- Let $\psi(x)=e^{-a|x|^2}$, with $ x \epsilon \textbf{R}$. Evaluate $\frac{(\| |x|^m \psi \|_2)(\| |y|^m \hat{\psi} \|_2)}{{\| \psi \|}_2^2}$
- Show that the inequality $(\| |x| \psi \|_2)(\| |y| \hat{\psi} \|_2) \geq \frac{(2 \pi)^\frac{1}{2}}{2} \| \psi \|_2^2$ is exact for these $\psi$'s.
- Can the inequality $(\| |x|^m \psi \|_2)(\| |y|^m \hat{\psi} \|_2) \geq C \| \psi \|_2^2$ be valid for every $\psi \epsilon L^2(\textbf{R})$, with C independent of $\psi$?
Note $\psi(x)$ is equal to the Gaussian, which we know the fourier transformation of the Gaussian is $\sqrt{\frac{\pi}{a}}e^\frac{-\pi^2y^2}{a}$.
Evaluate $\frac{(\| |x|^m \psi \|_2)(\| |y|^m \hat{\psi} \|_2)}{{\| \psi \|}_2^2}$
\begin{equation*} \begin{split} \| |x|^m \psi \|_2 & =(\int_\textbf{R} (|x|^me^{-a|x|^2})^2 \,d|x|)^\frac{1}{2} \\ & =(\int_\textbf{R} |x|^{2m}e^{-a|x|^4} \,d|x|)^\frac{1}{2}=\ldots \\ \end{split} \end{equation*}
How do I continue?
\begin{equation*} \begin{split} \| |y|^m \hat{\psi} \|_2 & =(\int_\textbf{R} (|y|^m \sqrt{\frac{\pi}{a}}e^\frac{-\pi^2y^2}{a})^2 \,d|x|)^\frac{1}{2} \\ & =(\int_\textbf{R} |y|^2m \frac{\pi}{a}e^\frac{-2\pi^2y^2}{a} \,d|x|)^\frac{1}{2} \\ \end{split} \end{equation*}
How do I continue?
\begin{equation*} \begin{split} \| \psi \|_2^2 & =\int_\textbf{R} (e^{-a|x|^2})^2 \,dx \\ & =\int_\textbf{R} e^{-2a|x|^2} \,dx \\ & =\sqrt{\frac{\pi}{2a}} \\ \end{split} \end{equation*}
So, $\frac{(\| |x|^m \psi \|_2)(\| |y|^m \hat{\psi} \|_2)}{{\| \psi \|}_2^2}=\frac{}{\sqrt{\frac{\pi}{2a}}}$
Show that the inequality $(\| |x| \psi \|_2)(\| |y| \hat{\psi} \|_2) \geq \frac{(2 \pi)^\frac{1}{2}}{2} \| \psi \|_2^2$ is exact for these $\psi$'s.
I believe that this result will come out as soon as we prove number one.
Can the inequality $(\| |x|^m \psi \|_2)(\| |y|^m \hat{\psi} \|_2) \geq C \| \psi \|_2^2$ with C independent of $\psi$? I am not sure if it is sure or not.