1. ## Polarization Identity

I'm trying to derive the $L^2(\mathbb{R} )$ inner product using the polarization identity but I'm getting stuck. This is how far I get

$4(f,g) = \Vert f+g\Vert ^2 - \Vert f-g\Vert ^2
= \int \vert f+g\vert ^2dx -\int \vert f-g \vert ^2dx$

$= \int (f+g)(\bar{f} +\bar{g})-(f-g)(\bar{f} -\bar{g})dx
= 2\int \bar{f} g +f\bar{g}dx$

But I want to end up with $4\int f\bar{g} dx$. I assume I've gone wrong somewhere???

2. Originally Posted by markwolfson16900
I'm trying to derive the $L^2(\mathbb{R} )$ inner product using the polarization identity but I'm getting stuck. This is how far I get

$4(f,g) = \Vert f+g\Vert ^2 - \Vert f-g\Vert ^2
= \int \vert f+g\vert ^2dx -\int \vert f-g \vert ^2dx$

$= \int (f+g)(\bar{f} +\bar{g})-(f-g)(\bar{f} -\bar{g})dx
= 2\int \bar{f} g +f\bar{g}dx$

But I want to end up with $4\int f\bar{g} dx$. I assume I've gone wrong somewhere???
I think you're mixing up the real and complex versions of the polarisation identity. In the real case you don't need the bars over the f and g (because they are real!).

For complex functions you would use the complex polarisation identity $4(f,g) = \|f+g\| ^2 + i\|f+ig\| ^2 - \| f-g\| ^2 - i\|f-ig\| ^2$.

3. Ah of course. Thank you.