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?