4
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $\int_{\mathbb{R}^n} |\frac{\partial u}{\partial y} (x,y)|^2 dx = \int_{\mathbb{R}^n} \left( \int_{\mathbb{R}^n} 2\pi |t|\hat{f}(t) e^{-2\pi i t \cdot x}e^{-2\pi|t|y}\right)^2 dx$...

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Eric Thoma Sep 17 '16 at 21:08
  • $\begingroup$ Ok I'll try. Thanks for your help. $\endgroup$ – David Tewodrose Sep 18 '16 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.