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Suppose we have a convex set C defined using the following half-space representation: $$ C= \{ x \in \mathbb{R}^{m}: \langle b , x \rangle \geq f(b), \forall b \in \Delta_m\},$$

where $\Delta_m$ is the probability simplex in $\mathbb{R}^{m}$ and $f(.)$ is a concave non-negative valued function defined on $\Delta_m$. Let $ y \notin C$, then there exists a unique projection of $y$ onto $C$ (since C is closed and convex) denoted by $\prod_C(y)$. Then is the following true: $y-\prod_C(y)=-\lambda b_0$ for some $\lambda>0$ and $b_0 \in \Delta_m$ and that $\langle b_0, \prod_C(y) \rangle =f(b_0)$ for some $b_0 \in \Delta_m$? If yes, how to show it rigorously?

Intuitively it feels correct. I know that $y-\prod_C(y)$ is in the normal cone of $\prod_C(y)$ but am unable to proceed further.

Sushant Vijayan
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1 Answers1

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I think this is a correct proof:

First define the extended function $$ \tilde{f}(b)= \begin{cases} f(b) & b \in \Delta_m,\\ -\infty & b \notin \Delta_m. \end{cases}$$

Now, we observe that: $$C= \{ x : \inf_{b \in \Delta_m} \langle b, x \rangle -f(b) \geq 0 \},$$ or $$C= \{ x : \tilde{f}^{*}(x) \geq 0\},$$ where $\tilde{f}^{*}$ denotes the Fenchel or concave conjugate.

Now, from this answer, we know that: $$\partial \tilde{f}^{*}(x)= -\underset{b \in \Delta_m}{\arg\min} \{\langle b,x \rangle -f(b)\}.$$

Now, as the set $C$ has been expressed above as the sublevel set of $\tilde{f}^{*}$, we have from this answer that the Normal cone at a boundary point of $C$ (that is, $x$ such that $\tilde{f}^{*}(x)=0$) is given by $\operatorname{Cone} \partial \tilde{f}^{*}(x)$. But as we have characterized this subdifferential and we know the projection point $\prod_C(y)$ lies on this boundary, we must have: $y-\prod_C(y)=-\lambda b_0$ for some $b_0 \in \Delta_m$. From above, we know that this $b_0$ is the minimizer when evaluating $\tilde{f}^{*}(\prod_C(y))$, and hence must satisfy: $$\langle b_0 ,\prod_c(Y) \rangle = f(b_0).$$

Rócherz
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Sushant Vijayan
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