Given the optimization problem
\begin{align} \boldsymbol{ x}^*, \boldsymbol{y}^* &= \min_{ \boldsymbol{x}, \boldsymbol{y}}( V( \boldsymbol{x}, \boldsymbol{y}))\\ s.t: \quad & \boldsymbol{A x} + \boldsymbol{I y }= \boldsymbol{0} \end{align} where \begin{align} V( \boldsymbol{x}, \boldsymbol{y})=& g( \boldsymbol{x}) + g_y( \boldsymbol{y})\\ g_y( \boldsymbol{y})=& \boldsymbol{y}^T \boldsymbol{y}\\ g( \boldsymbol{x})=& \sum_{i=1}^{n } g_i(x_i), \text{ for all } i \in \{1,\cdots, n\} \end{align}
with $ \boldsymbol{x} \in \mathcal{R} ^{n \times 1} $, $ \boldsymbol{y} \in \mathcal{R} ^{m \times 1} $, $ \boldsymbol{A} \in \mathcal{R} ^{m \times n} $ has full row-rank, $ m<n $ and $ \boldsymbol{I }\in \mathcal{R} ^{m \times m} $ is the identiy matrix, and the problem is strictly convex with one unique minimum (satisfying the KKT conditions).
Given we know an upper bound on $\nabla g_i(x_i) $:
\begin{align}
|\nabla g_i(x_i^*) |& \le c \text{ for all } i \in \{1,\cdots, n\}\\
\implies |\nabla_x g( \boldsymbol{x^*})|_{\infty}& \le c,
\end{align}
can we limit $ |\boldsymbol{y}^*|_{\infty} $?
My idea was to use that the directional derivative of $ V( \boldsymbol{x}, \boldsymbol{y}) $, in direction $ \boldsymbol{d} $, at the solution (where $ \boldsymbol{d} $ is any feasible direction such that $\boldsymbol{Ax } + \boldsymbol{I y }$ remains zero). \begin{equation}\label{key} \boldsymbol{d} \cdot \nabla V( \boldsymbol{x}^*, \boldsymbol{y}^*) = \boldsymbol{d} \cdot {\left[\begin{matrix}\nabla_x g( \boldsymbol{x}^*) \\ \nabla_y g( \boldsymbol{y}^*) \end{matrix} \right]} =0 \end{equation} and then get a limit for $|\nabla g_y(\boldsymbol{y}) |$, which would give a limit on $|\boldsymbol{y}^*|_{\infty}$
But I did not manage to proceed, and am currently stuck. Any help or tips on how to proceed is welcome.