Fourier transform of the Haar system

91 Views Asked by At

The Haar system form an orthonormal basis of $L^2(\mathbb{R})$. Being the Fourier transform an isometry of $L^2(\mathbb{R})$ onto itself, the Fourier transform of the Haar system is another orthonormal basis of $L^2(\mathbb{R})$. Can someone provide a source where to find the explicit form for the transformed basis?

1

There are 1 best solutions below

0
On BEST ANSWER

No reference, sorry, but it is straightforward to evaluate the transform: $$\psi_{n, k}(t) = 2^{n/2}([0 < 2^n t - k < 1/2] - [1/2 < 2^n t - k < 1]), \\ \int_{-\infty}^\infty \psi_{n, k}(t) e^{i p t} dt = \frac {2^{n/2 + 2} \sin^2(2^{-n - 2} p)} {i p} \exp(2^{-n - 2} (4 k + 2) i p).$$