Consider a multiresolution analysis (MRA) $\{V_j\}$ in $L^2(\mathbb R)$ generated by the scaling function $\phi(x)$ and such that for each $j\in\mathbb Z$, $V_j\subset V_{j-1}$. For a given function $f$ I will use the notation $f_{j,k}(x) = 2^{-j/2}f(2^{-j}x-k)$.

Let $\psi(x)$ be the wavelet corresponding to $\phi(x)$, then $\{\psi_{j,k}\}_{k\in\mathbb Z}$ spans the orthogonal complement $W_j$ to $V_j$ in $V_{j-1}$, namely, $W_j = \overline{span}\{\psi_{j,k}\}_{k\in\mathbb Z}$ is such that $V_{j-1} = V_j\oplus W_j$ and $V_j\perp W_j$. Thus for a fixed $J<0$ one can define the projection of a function $f\in L^2(\mathbb R)$ into $V_j$ through the expansion $$ P_J f(x) = \sum_{k\in\mathbb Z} \langle f,\phi_{0,k}\rangle\phi_{0,k}(x) + \sum_{j=0}^{J+1}\sum_{k\in\mathbb Z}\langle f,\psi_{j,k}\rangle \psi_{j,k}(x)\qquad (1) $$ where the equality is in the $L^2$-sense. That makes sense to me due to the orthogonality of $V_0$ and $W_0\oplus W_1\oplus ... \oplus W_{J+1}$.

Consider now a biorthogonal MRA. In this case one consider two dual MRAs $\{V_j\}$ and $\{\tilde V_j\}$ generated by the scaling functions $\phi(x)$ and $\tilde \phi(x)$. In a similar way as in the orthonormal case, one defines the dual wavelets $\psi(x)$ and $\tilde \psi(x)$ which span repsectively the (non orthogonal) complements $W_j$ of $V_j$ in $V_{j-1}$ and $\tilde W_j$ of $\tilde V_j$ in $\tilde V_{j-1}$. Namely one has $V_{j-1}=V_j+W_j$ and $\tilde V_{j-1} = \tilde V_j+\tilde W_j$ with $W_j = \overline{span}\{\psi_{j,k}(x)\}_{k\in\mathbb Z}$ and $\tilde W_j = \overline{span}\{\tilde \psi_{j,k}(x)\}_{k\in\mathbb Z}$.

In this case, for fixed $J$, one can write $$ P_J f(x) = \sum_{k}\langle f, \tilde \phi_{J,k}\rangle\phi_{J,k}(x) $$ and an analogous expansion exists for $\tilde P_j$. My question is: does an expression of the kind of (1) hold also in the biorthogonal case? i.e. can I write $$ P_J f(x) = \sum_{k\in\mathbb Z} \langle f,\tilde\phi_{0,k}\rangle\phi_{0,k}(x) + \sum_{j=0}^{J+1}\sum_{k\in\mathbb Z}\langle f,\tilde\psi_{j,k}\rangle \psi_{j,k}(x)\;? $$ It is not clear to me that this is possible since it is not true that $V_0$ is orthogonal to $W_{0}+...+W_{J+1}$. However the following argument appears to hold:

Let $Q_j := P_{j-1} - P_{j}$ (the detail operator) and define the dilation operator as $D_j f(x) := 2^{-j/2}f(2^{-j}x)$. We have $$P_{j+m}f(x) = D_{j} P_m (D_{-j} f(x))$$ and hence \begin{align*} Q_jf(x) &= P_{j-1}f(x) - P_{j}f(x) = D_j (P_{-1} - P_0) \big(D_{-j} f(x)\big)\\&= D_j Q_0\big(D_{-j}f(x)\big) \end{align*} Moreover it can be shown that $Q_0 f(x) \in W_0$ and from the above relation one obtain, for each $j\in\mathbb Z$ $$ Q_j f(x) = D_j\sum_k \langle D_{-j}f,\tilde \psi_{0,k}\rangle\psi_{0,k}(x) = \sum_{k} \langle f,\tilde \psi_{j,k}\rangle \psi_{j,k}(x) $$ Thus by definition of $Q_j$ \begin{align*} P_{j}f(x) &= P_{j+1}f(x) + Q_{j+1} f(x) = P_{j+1}f(x) + \big(P_{j+2}f(x)-P_{j+2} f(x)\big)+Q_{j+1} f(x)\\ &= P_{j+2}f(x) + Q_{j+2} f(x) + Q_{j+1}f(x) \\&\quad\vdots\\&=P_0 f(x) + \sum_{i=0}^{j+1} Q_i f(x) \end{align*} and thus (1) holds.

