HW2. Due September 27

Exercise: Show that if L is the Laplace operator on \mathbb{R}^n, then for t > 0, \mathbf{P}_t f (x)=\frac{1}{(4\pi t)^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-\frac{\|x-y\|^2}{4t} } f(y)dy. 

Exercise: Let L be an essentially self-adjoint diffusion operator on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}). Show that if the constant function 1 \in \mathcal{D}(L) and if L1=0, then \mathbf{P}_t 1=1. 

Exercise: Let L be an essentially self-adjoint diffusion operator on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})

  • Show that for every \lambda > 0, the range of the operator \lambda \mathbf{Id}-L is dense in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}).
  • By using the spectral theorem, show that the following limit holds for the operator norm on \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}), \mathbf{P}_t=\lim_{n \to +\infty} \left( \mathbf{Id} -\frac{t}{n} L\right)^{-n}.

Exercise: As usual, we denote by \Delta the Laplace operator on \mathbb{R}^n. The Mac-Donald’s function with index \nu \in \mathbb{R} is defined for x \in \mathbb{R} \setminus \{ 0 \} by K_\nu (x)=\frac{1}{2} \left( \frac{x}{2} \right)^\nu \int_0^{+\infty} \frac{e^{-\frac{x^2}{4t} -t}}{t^{1+\nu}} dt.

  • Show that for \lambda \in \mathbb{R}^n and \alpha > 0, \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n} e^{i \langle \lambda , x \rangle} \left( \frac{\| x \| }{\sqrt{\alpha}} \right)^{1-\frac{n}{2}} K_{\frac{n}{2}-1} (\sqrt{\alpha} \| x \| ) dx=\frac{1}{\alpha +\| \lambda \|^2}.
  • Show that for \nu \in \mathbb{R}, K_{-\nu}=K_\nu.
  • Show that K_{1/2}(x)=\sqrt {\frac{\pi}{2x}} e^{-x}.
  • Prove that for f \in \mathbf{L}^2 (\mathbb{R}^n,\mathbb{R}) and \alpha > 0, (\alpha\mathbf{Id}-\Delta)^{-1} f (x)=\int_{\mathbb{R}^n} G_\alpha (x-y) f(y) dy, where G_\alpha(x)=\frac{1}{(2\pi)^{n/2}} \left( \frac{\| x \| }{\sqrt{\alpha}} \right)^{1-\frac{n}{2}} K_{\frac{n}{2}-1} (\sqrt{\alpha} \| x \| ). (You may use Fourier transform to solve the partial differential equation \alpha g -\Delta g=f).

Exercise: By using the previous exercise, prove that for f \in \mathbf{L}^2 (\mathbb{R}^n,\mathbb{R}), \lim_{n \to + \infty} \left(\mathbf{Id} -\frac{t}{n} \Delta \right)^{-n} f =\frac{1}{(4\pi t)^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-\frac{\|x-y\|^2}{4t} } f(y)dy, the limit being taken in \mathbf{L}^2 (\mathbb{R}^n,\mathbb{R}). Conclude that almost everywhere, \mathbf{P}_t f (x)=\frac{1}{(4\pi t)^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-\frac{\|x-y\|^2}{4t} } f(y)dy.

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