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

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The discussion centers on understanding the proof of Proposition 11.3.1 from Garling's book on metric and topological spaces. The main confusion arises from the treatment of the term $$\langle x, y \rangle^2$$ and its relationship to the complex inner product. It is clarified that Garling's proof uses the product of $$\langle x, y \rangle$$ and its complex conjugate, resulting in $$|\langle x, y \rangle|^2$$, rather than treating $$\langle x, y \rangle^2$$ as a standalone term. This distinction resolves the issue regarding the sign and value of the expression in the proof. Overall, the conversation emphasizes the importance of careful notation in complex inner product spaces.
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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
 

Attachments

  • Garling - Proposition 11.3.1... Cauchy-Schwartz Inequality.png
    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
    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
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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