# Dual wavelet

In mathematics, a dual wavelet is the dual to a wavelet. In general, the wavelet series generated by a square integrable function will have a dual series, in the sense of the Riesz representation theorem. However, the dual series is not in general representable by a square integral function itself.

## Definition

Given a square integrable function $\psi\in L^2(\mathbb{R})$, define the series $\{\psi_{jk}\}$ by

$\psi_{jk}(x) = 2^{j/2}\psi(2^jx-k)$

for integers $j,k\in \mathbb{Z}$.

Such a function is called an R-function if the linear span of $\{\psi_{jk}\}$ is dense in $L^2(\mathbb{R})$, and if there exist positive constants A, B with $0 such that

$A \Vert c_{jk} \Vert^2_{l^2} \leq \bigg\Vert \sum_{jk=-\infty}^\infty c_{jk}\psi_{jk}\bigg\Vert^2_{L^2} \leq B \Vert c_{jk} \Vert^2_{l^2}\,$

for all bi-infinite square summable series $\{c_{jk}\}$. Here, $\Vert \cdot \Vert_{l^2}$ denotes the square-sum norm:

$\Vert c_{jk} \Vert^2_{l^2} = \sum_{jk=-\infty}^\infty \vert c_{jk}\vert^2$

and $\Vert \cdot\Vert_{L^2}$ denotes the usual norm on $L^2(\mathbb{R})$:

$\Vert f\Vert^2_{L^2}= \int_{-\infty}^\infty \vert f(x)\vert^2 dx$

By the Riesz representation theorem, there exists a unique dual basis $\psi^{jk}$ such that

$\langle \psi^{jk} \vert \psi_{lm} \rangle = \delta_{jl} \delta_{km}$

where $\delta_{jk}$ is the Kronecker delta and $\langle f\vert g \rangle$ is the usual inner product on $L^2(\mathbb{R})$. Indeed, there exists a unique series representation for a square integrable function f expressed in this basis:

$f(x) = \sum_{jk} \langle \psi^{jk} \vert f \rangle \psi_{jk}(x)$

If there exists a function $\tilde{\psi} \in L^2(\mathbb{R})$ such that

$\tilde{\psi}_{jk} = \psi^{jk}$

then $\tilde{\psi}$ is called the dual wavelet or the wavelet dual to ψ. In general, for some given R-function ψ, the dual will not exist. In the special case of $\psi = \tilde{\psi}$, the wavelet is said to be an orthogonal wavelet.

An example of an R-function without a dual is easy to construct. Let $\phi$ be an orthogonal wavelet. Then define $\psi(x) = \phi(x) + z\phi(2x)$ for some complex number z. It is straightforward to show that this ψ does not have a wavelet dual.
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