• I
• fsonnichsen
In summary: Therefore the Schwarz inequality is also called the Cauchy-Schwarz inequality.In summary, the conversation discusses the Schwarz inequality in Margenau and Murphy and how the sign in the usual "quadratic form" solution may be incorrect. The proof of the inequality for complex numbers in N dimensions is also briefly discussed. Overall, the conversation concludes that there may be a small error in the book and the proof of the inequality may not be intuitively obvious.

#### fsonnichsen

TL;DR Summary
From Margenau and Murphy--quadratic form use not clear
Looking at the proof of the Schwarz inequality in Margenau and Murphy, you will see what I attached. Gamma is asserted to be positive (OK). Given that the usual "quadratic form" solution would read "-(B+B*) .....". The sign does not seem correct to me as shown. In a fact B+B* = 2Re(B) and would be positive in this case given the integrals shown.
What am I missing here?

Thanks
Fritz

#### Attachments

• Margenau .jpg
33.2 KB · Views: 62
BvU
I think you found an error in the book. However, the values of ##\lambda## don't play a role in the argument (only the sign of the expression under the square root does), so I would call it a small error.

##\ ##

OK and thanks! The authors assumptions make sense but they must be evaluated carefully so I thought I may have missed something 49 years ago when I read the book the 1st time-it was quite a famous book back then.

I find the Cauchy-Schwarz ineq. for complex numbers in N dimensions somewhat tricky-the proof is not intuitively obvious to me upon brief examination.

Take care
Fritz

WWGD and BvU
Writing
$$\langle f|g \rangle=\int_{\mathbb{R}} \mathrm{d} x f^*(x) g(x),$$
you have from positive definiteness of the scalar product in ##L^2##
$$\langle f+\lambda g|f+\lambda g \rangle \geq 0,$$
and thus for all ##\lambda \in \mathbb{C}##
$$\langle f|f \rangle + \lambda^* \langle g|f \rangle+\lambda \langle f|g \rangle + |\lambda|^2 \langle g|g \rangle \geq 0.$$
Now set ##\lambda=-\langle g|f \rangle/\langle g|g \rangle,##
where we assume that ##g \neq 0## (otherwise the Schwarz inequality holds with the equality sign anyway). With this ##\lambda## the inequality reads
$$\langle f|f \rangle-\frac{|\langle f|g \rangle|^2}{\langle{g} | g\rangle} \geq 0.$$
This obviously is equivalent to
$$|\langle f|g \rangle| \leq \|f \| \|g \|, \quad \text{where} \quad \|f \|=\sqrt{\langle f|f \rangle}.$$
Further, due to the positive definiteness the equality sign holds if and only if there's a ##\lambda## such that ##|f \rangle+\lambda g \rangle=0##, i.e., if ##f \rangle## and ##|g \rangle## are linearly dependent.

BvU