Is the integral form of the polarisation identity useful for anything?
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It is well-known that the polarisation identity for real vector spaces is
$$ langle a,b rangle =frac{1}{4}sum_{k=0}^1 (-1)^klVert a+(-1)^k b rVert^2, $$
and the complex generalisation is
$$ langle a,b rangle =frac{1}{4}sum_{k=0}^3 i^{-k} lVert a+i^k b rVert^2. $$
There are two generalisations of this: take $omega$ a complex primitive $n$th root of unity. Then
$$ langle a,b rangle =frac{1}{n}sum_{k=0}^n omega^{-k}lVert a+omega^k b rVert^2, $$
because $ frac{1}{n}sum_{k=0}^n omega^{mk} = 0 $ unless $m=0$, when it's $1$. (Note we need to be careful with $n=2$, which is not enough to extract the imaginary part of the inner product: we instead have
$$ frac{1}{2}sum_{k=0}^1 (-1)^{k} lVert a+(-1)^k b rVert^2 = langle a,b rangle + langle b,a rangle = 2Relangle a,b rangle.) $$
But there's another generalisation: using Fourier series, we have
$$ langle a,b rangle = frac{1}{2pi}int_0^{2pi} e^{-itheta}lVert a+e^{itheta} b rVert^2 , dtheta left( = frac{1}{2pi i}int_{|z|=1} frac{lVert a+zb rVert^2}{z^2} , dz right) $$
My question is: is this just a pretty identity, or are there situations where this is the "right"/nicest form to use in proofs requiring a polarisation identity? Obviously it adds no new mathematical content, since the $n=4$ (indeed, $n=3$) case is sufficient to determine the inner product.
integration hilbert-spaces norm
asked Oct 13 '15 at 9:43
ChappersChappers
55.8k74192
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add a comment |
$begingroup$
It is well-known that the polarisation identity for real vector spaces is
$$ langle a,b rangle =frac{1}{4}sum_{k=0}^1 (-1)^klVert a+(-1)^k b rVert^2, $$
and the complex generalisation is
$$ langle a,b rangle =frac{1}{4}sum_{k=0}^3 i^{-k} lVert a+i^k b rVert^2. $$
There are two generalisations of this: take $omega$ a complex primitive $n$th root of unity. Then
$$ langle a,b rangle =frac{1}{n}sum_{k=0}^n omega^{-k}lVert a+omega^k b rVert^2, $$
because $ frac{1}{n}sum_{k=0}^n omega^{mk} = 0 $ unless $m=0$, when it's $1$. (Note we need to be careful with $n=2$, which is not enough to extract the imaginary part of the inner product: we instead have
$$ frac{1}{2}sum_{k=0}^1 (-1)^{k} lVert a+(-1)^k b rVert^2 = langle a,b rangle + langle b,a rangle = 2Relangle a,b rangle.) $$
But there's another generalisation: using Fourier series, we have
$$ langle a,b rangle = frac{1}{2pi}int_0^{2pi} e^{-itheta}lVert a+e^{itheta} b rVert^2 , dtheta left( = frac{1}{2pi i}int_{|z|=1} frac{lVert a+zb rVert^2}{z^2} , dz right) $$
My question is: is this just a pretty identity, or are there situations where this is the "right"/nicest form to use in proofs requiring a polarisation identity? Obviously it adds no new mathematical content, since the $n=4$ (indeed, $n=3$) case is sufficient to determine the inner product.
integration hilbert-spaces norm
asked Oct 13 '15 at 9:43
ChappersChappers
55.8k74192
$endgroup$
add a comment |
$begingroup$
It is well-known that the polarisation identity for real vector spaces is
$$ langle a,b rangle =frac{1}{4}sum_{k=0}^1 (-1)^klVert a+(-1)^k b rVert^2, $$
and the complex generalisation is
$$ langle a,b rangle =frac{1}{4}sum_{k=0}^3 i^{-k} lVert a+i^k b rVert^2. $$
There are two generalisations of this: take $omega$ a complex primitive $n$th root of unity. Then
$$ langle a,b rangle =frac{1}{n}sum_{k=0}^n omega^{-k}lVert a+omega^k b rVert^2, $$
because $ frac{1}{n}sum_{k=0}^n omega^{mk} = 0 $ unless $m=0$, when it's $1$. (Note we need to be careful with $n=2$, which is not enough to extract the imaginary part of the inner product: we instead have
$$ frac{1}{2}sum_{k=0}^1 (-1)^{k} lVert a+(-1)^k b rVert^2 = langle a,b rangle + langle b,a rangle = 2Relangle a,b rangle.) $$
But there's another generalisation: using Fourier series, we have
$$ langle a,b rangle = frac{1}{2pi}int_0^{2pi} e^{-itheta}lVert a+e^{itheta} b rVert^2 , dtheta left( = frac{1}{2pi i}int_{|z|=1} frac{lVert a+zb rVert^2}{z^2} , dz right) $$
My question is: is this just a pretty identity, or are there situations where this is the "right"/nicest form to use in proofs requiring a polarisation identity? Obviously it adds no new mathematical content, since the $n=4$ (indeed, $n=3$) case is sufficient to determine the inner product.
integration hilbert-spaces norm
asked Oct 13 '15 at 9:43
ChappersChappers
55.8k74192
$endgroup$
It is well-known that the polarisation identity for real vector spaces is
$$ langle a,b rangle =frac{1}{4}sum_{k=0}^1 (-1)^klVert a+(-1)^k b rVert^2, $$
and the complex generalisation is
$$ langle a,b rangle =frac{1}{4}sum_{k=0}^3 i^{-k} lVert a+i^k b rVert^2. $$
There are two generalisations of this: take $omega$ a complex primitive $n$th root of unity. Then
$$ langle a,b rangle =frac{1}{n}sum_{k=0}^n omega^{-k}lVert a+omega^k b rVert^2, $$
because $ frac{1}{n}sum_{k=0}^n omega^{mk} = 0 $ unless $m=0$, when it's $1$. (Note we need to be careful with $n=2$, which is not enough to extract the imaginary part of the inner product: we instead have
$$ frac{1}{2}sum_{k=0}^1 (-1)^{k} lVert a+(-1)^k b rVert^2 = langle a,b rangle + langle b,a rangle = 2Relangle a,b rangle.) $$
But there's another generalisation: using Fourier series, we have
$$ langle a,b rangle = frac{1}{2pi}int_0^{2pi} e^{-itheta}lVert a+e^{itheta} b rVert^2 , dtheta left( = frac{1}{2pi i}int_{|z|=1} frac{lVert a+zb rVert^2}{z^2} , dz right) $$
My question is: is this just a pretty identity, or are there situations where this is the "right"/nicest form to use in proofs requiring a polarisation identity? Obviously it adds no new mathematical content, since the $n=4$ (indeed, $n=3$) case is sufficient to determine the inner product.
integration hilbert-spaces norm
integration hilbert-spaces norm
asked Oct 13 '15 at 9:43
ChappersChappers
55.8k74192
asked Oct 13 '15 at 9:43
ChappersChappers
55.8k74192
edited Dec 10 '18 at 2:29
Chappers
asked Oct 13 '15 at 9:43
ChappersChappers
55.8k74192
asked Oct 13 '15 at 9:43
ChappersChappers
55.8k74192
asked Oct 13 '15 at 9:43
ChappersChappers
55.8k74192
55.8k74192
add a comment |
add a comment |
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