum{(Ba_j - Cb_j)(overline{Ba_j} - overline{Cb_j})}$
However, I don't understand why $overline{Ba_j} = Boverline{a_j}$, i.e, why the conjugate doesn't apply to $B$ in Rudin's proof.
In this proof $B = sum{|b_j|^2}$ and $C = sum{a_joverline{b_j}}$, where $a_1, ..., a_n$ and $b_1, ..., b_n$ are complex numbers and in the sums j runs over the values $1, ..., n$
real-analysis complex-analysis analysis complex-numbers
real-analysis complex-analysis analysis complex-numbers
asked Jan 23 at 14:58
jlocksjlocks
1
asked Jan 23 at 14:58
jlocksjlocks
1
asked Jan 23 at 14:58
jlocksjlocks
1
asked Jan 23 at 14:58
jlocksjlocks
1
asked Jan 23 at 14:58
jlocksjlocks
1
1
1
$begingroup$
$B$ is a real number, so $overline B=B$.
$endgroup$
– Aweygan
Jan 23 at 15:03
add a comment |
1
$begingroup$
$B$ is a real number, so $overline B=B$.
$endgroup$
– Aweygan
Jan 23 at 15:03
1
1
$begingroup$
$B$ is a real number, so $overline B=B$.
$endgroup$
– Aweygan
Jan 23 at 15:03
$begingroup$
$B$ is a real number, so $overline B=B$.
$endgroup$
– Aweygan
Jan 23 at 15:03
add a comment |
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$begingroup$
$B$ is a real number, so $overline B=B$.
$endgroup$
– Aweygan
Jan 23 at 15:03