Proof of the Cauchy-Schwarz Iequality .... Garling, Proposition 11.3.1 .... ....

[Math Amateur]

I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable" ... ...

I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...

I need some help to fully understand the proof of Proposition 11.3.1 ...

Garling's statement and proof of Proposition 11.3.1 reads as follows:View attachment 8956I need help with exactly how Garling concluded that with his substitution for \(\displaystyle \lambda\) ... we have

\(\displaystyle \lambda^2 \langle x, y \rangle = \frac{ \langle x, y \rangle^2 }{ \mid \langle x, y \rangle \mid^2 } \frac{ \| x \|^2 }{ \| y \|^2 } \| y \|^2 = \| x \|^2\) ... ... My problem is what sign (plus or minus) and value do we give to \(\displaystyle \langle x, y \rangle^2\) ... Garling seems to treat \(\displaystyle \langle x, y \rangle^2\) as if it were equal to \(\displaystyle \mid \langle x, y \rangle \mid^2\) ... and cancels with the denominator ... or so it seems ...?

But ... \(\displaystyle \langle x, y \rangle\) is a complex number, say \(\displaystyle z\) ... and so we are dealing with a complex number \(\displaystyle z^2 = \langle x, y \rangle^2\) ... and, of course, \(\displaystyle z^2\) is neither positive or negative ... ... ? ... so how do we end up with

\(\displaystyle \lambda^2 \langle x, y \rangle = \| x \|^2\)Hope that someone can help ...

Peter=========================================================================================It may help readers of the above post to have access to Garling's introduction to inner product spaces where he gives the relevant definitions and notation ... so I am providing access to the relevant text as follows:
View attachment 8957
View attachment 8958
Hope that helps ...

Peter

 

[Opalg]

My problem is what sign (plus or minus) and value do we give to \(\displaystyle \langle x, y \rangle^2\) .

In fact, \(\displaystyle \langle x, y \rangle^2\) does not occur anywhere in Garling's proof. The proof contains the terms $\overline{\lambda}\langle x, y \rangle$ and $\lambda\langle y, x \rangle$. In each of those cases, $\langle x, y \rangle$ gets multiplied by its complex conjugate $\langle y,x \rangle$, so that the product is equal to $|\langle x, y \rangle|^2$.

 

[Math Amateur]

In fact, \(\displaystyle \langle x, y \rangle^2\) does not occur anywhere in Garling's proof. The proof contains the terms $\overline{\lambda}\langle x, y \rangle$ and $\lambda\langle y, x \rangle$. In each of those cases, $\langle x, y \rangle$ gets multiplied by its complex conjugate $\langle y,x \rangle$, so that the product is equal to $|\langle x, y \rangle|^2$.

Hmmm ... yes ... you're right of course...

I should have been more careful when I was writing out the details of the proof ...

Thanks for the help ...

Peter