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
$\widehat{f}(\xi) :=(2\pi)^{-3/2}\int_{\mathbf R^3} e^{-ix\xi} f(x)dx.$
$\|\widehat{f}\|_{L^2(\mathbf R^3)}=\|f\|_{L^2(\mathbf R^3)}.$
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
$\|\widehat{fd\sigma}\|^2_{L^2 (\mathbf{R}^3)}\le (2\pi)^{-3/2}(4\pi)^{1/2}\|f\|^2_{L^2(S^2)}=2^{-1/2}\pi^{-1/2}\|f\|^2_{L^2(S^2)}.$
}