I've encountered an odd dual function for my problem. It seems non-concave, which of course cannot be true, but the construction also seems straightforward. I'll describe a bare-bones version of the problem I'm considering:
\begin{align} &\min{A\in \mathbb R^{m\times n}~,t\in \mathbb R}&&t\\ &\text{s.t.}&& \Vert e_i^T A\Vert_2^2\le t&&\forall i\in[m]\\ &&& AW = J\\ \end{align} where $e_i$ is the $i$th standard basis vector, $W\in \mathbb R^{d\times m}$ is a matrix fixed in advanced, and $J\in \mathbb R^{m\times d}$ is the all-ones matrix. So, the Lagrangian is just: \begin{align} L(A,t; \beta, \Gamma)&=t\left(1-\sum_{i}\beta_i\right) + \langle \text{diag}(\beta), AA^T\rangle + \langle \Gamma, J-AW\rangle, \end{align} where $\beta \ge 0$. It's clear that we'll need $\sum_i \beta_i=1$, so assume that to be the case. Now we just want to describe $g(\beta, \Gamma)=\min_{A}L(A,t;\beta, \Gamma)$. Since $L$ is convex with respect to $A$, we can just take the derivative. Notice that we will need to have each $\beta_i>0$, since otherwise we can take the norm of $e_i^TA$ to be arbitrarily large, making the inner product $\langle \Gamma, A\rangle$ approach $-\infty$. We have \begin{align} \frac{\partial }{\partial A}L= 2\text{diag}(\beta)A -\Gamma W^T, \end{align} implying that we should set $A=\frac{1}{2}\text{diag}(\beta)^{-1}\Gamma W^T$. Then, we have \begin{align} g(\beta, \Gamma)&=\frac{1}{4}\langle\text{diag}^{-1}(\beta), \Gamma W^TW\Gamma^T\rangle + \langle\Gamma, J\rangle - \frac{1}{2}\langle\text{diag}^{-1}(\beta), \Gamma W^TW\Gamma^T\rangle \\ &=\langle \Gamma, J\rangle - \frac{1}{4}\Vert\text{diag}^{-1/2}(\beta)\Gamma W^T\Vert_F^2. \end{align} But, this is not concave with respect to $\beta$ and $\Gamma$. Am I interpreting this incorrectly, or is there another way I should go about taking the dual of this problem?