I can' t show the following equality. Could you give some hints? $$ A^T\big( \psi(t) \otimes I \big) Q \big(\psi^T(t) \otimes I\big) A= A^T\big( \psi(t) \psi^T(t) \otimes Q\big) A$$ where
$\otimes$ is the Kronecker product,
$Q$ is a positive semi-definite matrix, $A$ is a real matrix,
$I$ is the identity matrix.
$\psi(t)$ is the Legendre wavelet function,
$\psi^T(t)$ is the transpose of $\psi(t)$
From the definition of the Kronecker product, for all $A$ matrices, we have $$ A = I_1 \otimes A = A \otimes I_1,\tag{$\heartsuit$} $$ where $I_n$ is the $n \times n$ identity matrix. Specially $I_1 = {\begin{bmatrix}1\end{bmatrix}}$.
The mixed-product property of the Kronecker product states that if $A,B,C$ and $D$ are matrices of such size that we can form the matrix products $AC$ and $BD$, then $$ (A \otimes B)(C \otimes D) = (AC) \otimes (BD). $$ In general, if $A_1,\dots, A_p$ and $B_1,\dots,B_p$ are matrices of such size that the following matrix products exist, then we have $$ (A_1 \otimes B_1)(A_2 \otimes B_2)\cdots(A_p \otimes B_p) = (A_1A_2\cdots A_p) \otimes (B_1B_2 \cdots B_p).\tag{$\spadesuit$} $$
Since $\psi$ is a Legendre wavelet function, we know that $\psi(t) \in \mathbb{R}^{n \times 1}$ for all $t$ in the domain of $\psi$, for some $n \in \mathbb{N}$. First we decompose $Q \in \mathbb{R}^{n \times n}$ by using $(\heartsuit)$ as the following. $$ \left( \psi(t) \otimes I_n\right) Q \left(\psi^T(t) \otimes I_n\right) = \left( \psi(t) \otimes I_n \right) \left(I_1 \otimes Q\right) \left(\psi^T(t) \otimes I_n\right). $$ Then by using $(\spadesuit)$, we find that $$ \left( \psi(t) \otimes I_n\right) \left(I_1 \otimes Q\right) \left(\psi^T(t) \otimes I_n\right) = (\psi(t)\psi^T(t)) \otimes Q.\tag{$\diamondsuit$} $$ If $A \in \mathbb{R}^{n^2 \times n^2}$, then by multiplying both sides of $(\diamondsuit)$ by $A^T$ from left and by $A$ from right, we have the desired result.