Suppose that $A\succeq 0,$$0\le y$ and $\|y\|\le \|x_\rho\|, \mu \in \mathbb R^n,$ and $e$ be the vector of ones. My goal is to show $\{x_\rho\}$ is bounded when large enough $\rho$ goes to $+\infty$, where\begin{equation}x_\rho= \frac{1}{2}(A+\rho I)^{-1}\left(\mu +2 \rho y+\frac{1-0.5e^T(A+\rho I)^{-1}(\mu +2\rho y)}{0.5e^T(A+\rho I)^{-1}e}e\right).\end{equation}
Notice that $x_\rho$ is the solution of the following problem:\begin{equation}\min_{x\in \mathbb R^n} x^TAx-\mu^Tx+\rho\|x-y\|^2 \quad \mbox{s.t.} \quad e^Tx=1,\end{equation}which is related to the mean-variance portfolio optimization problem. I am trying to give an ADMM algorithm for solving this problem and that is why I need to prove the boundedness of $\{x_\rho\}$.
My current efforts are as follows:
In virtue of $A\succeq 0$ and $\rho \ge 1,$ we see that$$\|(A+\rho I)^{-1}\|={\lambda_{\max}((A+\rho I)^{-1}})={1}/{{\lambda_{\min}(A+\rho I)}}={1}/{({\lambda_{\min}(A)+\rho)}}\le 1/{\rho},$$and$${n}/{(\lambda_{\max}(A)+\rho)}={\|e\|^2}/{\lambda_{\max}(A+\rho I)}=\|e\|^2\lambda_{\min}((A+\rho I)^{-1})\le {e^T(A+\rho I)^{-1}e}.$$Thus, since ($\|e\|=\sqrt{n}\ge 1$), we have\begin{equation*}\begin{aligned}\left\|\frac{1-0.5e^T(A+\rho I)^{-1}( \tau \mu +2\rho y)}{0.5e^T(A+\rho I)^{-1}e}e\right\|& \le \frac{1+0.5\|e\|\lambda_{\max}((A+\rho I)^{-1})(\tau \|\mu\|+2\rho \|y\|)}{0.5{n}/({\lambda_{\max}(A)+\rho})}\|e\|\\& \le\frac{\|e\|(\lambda_{\max}(A)+\rho)}{0.5n}+{(\lambda_{\max}(A)+\rho)}\|e\|\frac{\|e\|(\tau \|\mu\|+2\rho \|y\|)}{{n(\lambda_{\min}(A)+\rho)}}\\& \le(\lambda_{\max}(A)+\rho)+{(\lambda_{\max}(A)+\rho)}\|e\|\frac{\|e\|(\tau \|\mu\|+2\rho \|y\|)}{{n(\lambda_{\min}(A)+\rho)}}\\& \le{(\lambda_{\max}(A)+\rho)}+{(\lambda_{\max}(A)+\rho)}\frac{(\tau \|\mu\|+2\rho \|y\|)}{{\rho}}\\& \le{(\lambda_{\max}(A)+\rho)}+{(\lambda_{\max}(A)+\rho)}\frac{(\tau \|\mu\|\rho+2\rho \|y\|)}{{\rho}}\\& \le (\lambda_{\max}(A)+\rho)(1+\tau \|\mu\|+2\|y\|).\end{aligned}\end{equation*}Hence, the above and the equation for $x_\rho$ together with $\|y\|=\|H_k(x^+)\|\le \|x^+\|\le \|x\|$ lead to\begin{equation*}\begin{aligned}\|x\| &\le\frac{1}{2} \|(A+\rho I)^{-1}\|(\tau \|\mu\|+2\rho\|y\|+(\lambda_{\max}(A)+\rho)(1+\tau \|\mu\|+2\|y\|)) \\& \le\frac{1}{2\rho}(\tau\|\mu\| + 2\rho \|y\|+(\lambda_{\max}(A)+\rho)(1+\tau \|\mu\|+2\|y\|))\\& \le\frac{1}{2\rho}(\tau\|\mu\| + 2\rho \|y\|+(\lambda_{\max}(A)+\rho)(1+\tau \|\mu\|+2\|y\|))\\& \le\frac{1}{2\rho}(\tau \|\mu\| +(\lambda_{\max}(A)+\rho)(1+\tau \|\mu\|)+2(\lambda_{\max}(A)+2\rho)\|y\|)\\& \le \frac{1}{2\rho}(\tau \|\mu\| +(\lambda_{\max}(A)+\rho)(1+\tau \|\mu\|)+2(\lambda_{\max}(A)+2\rho)\|x\|)\\& \le \frac{1}{2}(\tau \|\mu\| +(\lambda_{\max}(A)+1)(1+\tau \|\mu\|))+\frac{1}{\rho}(\lambda_{\max}(A)+2\rho)\|x\|\end{aligned}\end{equation*}
Unfortunately, the coefficient for $\|x\|$ on the right-hand side is not less than 1, so things do not go through!
I am okay with putting more assumptions on the problem parameters to be able to find a bound for this sequence.