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Baby Rudin Proof of Theorem 1.33 (e) - Triangle Inequality

  1. Jun 7, 2012 #1
    Hi everyone,

    I have a question on Rudin's proof of Theorem 1.33 part e. Here he prove the following statement:

    The absolute value of z+w is equal or smaller than the absolute value of z plus the absolute value of w -Yes, is the triangle inequality, where z and w are both complex numbers-
    |z+w| $\leqslant$ |z| + |w|

    In the proof, the key is that he points out that
    $$2Re(z\overline{w}) \leqslant 2|z\overline{w}|$$

    which obviously implies that
    $$Re(z\overline{w}) \leqslant |z\overline{w}|$$

    Why is that so? How does he knows this inequality is satified? If you can help me I would appreciate it a lot.
  2. jcsd
  3. Jun 7, 2012 #2
    It's from part (d)...
  4. Jun 7, 2012 #3
    I think I got it now! We know that

    $$|Re(x)| \leqslant |x|$$

    $$ [Re(x)Re(x)]^{1/2} \leqslant |x|$$

    $$ Re(x) \leqslant |x| $$ which is what we wanted to prove. Right?
  5. Jun 11, 2012 #4
    I think that you are slightly overthinking this, the real part of a complex number is just that - a real number.

    And the modulus of a real number is just its absolute value, and a real number is always less than its absolute value.
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