Is this correct?

  • $\begingroup$ You're making this much more complicated than necessary by using two different bilinear forms where one is sufficient. Consider the two basis sets as duals under a single bilinear form. Then talking about the orthogonality between two vectors in the same basis always involves using the canonical inner product with that dual. The canonical inner product between two vectors from the same basis does not exist. So your statement that $V_0$ is not orthogonal to $\sum W_n$ is irrelevant and meaningless because it refers to the canonical inner product. $\endgroup$
    – Jazzmaniac
    Oct 1, 2016 at 10:18
  • $\begingroup$ Thanks for comment, honestly I'm not sure to have understood your comment completely, but that makes sense to me. Thus my conclusion is correct? $\endgroup$
    – LJSilver
    Oct 1, 2016 at 10:50

1 Answer 1


Let me try to give you a short introduction to the concept of bi-orthogonal bases from a mathematical point of view. You are already using the concept, but your notation and nomenclature is obfuscating what is really going on. I will make this answer independent of the wavelet context you have, so that it is more general. You will certainly be able to apply it to your situation however.

You have a non-orthogonal basis, i.e. a set of linearly independent vectors spanning your $L^2(K)$ Hilbert space $\mathcal{H}$. That vector space induces another vector space, that of all linear forms $F:\mathcal{H}\to K$. We can make this space a dual space of $\mathcal{H}$ if we take the Hilbert space inner product $\langle \cdot, \cdot\rangle : \mathcal{H} \times \mathcal{H} \to K $ to generate a linear form $\langle \phi, \cdot \rangle : \mathcal{H} \to K$ for every $\phi \in \mathcal{H}$.

In an effort to simplify the notation a bit and make things more intuitive, we can introduce Dirac's Bra-Ket notation for vectors and their duals. Instead of $\phi \in \mathcal{H}$, we write the Ket $\left| \phi \right\rangle$ and instead of $\langle \phi , \cdot \rangle \; \phi \in \mathcal{H}$ we write the Bra $\left\langle \phi \right|$. Applying the linear form $\left\langle \psi\right|$ to $\left| \phi \right\rangle$ is then simply the notational fusion to a braket standing for the inner product $\left\langle \psi | \phi \right\rangle$.

In the bi-orthogonal basis business, this distinction is often ignored, making everything more confusing that it should be. Consider a basis that is not orthogonal with respect to the canonical inner product. We can of course use the usual knowledge of linear algebra to still construct everything of interest and expand vectors with respect to this basis, but it tends to get messy quickly.

That's where a neat trick comes in. We simply invent a new inner product, i.e. a non-degenerate, non-negative bilinear form $B: \mathcal{H} \times \mathcal{H} \to K$, so that $B( \phi_n, \phi_m ) = \delta_{n,m}$ for your basis $\{ \phi_n, n\in\mathbb{N}\}$ and the Kronecker symbol $\delta$. Because the basis is complete, $B$ is entirely determined by this condition. We can also write this bilinear form using a linear operator $B:\mathcal{H}\to\mathcal{H}$ with the same name. Then we get $$\left\langle \phi_n | B | \phi_m \right\rangle := \left\langle \phi_n | B \phi_m \right\rangle = B( \phi_n, \phi_m ) = \delta_{n,m}$$.

