<s>[INST] Verification of a certain identity in wavelet basis lemma. This is from Lemma 7.1 in Mallat's Wavelet Tour 2nd edition.
I am trying to show that
$$
b(2x)h(x) + c(2x)g(x) = a(x)
$$
when
\begin{align*}
b(2x) &= \frac{1}{2}\left[ a(x)h(x)^* + a(x+\pi)h(x+\pi)^* \right] \\
c(2x) &= \frac{1}{2}\left[ a(x)g(x)^* + a(x+\pi)g(x+\pi)^* \right]
\end{align*}
where $z^*$ denotes complex conjugation.
Furthermore, in addition to the fact that these are all $2\pi$ periodic functions, we have the following identities to make use of:
\begin{align}
|g(x)|^2 + |g(x+\pi)|^2 &= 2 \\
|h(x)|^2 + |h(x+\pi)|^2 &= 2 \\
g(x)h(x)^* + g(x+\pi)h(x+\pi)^* &= 0
\end{align}
I thought this was supposed to be a straightforward calculation, but apparently I'm missing something.
My attempt:
In the interest of things not spilling over margins, let me denote $g(x) = g$ and $g(x+\pi) = g_{\pi}$ for all functions. Now, I just plug away...
\begin{align*}
& b(2x)h(x) + c(2x)g(x) \\
&= \frac{1}{2}a\left[|g|^2 + |h|^2\right] + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right] \\
&= \frac{1}{2}a\left[4-(|g_{\pi}|^2 + |h_{\pi}|^2)\right] + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right] \\
&= 2a-\frac{1}{2}a\left[|g_{\pi}|^2 + |h_{\pi}|^2\right] + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right] \\
&= a + \left\{a - \frac{1}{2}a\left[|g_{\pi}|^2 + |h_{\pi}|^2\right] + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right]\right\}
\end{align*}
I've used the first and second identities in the second line. The last line implies necessarily that
\begin{align*}
0 &= a - \frac{1}{2}a\left[|g_{\pi}|^2 + |h_{\pi}|^2\right] + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right] \\
&= \frac{1}{2}a \left(2 - \left[|g_{\pi}|^2 + |h_{\pi}|^2\right] \right) + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right] \\
&= \frac{1}{2}a \left(|h|^2 - |g_{\pi}|^2 \right) + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right] \\
&= \frac{1}{2}a \left(|g|^2 - |h_{\pi}|^2 \right) + \frac{1}{2}a_{\pi}\left[gg_{\pi}^* + hh_{\pi}^*\right] \\
\end{align*}
I'm not really sure what to do with the $a_{\pi}$ term, or what do at all from here except run in circles. I appreciate any help.
 [/INST] I figured it out. I'll provide the verification (for future googlers) in a way that elucidates more clearly where the definition of $b(2x)$ and $c(2x)$ arise from.
We have the equation:
$$
a(x) = b(2x)h(x) + c(2x)g(x)
$$
Since everyone's $2\pi$ periodic, we also have the equation:
$$
a(x+\pi) = b(2x)h(x+\pi) + c(2x)g(x+\pi)
$$
Thus, using the notation introduced in the question, we have a 2D linear system:
$$
\begin{bmatrix}
a \\
a_{\pi}
\end{bmatrix}
=
\begin{bmatrix}
h & g \\
h_{\pi} & g_{\pi}
\end{bmatrix}
\begin{bmatrix}
b \\
c
\end{bmatrix}
$$
So that:
$$
\begin{bmatrix}
b \\
c
\end{bmatrix}
=
\frac{1}{hg_{\pi} - a_{\pi}g}
\begin{bmatrix}
g_{\pi} & -g \\
-h_{\pi} & h
\end{bmatrix}
\begin{bmatrix}
a \\
a_{\pi}
\end{bmatrix}
$$
Focusing on $b$ (c follows similarly):
$$
b = \frac{ag_{\pi} - a_{\pi}g}{hg_{\pi} - h_{\pi}g} = \frac{ag_{\pi}}{hg_{\pi} - h_{\pi}g} - \frac{a_{\pi}g}{hg_{\pi} - h_{\pi}g}
$$
where
\begin{align*}
\frac{g_{\pi}}{hg_{\pi} - h_{\pi}g} &= \frac{h^*g_{\pi}}{|h|^2g_{\pi} - h_{\pi}gh^*} \\
&= \frac{h^*g_{\pi}}{|h|^2g_{\pi} + h_{\pi}g_{\pi}h_{\pi}^*} \tag{1} \\
&= \frac{h^*g_{\pi}}{g_{\pi}\left(|h|^2 + |h_{\pi}|^2\right)} \\
&= \frac{h^*}{2} \tag{2} \\
\end{align*}
with $(1)$ using $gh^* + g_{\pi}h_{\pi}^* = 0$ and $(2)$ using $|h|^2 + |h_{\pi}|^2 = 2$.
Similarly,
\begin{align*}
\frac{-g}{hg_{\pi} - h_{\pi}g} &= \frac{-h_{\pi}^*g}{hg_{\pi} h_{\pi}^* - |h_{\pi}|^2g} \\
&= \frac{-h_{\pi}^*g}{-hg h^* - |h_{\pi}|^2g} \tag{3} \\
&= \frac{-h_{\pi}^*g}{-g \left( |h|^2 + |h_{\pi}|^2 \right)} \\
&= \frac{h_{\pi}^*}{2} \tag{4}
\end{align*}
where $(3), (4)$ use identities as with $(1),(2)$.
SO FINALLY...
$$
b(2x) = \frac{1}{2} \left( a(x)h(x)^* + a(x+\pi)h(x+\pi)^* \right)
$$
as defined.
</s>