I have a very precise question concerning p. 82-83 of Stein's book "Singular integrals and differentiability properties of functions". Actually it is a calculation problem. For $f \in L^{2}(\mathbb{R}^n)$, denote by $u(x,y)$ the Poisson integral of $f$ $$u(x,y)=\int_{\mathbb{R}^n} P_{y}(t) f(x-t) dt.$$ Set $$|\nabla u(x,y)|^2=\left| \frac{\partial u}{\partial y}\right|^2 + \sum_{j=1}^{n} \left| \frac{\partial u}{\partial x_{j}} \right|^2.$$ After a calculation, it comes: $$\frac{\partial u}{\partial y}=\int_{\mathbb{R}^n} - 2 \pi |t| \hat{f}(t) e^{-2\pi i t \cdot x}e^{-2 \pi |t| y} dt$$ and $$\frac{\partial u}{\partial x_{j}}=\int_{\mathbb{R}^n} - 2 \pi i t_{j} \hat{f}(t) e^{-2\pi i t \cdot x}e^{-2 \pi |t| y} dt.$$ Ok. But then, it is written that $$\int_{\mathbb{R^n}} |\nabla u (x,y)|^2 dx = \int_{\mathbb{R}^n} 8 \pi^2 |t|^2 |\hat{f}(t)|^{2}e^{-4 \pi |t|y}dy.$$ I don't understand how to get that. Especially I don't see how the square outside the integral in for instance $\left| \frac{\partial u}{\partial y} \right|$ manages to come inside the integral. Any hint is welcome.
2 Answers
Hint: write $u(\cdot,y) = P_y \ast f$. Now for the Fourier transform of a convolution, we have a nice identity.
Slightly more detailed answer follows.
Denote by $\mathcal{F}_x$ the Fourier transform in the variable x. We have by Plancherel's identity and $\mathcal{F}(P_y \ast f) = \hat{P_y} \hat{f}$ that $$ \int \lvert \nabla u(x,y) \rvert^2 dx = \int \lvert \xi \rvert^2 \lvert \hat{u}(\xi, y)\rvert^2 d\xi = \int \lvert \xi \rvert^2 \lvert \hat{P_y}(\xi) \rvert^2 \lvert \hat{f}(\xi) \rvert^2 d\xi. $$ One needs compute the Fourier transform of the Poisson kernel to finish.

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Thanks for your answer. However I'm still confused. Considering the very definition of $|\nabla u (x,y)|^2$, $$\int_{\mathbb{R}^n} \left|\nabla u (x,y)\right|^2 dx = \int_{\mathbb{R}^n} \left|\frac{\partial u}{\partial y} (x,y)\right|^2 dx + \sum_{j=1}^n \int_{\mathbb{R}^n} \left|\frac{\partial u}{\partial x_{j}} (x,y)\right|^2 dx.$$ Now I agree on the fact that $$\sum_{j=1}^n \int_{\mathbb{R}^n} \left|\frac{\partial u}{\partial x_{j}} (x,y)\right|^2 dx = \int_{\mathbb{R}^n} |\xi|^2|\hat{P_{y}}(\xi)|^2|\hat{f} (\xi)|^2 d\xi.$$ But my main problem is precisely how to deal with $\int_{\mathbb{R}^n} |\frac{\partial u}{\partial y} (x,y)|^2 dx = \int_{\mathbb{R}^n} \left( \int_{\mathbb{R}^n} 2\pi |t|\hat{f}(t) e^{-2\pi i t \cdot x}e^{-2\pi|t|y}\right)^2 dx$...

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You're right. I haven't really thought this through, but we can write (probably) $\partial_y u(x,y) = ((\partial_y P_y) \ast f)(x)$. Now we can do the same sort of thing hopefully. – Eric Thoma Sep 17 '16 at 21:08
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Ok I'll try. Thanks for your help. – David Tewodrose Sep 18 '16 at 08:29