How can I calculate the proximal operator of mixed norm $ {L}_{\infty,1} $ for any general matrix, $X\in R^{m\times n}$ i.e.,
$$ {X}^{\ast} = \arg \min_X {\left\| X \right\|}_{\infty, 1} + \frac{1}{2 \tau} {\left\| X - Y \right\|}_{F}^{2} $$
Where $ {\left\| X \right\|}_{\infty, 1} = \max \left( {\left\| {X}_{i,:} \right\|}_{1}, \forall i = 1, \dots, m \right) $
Here is a partial answer. Let us write $x_i$ and $y_i$ for rows of $X$ and $Y$. Then,
\begin{align} \min_X \| X\|_{\infty,1} + \frac1{2\tau} \| X - Y\|_F^2 &= \min_X \Big[ \max_i \| x_i\|_1 + \sum_j \frac1{2\tau} \| x_j - y_j\|_2^2 \Big] \\ &= \min_X \max_i \Big[ \| x_i\|_1 + \sum_j \frac1{2\tau} \| x_j - y_j\|_2^2 \Big]\\ &\ge \max_i \min_X \Big[ \| x_i\|_1 + \sum_j \frac1{2\tau} \| x_j - y_j\|_2^2 \Big] \\ &= \max_i \Big[ \| S_\tau(y_i)\|_1 + \frac1{2\tau} \| S_\tau(y_i) - y_i\|_2^2 \Big] \end{align} where $S_\tau$ is the soft-thresholding operator (the proximal operator associated with the $\ell_1$ norm), and the inequality is by weak duality. Let $i_*$ be an index that is the maximizer of the last expression. Let $Y^*_{i,:} = y_i$ if $i\neq i_*$ and $Y^*_{i,:} = S_\tau(y_{i_*})$. If $\|y_i\|_1 \le \|S_\tau(y_{i_*})\|_1$ for $i \neq i_*$, then $Y^*$ is the desired solution (if I haven't made mistakes!) What happens otherwise seems to be more complicated.