Problem
Given \begin{equation}\label{4} \boldsymbol{x} = \boldsymbol{y } \boldsymbol{A} \quad \text{(i.e., Elementwise equality)} \end{equation}
with $ | \boldsymbol{x}|_{\infty} \le c $, $ \boldsymbol{x} \in \mathbb{R}^{1 \times n}, \boldsymbol{y} \in \mathbb{R}^{ 1 \times m}, \boldsymbol{A} \in \mathbb{R}^{m \times n} $, and $ \boldsymbol{A} $ full row-rank.
Find a bound on the form: \begin{equation}\label{5} |\boldsymbol{y} | _{\infty} \le f( c, \boldsymbol{A}) \end{equation}
My thoughts (feel free to ignore if you have another approach)
1: Multiply both from the right $ \boldsymbol{H} $, where $ \boldsymbol{H} $ is any generalized inverse of $ \boldsymbol{A} $ such that $ \boldsymbol{A } \boldsymbol{H} \boldsymbol{z} = \boldsymbol{ z}\text{ for all } \boldsymbol{z} \in \mathbb{R}^{m\times 1} $.
\begin{equation}\label{key} \boldsymbol{x} \boldsymbol{H} = \boldsymbol{y} \end{equation}
2: Now let $ \boldsymbol{p} \in \mathbb{R}^m$ be the $ k^{th} $ coordinate vector, that only has one nonzero element $p_k= 1 $. Multiplying both sides with $ \boldsymbol{p}$ and taking the absolute norm, we get \begin{equation}\label{1} |\boldsymbol{x} \boldsymbol{H} \boldsymbol{p} |= |{y} _{i}| \end{equation}
3: Use \begin{equation}\label{2} | \boldsymbol{x} \boldsymbol{H} \boldsymbol{p}| \le |\boldsymbol{x}|_{\infty} |\boldsymbol{H} \boldsymbol{p} |_{1} \le |\boldsymbol{x}|_{\infty} ||\boldsymbol{H} ||_{1} \end{equation} Where $ ||\cdot||_{1} $ is the matrix induced norm (since $ | \boldsymbol{p}|_1 = 1 $).
\begin{equation}\label{a} \implies {y}_i \le||\boldsymbol{H} ||_{1} |\boldsymbol{x}|_{\infty} \end{equation}
4: Since it must hold for any generalized inverse we can replace $ \boldsymbol{H} $ with the generalized inverse $ \boldsymbol{H}_{\nu1} $ of $\boldsymbol{A}$ that minimize the matrix induced norm of $ \boldsymbol{H} $.
\begin{equation}\label{3} {y}_i \le|| c \boldsymbol{H}_{\nu1} ||_{1} \end{equation}
\begin{equation}\label{3s} \implies |\boldsymbol{y} | _{\infty} \ \le c|| \boldsymbol{H}_{\nu1} ||_{1} \end{equation}
For simplicity, one may use the pseudo-inverse instead, i.e., $ \boldsymbol{H}= \boldsymbol{A}^{\dagger} $ to get
\begin{equation}\label{3ss} \implies |\boldsymbol{y} | _{\infty} \ \le c|| \boldsymbol{A}^{\dagger} ||_{1} , \end{equation} which will yield a higher bound
To me it seems like the above holds. However.
-I am in general interested in other strategies that (especially if they were to yield lower bounds).
-If possible, it would be nice to get an simpler answer avoiding the use of the matrix induced norm on the inverse.
Any input is appreciated.