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In Luenberger book Cauchy-Schwarz Inequality is defined like this: For all $x,y$ in an inner product space $|(x|y)| \le \|x\|\|y\|$. Equality holds if and only if $x = \lambda y$ or $y = \theta$.

Proof starts for all scalars $\lambda$,

$$ 0 \le (x-\lambda y | x-\lambda y) = (x|x) - \lambda(y|x) - \bar{\lambda}(x|y) + |\lambda|^2 (y|y) $$

I understand this expansion. But then, it will select a particular $\lambda = (x|y)/(y|y)$, and obtains

$$ 0 \le (x|x) - \frac{|(x|y)|^2}{(y|y)} $$

I dont understand how he chose that particular $\lambda$. I guess I understand why, he chose it to get rid of it in the main equation, but is it okay to chose any $\lambda$ that will clean up the equation like this?

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I find the notation in this question a bit hard to read. (Indeed, I think it has induced some typos.) Is that the notation from the source you cite? Using $|$ both as a separator in the notation for an inner product and for the standard modulus $|\cdot|$ makes things more difficult to parse than need be. I would suggest either $\langle \cdot,\cdot\rangle$ or, if you insist on parentheses, $(\cdot,\cdot)$. Cheers. –  cardinal Feb 7 '12 at 14:37
@cardinal It seems that the author of the book really uses this notation, see the Google Books preview: books.google.com/books?id=lZU0CAH4RccC&pg=PA47 –  Martin Sleziak Feb 7 '12 at 14:41
@Martin: (+1) Yikes. –  cardinal Feb 7 '12 at 14:43
Is the condition for equality to hold stated correctly in Luenberger's book? What the OP has posted above is what is in the book in a chapter titled Hilbert Space, but it is in a section titled Pre-Hilbert Spaces. I believe that in Hilbert space, $x$ need not equal $\lambda y$ for equality to hold but rather only the slightly weaker condition $||x - \lambda y|| = 0$ is needed. But perhaps the condition is intended for spaces in which $||z||=0$ exactly when $z$ is identically $0$, and the more general version is proved later. I don't have the book, only the couple of pages in the link. –  Dilip Sarwate Feb 7 '12 at 15:12
Aimed at physics students, I guess, where $(x|y)$ or $\langle x | y \rangle$ is the standard. –  GEdgar Feb 7 '12 at 15:31
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3 Answers 3

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I dont understand how he chose that particular $\lambda$. I guess I understand why, he chose it to get rid of it in the main equation, but is it okay to chose any $\lambda$ that will clean up the equation like this?

Yes, the inequality that you display holds for all $\lambda$ and so "it is okay to choose any $\lambda$" that you like. Luenberger's choice (it might well be the one used originally by Cauchy and/or Schwarz) "cleans up the equation" as you note, and provides motivation for its use. But if you have another value for $\lambda$ in mind that allows you to reach the conclusion $$0 \le (x|x) - \frac{|(x|y)|^2}{(y|y)},$$ (which is just a re-arrangement of the Cauchy-Schwarz Inequality), by all means, go for it.

See Appendix B of this Lecture Note for a more prolix proof of the Cauchy-Schwarz Inequality than the one in the Luenberger book.

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As said before, we always have $$0\leq\|x-\lambda y\|^2=(x-\lambda y|x-\lambda y)$$ Of course the choice $\lambda=\frac{(x|y)}{\|y\|^2}$ is neither arbitrary or accidental luck. Geometrically the specific $\lambda$ chosen is the one that gives the orthogonal projection of $x$ along the vector $y$.

The product $(x|y)$ removes the part of $x$ that is orthogonal to $y$, and dividing by $\|y\|^2$ normalizes $y$.

Consider the picture below where

  • the black arrow is to picture $x$

  • the blue arrow is to picture $y$

  • the red arrow is to picture $\lambda y$

  • the green arrow is to picture $x-\lambda y$


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@JonasMeyer Yes, of course. Thanks Jonas :) –  AD. Feb 12 '12 at 7:50
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You have a wrong idea. Actually, you don't need to select $\lambda$. For every $\lambda$,

$0 \le (x-\lambda y | x-\lambda y)$, that means

$0 \le (x|x)-\lambda(y|x)-\bar{\lambda}(x|y) + |\lambda|^2 (y|y)$,Consider it a function with the independent variable $\lambda$. So if $\lambda$ is R

$\Delta= 4(x|y)^2-4(x|x)(y|y) \le 0$. If $\lambda is C$, you need to use other functions.

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