This is the problem 6.19 from the book Distributions and Operators, Gerd Grubb. I already have done parts (a) and (b).

The part (a) of this problem is proving that for $r\in(0,1]$, the sequence $$\{\frac{1}{2\pi}\sum_{n=-N}^{N}r^{|n|}e^{inx}\}_{N\in\mathbb{N}}$$ converges to a distribution $P_{r}$ in $D'((-\pi,\pi))$ and that $P_{1}=\delta$. Part (b) is just showing that $r\mapsto P_{r}$ is continuous.

Now, for part (c), I have to show that when $r$ converges to $1$ from the left, then $$\int_{-\pi}^{\pi}\frac{1-r^2}{1-2r\cos\theta+r^2}\;\varphi(\theta)\;d\theta$$ converges to $\varphi(0)$ for any $\varphi\in C_{0}^{\infty}((-\pi,\pi))$.

I thought about this for a while but I don't have a clue. Thanks for the help.

  • $\begingroup$ It is a rational function so you can apply partial fraction decomposition and integrate. It will converge to $\delta$ iff its primitive converges to $\frac12+\frac12\text{sign}(\theta)$. A common alternative is to expand in $e^{i \theta}$ to obtain its Fourier series. $\endgroup$ – reuns Oct 19 '18 at 2:03
  • $\begingroup$ Can you elaborate? Please. I think it should not be easy to find an antiderivative of this, because of the $\varphi$ multiplicating. It is also a difficult integral with the $e^{ik\theta}$. $\endgroup$ – mrde05 Oct 19 '18 at 2:27
  • $\begingroup$ This is the exact statement of the part (c). Let me edit the question and add the rest. $\endgroup$ – mrde05 Oct 19 '18 at 2:42
  • $\begingroup$ @mrde05 An antiderivative of $\frac{1-r^2}{1-2r\cos\theta+r^2}$ which is a rational function in $e^{i \theta}$ $\endgroup$ – reuns Oct 19 '18 at 3:06

Your Answer

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

Browse other questions tagged or ask your own question.