Fourier series and Parseval-Bessel formula

93 Views Asked by At

I parametrized the boundary $\Gamma$ of the upper half unit disc by the following manner:

$$\gamma(\theta)=\gamma(e^{i\theta})=e^{i\theta} \text{ if }\theta \in [0,\pi],$$

$$\gamma(t)=e^{i(t+3)\pi/2}\text{ if }t\in]-1,1[.$$

I am considering a function $f \colon \Gamma \rightarrow \mathbb{R}$ and I want to obtain a Poincaré inequality of the form

$$\int_\Gamma \vert f-\frac{1}{\vert \Gamma \vert} \int_\Gamma f \vert^2 \leq C\int_\Gamma \vert \nabla_\tau f \vert^2.$$

To do this I set

$$f=\sum_{k \in \mathbb{Z}}a_ke^{ik\theta}\chi_{[0,\pi]}(\theta)+\sum_{k \in \mathbb{Z}}a_ke^{ik(t+3)\pi/2}\chi_{]-1,1[}(t).$$

Calculating the left side of the inequality, I obtain

$$f-\frac{1}{\vert \Gamma \vert}\int_\Gamma f = \sum_{k \neq 0} a_ke^{ik\theta}\chi_{[0,\pi]}(\theta)+\sum_{k \neq 0}a_ke^{ik(t+3)\pi/2}\chi_{]-1,1[}(t)-c$$

with

$$c=-\frac{2i}{\pi}\frac{\pi-2}{\pi+2}\sum_{k \in \mathbb{Z}} \frac{a_{2k+1}}{2k+1}$$ is constant.

I don't know how to continue the calculation when taking the square. Indeed I would like to use Parseval formula, but the family

$$(e^{ik\theta}\chi_{[0,\pi]}(\theta)+e^{ik(t+3)\pi/2}\chi_{]-1,1[}(t))_{k \in \mathbb{Z}}$$

is not orthogonal...

If I don't mistake, for the right side of the inequality this problem disappears because the derivative makes a coefficient $\pi/2$ come out from the function on the straight part of the boundary, and the family

$$(e^{ik\theta}\chi_{[0,\pi]}(\theta)+\frac{\pi}{2}e^{ik(t+3)\pi/2}\chi_{]-1,1[}(t))_{k \in \mathbb{Z}}$$

is orthogonal, so I can use Parseval formula here.

Any idea for the left part of my inequality?