Valid proof of Cauchy-Schwarz inequality?

  • Thread starter phosgene
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  • #1
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Homework Statement



I was discussing the proof for the Cauchy-Schwarz inequality used in our lectures, and another student suggested an easier way of doing it. It's really, really simple. But I haven't seen it anywhere online or in textbooks, so I'm wondering if it's either wrong or is only valid in certain scenarios.

The quick version is this:

[itex]u \bullet v = ||u|| ||v|| cosθ[/itex] where θ is between 0 and [itex]\pi[/itex].

The upper and lower bounds of |cosθ| are 1 and 0, respectively, so [itex]|u \bullet v | \leq ||u|| ||v||[/itex]
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



I was discussing the proof for the Cauchy-Schwarz inequality used in our lectures, and another student suggested an easier way of doing it. It's really, really simple. But I haven't seen it anywhere online or in textbooks, so I'm wondering if it's either wrong or is only valid in certain scenarios.

The quick version is this:

[itex]u \bullet v = ||u|| ||v|| cosθ[/itex] where θ is between 0 and [itex]\pi[/itex].

The upper and lower bounds of |cosθ| are 1 and 0, respectively, so [itex]|u \bullet v | \leq ||u|| ||v||[/itex]

You need to tell us what is the definition of inner product that you start with. If it is ##u \cdot v = \sum_i u_i v_i,## then you first need to prove that ##u \cdot v = |u| |v| \cos \theta.## Actually, it is sometimes done in the opposite way: from ##u \cdot v = \sum_i u_i v_i,##, one proves the C_S inequality, and then notes that one can *define* ##\theta## from
[tex]\cos \theta =\frac{u \cdot v}{|u| |v|}.[/tex]
 
  • #3
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Thanks for the reply, so it seems that the proof in my original post would probably not be a good one to use in an exam? The course in question is multivariable and complex calculus. The proof that I've used myself relies on algebraically manipulating the square of the length of the difference of two vectors ( [itex] ||a \widehat{x} - b \widehat{y} ||^2 \geq 0 [/itex] ) using the properties of the dot product, rather than the cosine argument. It's the proof I've seen in the textbooks I have, but it's a bit longer to write out.
 

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