Let $\psi\in L_2(\mathbb{R})$ be an orthonormal wavelet and |$\widehat\psi$|be a continuous function. What is $\widehat\psi (2kπ)$?

Question

Let $\psi \in L_2(\mathbb{R})$ be an orthonormal wavelet and $|\widehat\psi|$ be a continuous function. Then it is known that $\widehat\psi(0)=0$.

Assume that $\psi\in L_2(\mathbb{R})$ and $\psi$ is continuous.

What is $\widehat\psi(2k\pi)$ for $k\in\mathbb{Z}$?

I am just computing that,

$\widehat\psi(2\pi)=\int_{\mathbb{R}}\psi(x)e^{-2\pi i x}dx$.

We know that $\widehat\psi(0)=\int_{\mathbb{R}}\psi(x)dx=0$.

I dont know how to proceed next.

I try to apply integration by parts formula. But it doesn't helping me.

Answer 1

Not an answer. About your condition for the orthonormality of the wavelet, Not really. Let $\psi_a(t) = \frac{1}{|a|^{1/2}}\psi(t/a)$.

The continuous wavelet transform is then $$X_a(b) = x \ast \overline{\psi_a}(b)$$ where $\ast $ is the convolution, and $\psi$ is an orthonormal wavelet means that the inverse WT is $$x(t)= \int_{-\infty}^\infty X_a \ast \psi_a(t) \frac{da}{a^2}$$

i.e. with $h(t) = \int_{-\infty}^\infty \overline{ \psi_a }\ast \psi_a(t) \frac{da}{a^2}$ then $x(t) = x \ast h(t)$ so that $h(t) = \delta(t)$ the Dirac delta. This is valid only in the sense of distributions, but taking the Fourier transform, we get that the Fourier transform of $ \overline{ \psi_a } \ast \psi_a(t)$ is $|\hat{\psi_a}(\xi)|^2 =|a|\, |\hat{\psi}(a\xi)|^2$ and $h(t) = \delta(t)$ becomes

$$1 = \hat{\delta}(\xi) = \hat{h}(\xi) = \int_{-\infty}^\infty |\hat{\psi}(a\xi)|^2 \frac{da}{|a|}$$ almost everywhere