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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 itself in general representable by a square-integrable function.
Given a square-integrable function ψ ∈ L 2 ( R ) {\displaystyle \psi \in L^{2}(\mathbb {R} )} , define the series { ψ j k } {\displaystyle \{\psi _{jk}\}} by
for integers j , k ∈ Z {\displaystyle j,k\in \mathbb {Z} } .
Such a function is called an R-function if the linear span of { ψ j k } {\displaystyle \{\psi _{jk}\}} is dense in L 2 ( R ) {\displaystyle L^{2}(\mathbb {R} )} , and if there exist positive constants A, B with 0 < A ≤ B < ∞ {\displaystyle 0<A\leq B<\infty } such that
for all bi-infinite square summable series { c j k } {\displaystyle \{c_{jk}\}} . Here, ‖ ⋅ ‖ l 2 {\displaystyle \Vert \cdot \Vert _{l^{2}}} denotes the square-sum norm:
and ‖ ⋅ ‖ L 2 {\displaystyle \Vert \cdot \Vert _{L^{2}}} denotes the usual norm on L 2 ( R ) {\displaystyle L^{2}(\mathbb {R} )} :
By the Riesz representation theorem, there exists a unique dual basis ψ j k {\displaystyle \psi ^{jk}} such that
where δ j k {\displaystyle \delta _{jk}} is the Kronecker delta and ⟨ f | g ⟩ {\displaystyle \langle f\vert g\rangle } is the usual inner product on L 2 ( R ) {\displaystyle L^{2}(\mathbb {R} )} . Indeed, there exists a unique series representation for a square-integrable function f expressed in this basis:
If there exists a function ψ ~ ∈ L 2 ( R ) {\displaystyle {\tilde {\psi }}\in L^{2}(\mathbb {R} )} such that
then ψ ~ {\displaystyle {\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 ψ = ψ ~ {\displaystyle \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 ϕ {\displaystyle \phi } be an orthogonal wavelet. Then define ψ ( x ) = ϕ ( x ) + z ϕ ( 2 x ) {\displaystyle \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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