This is a minor technical matter that I know to be true, I just don't know how to prove it. Help would be much appreciated.
Let:$$\psi\left(x\right)=\sum_{n=0}^{\infty}e^{-2^{n}v\left|x\right|}$$ where $v$ is a positive constant. $\psi$ is in both $L^{1}\left(\mathbb{R}\right)$ and $L^{2}\left(\mathbb{R}\right)$, and its Fourier transform is:
$$\Psi\left(\xi\right)=\int_{-\infty}^{\infty}\psi\left(x\right)e^{-2\pi i\xi x}dx=\frac{2}{v}\sum_{n=0}^{\infty}\frac{2^{-n}}{\left(\frac{2\pi\xi}{2^{n}v}\right)^{2}+1}$$
I am trying to prove that $2\Psi\left(2\xi\right)>\Psi\left(\xi\right)$ for all $\xi\in\mathbb{R}$. Any ideas, people?
Thanks in advance!
$$2\Psi\left(2\xi\right)=2\frac{2}{v}\sum_{n=0}^{\infty}\frac{2^{-n}}{\left(\frac{2\pi(2\xi)}{2^{n}v}\right)^{2}+1} = \frac{2}{v}\sum_{n=0}^{\infty}\frac{2^{-n+1}}{\left(\frac{2\pi\xi}{2^{n-1}v}\right)^{2}+1} = \frac{2}{v}\sum_{n=-1}^{\infty}\frac{2^{-n}}{\left(\frac{2\pi\xi}{2^{n}v}\right)^{2}+1} >\Psi\left(\xi\right)$$ since the first term of the series is positive and sum of the rest of the terms is equal to $\Psi\left(\xi\right)$.