Show that equation has solution (Fixed Point Theorem) Let $A,B\in\mathcal{M}_{n\times n}(\mathbb{R})$ be matrices such that $\|A\|_{F}\cdot\|B\|_{F}<\frac{9}{16}$, where $\|\cdot\|_F$ denotes the Frobenius norm. Show that in $B[0_{n\times n},1/2]$ (closed ball) the equation:
$$AX^{2}B-\frac{1}{3\sqrt{n}}I=X$$
has a solution using the Fixed Point Theorem.
I defined the aplication $X\mapsto\varphi(X)=AX^{2}B-\frac{1}{3\sqrt{n}}I$ to use the fixed point theorem. The ball is closed and I already prove that $B[0_{n\times n},1/2]\subseteq\varphi(B[0_{n\times n},1/2])$, then I just need to show that it is a contracting function:
Let $X,Y\in B[0_{n\times n},1/2]$, then:
$\|\varphi(X)-\varphi(Y)\|_F=\|AX^{2}B-AY^{2}B\|_F\leq\|A\|_{F}\|X^2-Y^2\|_F\|B\|_F<\frac{9}{16}\|X^2-Y^2\|_F$
I don't know how to make appear $\|X-Y\|_F$. Any hint?
 A: $$
\begin{aligned}
&\|AX^2B-AY^2B\|_F\\
&=\|A\,\left[\,X(X-Y)+(X-Y)Y\,\right]\,B\|_F\\
&\le\|A\|_F\|B\|_F\,\left(\|X\|_F+\|Y\|_F\right)\,\|X-Y\|_F\\
&\le\left(\frac{9}{16}\right)^2\left(\frac{1}{2}+\frac{1}{2}\right)\|X-Y\|_F.
\end{aligned}
$$
A: Let $X \in B(0, 1/2)$ be arbitrary. For all $H \in M(n \times n, \mathbb{R})$,
$$D\phi(X)H = A(HX + XH)B \leq 2\|A\|\|X\|\|B\|\|H\|.$$
So
$$\|D\phi(X)\| \leq 2\|A\|\|B\|\|X\| \leq 9/16.$$
Since $9/16 < 1$, it follows from the fundamental theorem of calculus that $\phi$ is a contraction.
A: Let $M=\max\{\|A\|_F,\|B\|_F\}$. We will show that the mapping $T(X)=AX^2B-\frac{1}{3\sqrt{n}}I$ is a contraction in $B([0_{n\times n},\frac{1}{2}])$, and this will imply that $T$ has a unique fixed point in $B([0_{n\times n},\frac{1}{2}])$, which will be the solution to the equation $AX^2B-\frac{1}{3\sqrt{n}}I=X$.
Given $X,Y\in B([0_{n\times n},\frac{1}{2}])$, we have
$$\|T(X)-T(Y)\|_F\leq\|A\|_F\|X^2-Y^2\|_F+\|B\|_F\|X-Y\|_F<\frac{1}{2}\|X+Y\|_F+\frac{1}{2}\|X-Y\|_F=\|X-Y\|_F$$
So, $T$ is a contraction, and by the Contraction Mapping Theorem, it has a unique fixed point in $B([0_{n\times n},\frac{1}{2}])$, which is the solution to the equation $AX^2B-\frac{1}{3\sqrt{n}}I=X$.
