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I have the primal least-squares problem

$$\min_{w \in \mathbb R^p} \quad \frac{1}{2} \| y - Xw \|_2^2 + \sum_{i=1}^{d} h_i (w_i)$$

where $w_i$ are partitions of $w$, $w_i \in \mathbb R^{p_i}$, and $X_i$ denotes the corresponding columns in $X$. Let

$$h_i(w_i) = \max_{v \in D_i} \langle v, w_i \rangle$$

where set $D_i \subseteq \mathbb R^{p_i}$ is convex and closed. I need to prove the dual of this primal is a best approximation problem, namely,

$$\min_{u \in \bigcap_{i=1}^d C_i} \| y - u \|_2^2$$

where $C_i$ are inverse image of $D_i$ under ${X_i}^T$, or ${X_i}^T c \in D_i, c \in C_i$.

I don't know exactly how to convert a set constraint to a dual so I have no idea how to tackle this. Any help is appreciated. Thanks!

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  • $\begingroup$ Usually one considers the dual of a problem where there are equality and inequality constraints. However I don't see any constraints in your primal (except that each $h_i(w_i)$ must be finite). $\endgroup$ – Gabriel Romon May 21 at 8:41
  • $\begingroup$ Yes that's the part also confuses me. I guess this may not be a direct dual conversion. Before that one may need to introduce a slack variable for $h_i$. My thought is replace $h_i$ with a slack variable $t_i$ and add in unequality constraint $t \geq <v, w_i>$ for all $v \in D_i$. Then one can apply the dual transform. That doesn't seem to solve the problem though. $\endgroup$ – JC Wang May 21 at 20:53
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I found the solution of this. One needs to introduce two set of slack variables. Let $z = Xw$ substituting in the square part, $w^{'}_i = w_i$ substituting in the $h_i$ part. Then one can introduce the corresponding Lagrange multipliers $u$ and $v_i$. Note that the $h_i$ here is the support function of set $D_i$, whose conjugate function is the indicator function $I_{D_i}$. Take derivative wrt to $z$ yields result of $u = y - Xw$, take derivative wrt to $w^{'}_i$ yields $v_i \in D_i$ and take derivative wrt to $w_i$ yields $X_i^T u = v_i$. This concludes the proof.

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