Posted by: Shuanglin Shao | February 28, 2009

A misleading argument for the adjoint Fourier restriction for the sphere in 3 dimension

Today I thought that I had a new proof of the adjoint Fourier transform for the sphere in three dimensions; but it turns out it was wrong. I would like to record it here, which I hope is useful at some point.

First let us define the notion of Fourier transform. Let
\widehat{f}(\xi) :=(2\pi)^{-3/2}\int_{\mathbf R^3} e^{-ix\xi} f(x)dx.
Then the Plancherel theorem reads,
\|\widehat{f}\|_{L^2(\mathbf R^3)}=\|f\|_{L^2(\mathbf R^3)}.
From this definition the convolution of two functions behaves under the Fourier transform like
\widehat{ f\ast g}= (2\pi)^{3/2} \widehat{f}\widehat{g}.

Now let us begin the argument. By Plancherel,
\|\widehat{fd\sigma}\|^2_{L^2(\mathbf{R}^3)} =(2\pi)^{-3/2} \|fd\sigma \ast fd\sigma \|_{L^2(\mathbf{R}^3)}.
In view of this, we may assume that f\ge 0.
We write the integrand out,
fd\sigma \ast fd\sigma (x) =\int_{S^2} [f\sigma](x-y)[fd\sigma](y).

(1) By Cauchy-Schwarz inequality, we have
f(x-y)f(y)\le \dfrac {|f(x-y)|^2+|f(y)|^2}{2}.
We observe that
\int_{S^2}f^2(x-y)d\sigma(x-y)d\sigma(y)=\int_{S^2}d\sigma(x-y)f^2(y)d\sigma(y).
Then
\|fd\sigma \ast fd\sigma \|_{L^2(\mathbf{R}^3)}
\le\|\int_{S^2}f^2(y)d\sigma(x-y)d\sigma(y) \|_{L^2(\mathbf{R}^3)}.

(2)Then by Minkowsik inequality,
\|\int_{S^2}d\sigma(x-y)f^2(y)d\sigma(y) \|_{L^2(\mathbf{R}^3)}
\le \|d\sigma\|_{L^2} \int_{S^2} f^2(y)d\sigma=(4\pi)^{1/2} \|f\|^2_{L^2(S^2)}.

Note that the best constant for Young’s inequality L^1\times L^2 \to L^2 is 1.

(I orginially thought that \|d\sigma\|_{L^2(\mathbf R^3)} =(\int_{S^2} d\sigma)^{1/2}=(4\pi)^{1/2}. However, it is wrong since d\sigma\neq d\sigma^2 or d\sigma\notin L^2(\mathbf R^3), which can be seen from the asympotics \widehat{d\sigma}(\xi)\sim (1+|\xi|)^{-1} for large \xi in \mathbf R^3. Incidentally, today I found that Terry remarked on his blog that the Dirac mass \delta^2 \notin L^1 by a trick of epsilon regularization (another way is to use Pancherel theorem and observe that \widehat{\delta} =1).)

At this point, there is no need to read on since the following is based on Step 2. I worte it down before I realized the mistake; so I backed it up as my notes.

{Then we can conclude that
\|\widehat{fd\sigma}\|^2_{L^2 (\mathbf{R}^3)}\le (2\pi)^{-3/2}(4\pi)^{1/2}\|f\|^2_{L^2(S^2)}=2^{-1/2}\pi^{-1/2}\|f\|^2_{L^2(S^2)}.
}

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