At this point, it makes sense to define the adjoint $A^\dagger$ of a linear operator $A$ like follows: $$\left\langle A^\dagger \phi_n | \phi_m \right\rangle = \left\langle \phi_n | A \phi_m \right\rangle$$

Of course it follows from the definitions above: $$\left\langle B^\dagger \phi_n \right| = B(\phi_n,\cdot)$$ and therefore $$\left\langle B^\dagger \phi_n | \phi_m \right\rangle=\delta_{n,m}$$ which means $\left\langle B^\dagger \phi_n \right|$ is the dual of $\left|\phi_n\right\rangle$ under $B$, whereas the dual under the canonical inner product is still $\left\langle \phi_n \right|$.

Now what you can often see, and I cannot stress enough how much I think this is a bad idea, is that the dual under $B$ is mapped back to $\mathcal{H}$ as a dual under the canonical inner product. That means you construct a second basis $\left| \tilde{\phi}_n\right\rangle =\left| B \phi_n \right\rangle$ which is bi-orthogonal to the first basis with respect to the canonical inner product: $$ \left\langle \tilde{\phi}_n | \phi_m \right\rangle = \delta_{n,m}$$ This obfuscates the real meaning of the second basis and puts both on the same footing and suggests that both bases live in the same space and transform in the same way, which they don't.

So, what is the alternative? Not taking the $B$-dual back to $\mathcal{H}$ makes things much more intuitive. You want to expand $\left|\psi\right\rangle$ in the basis $\phi_n$? Just use the $B$-dual: $\left|\psi\right\rangle=\sum_n \left| \phi_n \right\rangle \left\langle B^\dagger \phi_n | \psi \right\rangle $. Or you want to project onto a vector $\left| \kappa \right\rangle$? Just use the $B$-dual: $\left| \kappa \right\rangle \left\langle B^\dagger \kappa | \psi \right\rangle$. So whatever you do, just use $B$ as your inner product and you can work like with an orthogonal basis.

And to come back to your original question. You say that $V_0$ is not orthogonal to $\oplus_n W_n$. Yes, but only with respect to the canonical inner product. With respect to $B$, they are orthogonal. And so your argument for the orthogonal basis case works.

I hope all this does not confuse you. I've tried to be as clear as possible, but it's not a trivial topic, so don't be frustrated if you have to read the answer several times until it makes sense. If I can clarify anything, please let me know.

  • $\begingroup$ Thanks a lot for this really useful and interesting answer. That's incredible how complex things sometimes admits so simple descriptions. I think indeed that changing the inner product is actually the first thing that should come in mind, rather than disturb biorthogonal bases. So making a connection to biorthogonal bases, if $\{\phi_k\}$ is Riesz basis for $\mathcal H$ then there exists a bounded bijective operator $U$ on $\mathcal H$ such that $\phi_k = Ue_k$ with $\{e_k\}$ an orthonormal basis for $\mathcal H$. Thus the operator $B$ you are referring to is $U^{-1}$? $\endgroup$
    – LJSilver
    Oct 1, 2016 at 18:37
  • $\begingroup$ by the way, can I ask you to provide me some reference about this that you think could be suitable for the a quite poor background like mine? Thanks a lot. $\endgroup$
    – LJSilver
    Oct 1, 2016 at 18:39
  • $\begingroup$ I don't have a reference at hand. I can check my library after the weekend. And my $B$ relates to your $U$ by ${U^{-1}}^\dagger U^{-1} = B$, which also shows the non-uniqueness of $U$ by left-multiplication with a unitary operator. $\endgroup$
    – Jazzmaniac
    Oct 1, 2016 at 20:32
  • $\begingroup$ Yes sorry that was the right relation with U (I could not modify the comment). Anyway if you will remember to check for a reference Id really appreciate, if not, dont worry, you already help me a lot! Thanks again. $\endgroup$
    – LJSilver
    Oct 1, 2016 at 20:36

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