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We know that the Fourier transform of the Poisson kernel $P(x,t)$ \begin{equation} \frac{\Gamma(\frac{(n+1)}{2})}{\pi^{\frac{(n+1)}{2}}}\frac{t}{(t^2+\lvert x\rvert^2)^{\frac{(n+1)}{2}}} \end{equation} is the Abel kernel $K(x,t)$ \begin{equation} e^{-2\pi t \lvert \xi \rvert}. \end{equation} However, I have just seen one method of proving it from Stein's Introduction to Fourier Analysis on Euclidean Spaces. The key of the proof is to use \begin{equation} e^{-\beta}=\frac{1}{\sqrt \pi} \int_0^\infty \frac{e^{-u}}{\sqrt u} e^{-\frac{\beta^2}{4u}} \, \mathrm{d} u. \end{equation} And it start with the Abel kernel to Poisson kernel. But I feel that this proof is a little trick. So is there any other proof of it?

Thank you very much!

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1 Answer 1

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Here is a proof that uses PDE theory to bypass the extensive computation.

First note that for $f$ a Schwartz function on $\mathbb{R}^n$ there is a unique $u\in C^{\infty}(\mathbb{R}^n\times \mathbb{R}_+)$ satisfying $$ \begin{cases} \frac{\partial^2 u}{\partial t^2}+\Delta_x u=0\\ \lim_{t^2+|x|^2\to \infty} u=0\\ \lim_{t\to 0}||u(\cdot, t)-f(\cdot)||_{\infty}=0. \end{cases} $$ Uniqueness of the above solution follows from the maximum principle, and existence follows from choosing $u(x,t)=P(x,t)*f(x)$.

One can also show existence and uniqueness for the following problem. $$ \begin{cases} \frac{\partial^2 u}{\partial t^2}-4\pi|\xi|^2 u=0\\ \lim_{t^2+|\xi|^2\to \infty} u=0\\ \lim_{t\to 0}||u(\cdot, t)-f(\cdot)||_{\infty}=0. \end{cases} $$ In this case, uniquness follows from the fact that the first two conditions imply for each fixed $\xi_0$, $u(t,\xi_0)=Ce^{-2\pi t |\xi_0|}$. The last condition then implies that these constants are determined by $f$, so we have $u(t,\xi)=e^{-2\pi t |\xi|}\cdot f(\xi)$, and this formula gives existence.

Finally, if $u$ solves the first system with initial data $f$, then $\hat{u}$ solves the second system with initial data $\hat{f}$, so we have that by uniqueness of solutions, $$ \hat{f}\cdot e^{-2\pi t |\xi|}=(f*P(x,t))^\wedge=\hat{f}\cdot \hat{P}(x,t). $$ As this holds for all $f$ Schwartz functions, $\hat{P}(x,t)=e^{-2\pi t |\xi|}$.

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