= Solution
\b[<Absolute values> are missing in the printed formula.] For signed real blocks, the intended <block mixed norm> must be
$$
\|v\|_{q,1}=\sum_{i,j}\left(\sum_{r=1}^d|v^r_{i,j}|^q\right)^{1/q},\qquad q\ge1.
$$
Without them, at $q=1$ a block $(-1,1)$ has value zero although it is nonzero, and a negative <scalar> has negative value. For nonintegral $q$, a negative component may not even have a real power. The literal expression therefore is not a <norm> and does not define the claimed general <convex> problem. Even <integer> exponents do not fix all other $q$ in the stated range.
With the corrected <norm>, let $q'$ be the <Holder conjugate exponent>: $q'=q/(q-1)$ for $q>1$ and $q'=\infty$ for $q=1$. Blockwise <Holder inequality>, with equality from a norming block, gives
$$
\lambda\|Au\|_{q,1}=\sup_{p\in P_{\lambda,q'}}\langle Au,p\rangle,\qquad P_{\lambda,q'}=\{p\in X^d:\|p_{i,j}\|_{q'}\le\lambda\text{ for all }i,j\}.
$$
The <duality of block mixed norms> identifies the closed <compact> <convex set> and the projection residual from the <Moreau decomposition>
$$
\boxed{C= A^*P_{\lambda,q'},\qquad u=g-P_Cg.}
$$
For $q=2$, the dual blocks are <Euclidean balls>; for $q=1$, they are cubes given by componentwise bounds $|p^r_{i,j}|\le\lambda$. No injectivity or surjectivity of $A$ is required: <compactness> of the block product makes its <linear> image <compact>, and the <strictly convex> fidelity still gives a unique primal minimizer. For general $q'$, the Euclidean projection onto a dual block ball is not generally obtained by radial scaling, so the special clipping formula from the isotropic case should not be reused without further analysis.
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