Polarization Identity

**markwolfson16900** · May 20th 2010, 12:42 PM

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

$4(f,g) = \Vert f+g\Vert ^2 - \Vert f-g\Vert ^2 <br /> = \int \vert f+g\vert ^2dx -\int \vert f-g \vert ^2dx$
$= \int (f+g)(\bar{f} +\bar{g})-(f-g)(\bar{f} -\bar{g})dx<br /> = 2\int \bar{f} g +f\bar{g}dx$

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

**Opalg** · May 20th 2010, 12:49 PM

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

$4(f,g) = \Vert f+g\Vert ^2 - \Vert f-g\Vert ^2 <br /> = \int \vert f+g\vert ^2dx -\int \vert f-g \vert ^2dx$
$= \int (f+g)(\bar{f} +\bar{g})-(f-g)(\bar{f} -\bar{g})dx<br /> = 2\int \bar{f} g +f\bar{g}dx$

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 $4(f,g) = \|f+g\| ^2 + i\|f+ig\| ^2 - \| f-g\| ^2 - i\|f-ig\| ^2$ .

**markwolfson16900** · May 20th 2010, 01:30 PM

Ah of course. Thank you.

Thread: Polarization Identity

LinkBack

Thread Tools

Display

Polarization Identity

Similar Math Help Forum Discussions

tan identity

What identity should I use?

another identity

identity

Polarization online hw help

Search Tags