Let $A \in \mathcal M(n \times n; \mathbb R)$ with $\rho(A) < 1$. Suppose $X_1$ is the solution to following Lyapunov matrix equation \begin{align*} A^T X_1 A + Q_1 = X_1, \end{align*} where $Q_1 \in \mathcal M(n \times n; \mathbb R)$ and no particular structure of $Q_1$ is assumed. Suppose $X_2(t)$ is the solution to the Lyapunov matrix equation \begin{align*} A^T X_2 A + tQ_2 = X_2, \end{align*} where $Q_2 \succ 0$ is positive definite and $t$ is a free parameter.
I am wondering whether there exists some choice of $t$ such that $\|X_1\|_F \le \|X_2\|_F$? I think the choice of $t$ should exist since if $t \to \infty$, $X_2$ should be unbounded in the entries while $X_1$ should have a fixed norm. I am interested in getting conditions on $t$ in terms of $Q_1, Q_2$ (without $A$) if possible such that $\|X_1\|_F \le \|X_2\|_F$.
We stack the matrices row by row, and we put $q_i=vec(Q_i),x_i=vec(X_i)$. Then $x_1=(I-A^T\otimes A^T)^{-1}q_1,x_2=t(I-A^T\otimes A^T)^{-1}q_2$.
${||X_1||_F}^2={x_1}^Tx_1=tr(q_1^T(I-A\otimes A)^{-1}(I-A^T\otimes A^T)^{-1}q_1)=tr(q_1q_1^T(I-A\otimes A)^{-1}(I-A^T\otimes A^T)^{-1})\leq \sqrt{tr((q_1q_1^T)^2)}\sqrt{tr(((I-A\otimes A)^{-1}(I-A^T\otimes A^T)^{-1})^2)}$
Let $\sigma_1\leq\cdots\leq \sigma_n$ be the singular values of $I-A\otimes A$; they are $>0$.
then ${||X_1||_F}^2\leq \sqrt{tr((q_1q_1^T)^2)}\sqrt{\sum_i{\sigma_i}^{-4}} \leq \sqrt{tr((q_1q_1^T)^2)}\dfrac{\sqrt{n}}{{\sigma_1}^2}$.
${||X_2||_F}^2={x_2}^Tx_2=t^2{q_2}^T(I-A\otimes A)^{-1}(I-A^T\otimes A^T)^{-1}q_2$.
Let $(e_i)$ be an othonormal basis of eigenvectors of $(I-A\otimes A)(I-A^T\otimes A^T)$ associated to the eigenvalues $({\sigma_i}^2)$ and $q_2=\sum_i\alpha_ie_i$. Then ${||X_2||_F}^2=t^2\sum_i\dfrac{\alpha_i^2}{\sigma_i^2}\geq t^2\dfrac{\alpha_1^2}{\sigma_1^2}$.
Conclusion. We obtain (for $t$) a condition that does not depend on $A$ under the condition that $\alpha_1^2$ has a positive lower bound.