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Suppose $X$ and $Y$ are convex sets in $\mathbb{R}^d$ such that the origin is in each of their interiors. Then the dual of $X$, $X'$ is defined as the set of linear functionals $\alpha$ such that $\alpha(x) \leq 1$ for all $x \in X$. Is there a way to derive $(X+Y)'$ form $X'$ and $Y'$? I was thinking $(X+Y)' = X' \cap Y'$ but have been unable to prove this and am not sure if it is true.

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  • $\begingroup$ The guess is refuted by taking $X=Y$. I don't know the actual answer (and there might not be a simple formula.) $\endgroup$ – Noam D. Elkies May 1 '16 at 14:31
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Disappointingly, there is no simpler expression than the definition itself. in Combinatorial Convexity and Algebraic Geometry by Günter Ewald, page 105, the author states:

The polar body $(K+L)^*$ of a sum of convex bodies has, in general, no plausible interpretation in terms of $K^*, L^*$. Only in the case of direct sums do we present such an interpretation.

Namely: if $K$ and $L$ both contain $0$ and are contained in linear subspaces $A,B$ with $A\cap B=\{0\}$, then $(K+L)^*$ is the convex hull of $K^{*_A}\cup L^{*_B}$ where the subscripts on asterisks mean that the polar is taken within the indicated subspace.

An example: the polar of the Minkowski sum of line segments is the convex hull of (different) line segments: this matches the description of the unit ball of $\ell^1$.

The above does not apply in your case since you assume nonempty interior.

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