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I am stuck with this line in my reading of a book:

By the Plancherel formula we have:

$$\int \frac{\lvert u(x+y)-u(x)\rvert^2}{\lvert y\rvert^{2s+d}}dx = \int \frac{\lvert e^{i(y\xi)} -1\rvert ^2}{\rvert y\lvert^{2s+d}} \lvert \hat{u}(\xi)^2 d\xi$$

Here $u$ is $L^2_{loc}(\mathbb{R}^d)$ and it admits a fourier transform (in the sens of distribution) which is $L^1_{loc}$ Is there a plancherel formula in this context (there is a plancherel formula for $u$ in $L^2$ but what about if $u$ is in $L^2_{loc}$ and $\hat{u}$ in $L^1_{loc}$?

I mean for every $K$ compact if $f$ is $L^2_{loc}$ we do have $\int 1_{K}|f|^2 = \int |\hat{1_{K}f}|^2$, the lhs does converge to $\int |f|^2$ for growing K's but is there a mean to treat the rhs.

Or should I search the reason of my equality by other arguments ?

incas
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    just use that the Fourier transform with respect to $x$ of $u(x+y)$ is $e^{iy\xi}\hat{u}(\xi)$ and then apply Plancherel as usual. – Carlo Beenakker Mar 12 '14 at 18:39
  • My concern is not with the formula of the fourier transform of $u(x+y)$ but more with whether have we the right to use the plancherel formula since $u$ doesn't belong to $L^2$ –  Mar 12 '14 at 21:09
  • Which book are you referring to? and on which page is this stated? – Dunham Oct 24 '14 at 08:51

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