MHB Polarization Formulae for Inner-Product Spaces ....

[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 with the polarization formula for the complex case ...

Garling's statement of the polarization formulae reads as follows:https://www.physicsforums.com/attachments/7914In the above text from Garling we read the following:" ... ... in the complex case we have the polarization formula $$\langle x,y \rangle = \frac{1}{4} \left( \sum_{ j = 0 }^3 i^j \| x + i^j y \|^2 \right)$$ ... ... "
Can someone please demonstrate how to prove that $$\langle x,y \rangle = \frac{1}{4} \left( \sum_{ j = 0 }^3 i^j \| x + i^j y \|^2 \right)$$ ...?Help will be appreciated ...

Peter
==========================================================================================***NOTE***

It may help readers of the above post to know Garling's notation and approach to inner-product spaces ... ... so I am providing the same ... as follows:
https://www.physicsforums.com/attachments/7915
https://www.physicsforums.com/attachments/7916
https://www.physicsforums.com/attachments/7917
https://www.physicsforums.com/attachments/7918Hope that helps ...

Peter

 

[Opalg]

Can someone please demonstrate how to prove that $$\langle x,y \rangle = \frac{1}{4} \left( \sum_{ j = 0 }^3 i^j \| x + i^j y \|^2 \right)$$ ...?

You just need to expand the four terms in that sum, remembering that (i) the square of the norm of a vector is equal to the inner product of the vector with itself, and (ii) in a complex vector space, the inner product is linear in the first variable and conjugate-linear in the second variable.

So for example the second term in the above sum (the term given by $j=1$) is $i\|x+iy\|^2 = i\langle x+iy,x+iy \rangle$. But $$\langle x+iy,x+iy \rangle = \langle x,x \rangle + \langle iy,x \rangle + \langle x,iy \rangle + \langle iy,iy \rangle = \langle x,x \rangle + i\langle y,x \rangle - i\langle x,y \rangle + \langle y,y \rangle.$$ So $$i\|x+iy\|^2 = i\langle x,x \rangle - \langle y,x \rangle + \langle x,y \rangle + i\langle y,y \rangle.$$ When you do that for all four terms in the sum $$\frac{1}{4} \left( \sum_{ j = 0 }^3 i^j \| x + i^j y \|^2 \right)$$, you will find that the coefficients of $ \langle x,x \rangle$, $\langle y,x\rangle$ and $\langle y,y \rangle$ all cancel out, and you are just left with $\langle x,y \rangle.$

 

[Math Amateur]

You just need to expand the four terms in that sum, remembering that (i) the square of the norm of a vector is equal to the inner product of the vector with itself, and (ii) in a complex vector space, the inner product is linear in the first variable and conjugate-linear in the second variable.

So for example the second term in the above sum (the term given by $j=1$) is $i\|x+iy\|^2 = i\langle x+iy,x+iy \rangle$. But $$\langle x+iy,x+iy \rangle = \langle x,x \rangle + \langle iy,x \rangle + \langle x,iy \rangle + \langle iy,iy \rangle = \langle x,x \rangle + i\langle y,x \rangle - i\langle x,y \rangle + \langle y,y \rangle.$$ So $$i\|x+iy\|^2 = i\langle x,x \rangle - \langle y,x \rangle + \langle x,y \rangle + i\langle y,y \rangle.$$ When you do that for all four terms in the sum $$\frac{1}{4} \left( \sum_{ j = 0 }^3 i^j \| x + i^j y \|^2 \right)$$, you will find that the coefficients of $ \langle x,x \rangle$, $\langle y,x\rangle$ and $\langle y,y \rangle$ all cancel out, and you are just left with $\langle x,y \rangle.$

Thanks Opalg ...

Just working through your post now ...

Thanks again,

Peter