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

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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 $$\lambda$$ ... we have

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

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

$$\lambda^2 \langle x, y \rangle = \| x \|^2$$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
 

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  • Garling - Proposition 11.3.1... Cauchy-Schwartz Inequality.png
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  • Garling - 1 - Start of Secton 11.3 on Inner Product Spaces ... .PAGE 1 .png
    Garling - 1 - Start of Secton 11.3 on Inner Product Spaces ... .PAGE 1 .png
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    Garling - 2 - Start of Secton 11.3 on Inner Product Spaces ... .PAGE 2 .png
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Peter said:
My problem is what sign (plus or minus) and value do we give to $$\langle x, y \rangle^2$$ .
In fact, $$\langle x, y \rangle^2$$ 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 $|\langle x, y \rangle|^2$.
 
Opalg said:
In fact, $$\langle x, y \rangle^2$$ 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
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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