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.
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.