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It is with some degree of reluctance that I ask and answer my own question, but someone may find it curious. It is not particularly difficult to answer, but it took me an inordinate amount of time which I can only attribute to experience.

In Convex Analysis, Rockafellar defines the umbra $U$ and penumbra $P$ of a set $C$ with respect to a set $S$ to be $U=\cap_{x \in S} \{ \lambda y + (1-\lambda) x \mid \lambda \ge 1, y \in C \}$ and $P=\cup_{x \in S} \{ \lambda y + (1-\lambda) x \mid \lambda \ge 1, y \in C \}$.

Show that if $C$ is convex then $U$ is convex and if $S,C$ are convex then $P$ is convex.

A moment's thought shows that the question can be reduced to considering the sets $S=[s_1,s_2]$ and $C=[x_1,x_2]$ and showing that the resulting penumbra $P$ is convex. The notation $[x,y]$ means the convex hull of the two points $x,y$.

copper.hat
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1 Answers1

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Suppose $y_k = \lambda_k x_k + (1-\lambda_k) s_k$ with $\lambda_k \ge 1$ and some $\mu \in (0,1)$. We need to show that $y = \mu y_1 + (1-\mu)y_2 \in P$.

Expanding gives $y = \mu \lambda_1 x_1 + \mu(1-\lambda_1)s_1+ (1-\mu) \lambda_2 x_2 + (1-\mu)(1-\lambda_2)s_2$, note that $\mu \lambda_1 + \mu(1-\lambda_1)+ (1-\mu) \lambda_2 + (1-\mu)(1-\lambda_2) = 1$.

Let $\lambda = \mu \lambda_1 + (1-\mu) \lambda_2$ and note that $\lambda \ge 1$.

If $\lambda = 1$ then we must have $\lambda_1 = \lambda_2 = 1$ and the result is immediate, so assume that $\lambda >1$.

Note that we can write $y = \lambda {\mu \lambda_1 x_1 + (1-\mu) \lambda_2 x_2 \over \lambda} + (1-\lambda) {\mu(1-\lambda_1)s_1+ (1-\mu)(1-\lambda_2)s_2 \over 1-\lambda}$ and ${\mu \lambda_1 x_1 + (1-\mu) \lambda_2 x_2 \over \lambda} \in [x_1,x_2]$ and ${\mu(\lambda_1-1)s_1+ (1-\mu)(\lambda_2-1)s_2 \over \lambda-1} \in [s_1,s_2]$, hence $y \in P$.

copper.hat
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