Meyer wavelet

Example

Use the Fourier transform $\hat f(\omega) = \int_{\R} f(x) e^{-i\omega x} \, dx$. Fix an infinitely differentiable function $\nu \colon \R \to [0,1]$ with $\nu(x) = 0$ for $x \le 0$, $\nu(x) = 1$ for $x \ge 1$, and $\nu(x) + \nu(1-x) = 1$ for all $x$. The Meyer wavelet $\psi$ is the function in $L^2(\R)$ whose Fourier transform is $$\hat\psi(\omega) = \begin{cases} e^{i\omega/2} \sin\left(\frac{\pi}{2} \nu\left(\frac{3\abs{\omega}}{2\pi} - 1\right)\right) & \text{if } \frac{2\pi}{3} \le \abs{\omega} \le \frac{4\pi}{3}, \\ e^{i\omega/2} \cos\left(\frac{\pi}{2} \nu\left(\frac{3\abs{\omega}}{4\pi} - 1\right)\right) & \text{if } \frac{4\pi}{3} \le \abs{\omega} \le \frac{8\pi}{3}, \\ 0 & \text{otherwise.} \end{cases}$$ The functions $\psi_{j,k}(x) = 2^{j/2} \psi(2^j x - k)$, for $j, k \in \Z$, form an orthonormal basis of $L^2(\R)$. So $\psi$ is an orthonormal wavelet. It comes from a multiresolution analysis whose scaling function $\varphi$ has $\hat\varphi(\omega) = 1$ for $\abs{\omega} \le 2\pi/3$ and $\hat\varphi(\omega) = 0$ for $\abs{\omega} \ge 4\pi/3$.