Let $S \subset C \subset \mathbb{R}^d$ be two subsets of $\mathbb{R}^d$, one included in the other. For the sake of simplicity, assume that they are both compact and $\boldsymbol{0}$ belongs to both of their interiors.

My question is the following. If $C$ is convex and $S$ is strictly convex, is it true that for all $\lambda \in [0,1)$, their convex combination $$ (1-\lambda)S + \lambda C = \bigl\{(1-\lambda) \boldsymbol{s} + \lambda\boldsymbol{c} : \boldsymbol{s} \in S, \boldsymbol{c} \in C \bigr\} $$ is strictly convex?

It is known that the Minkowski sum $A+B = \{a+b:a\in A, b\in B\}$ of any two convex sets is itself convex, but my intuition (see pic. below) tells me that (at least under the assumptions above) the strict convexity of just one of them should imply the strict convexity of the convex combination.

enter image description here

  • 1
    $\begingroup$ What happens if you apply definitions? What definition of strict convexity are you using for sets? $\endgroup$
    – supinf
    Mar 6, 2020 at 12:06
  • $\begingroup$ I meant that $S$ is strictly convex iff for all $x,y\in S$ and all $\lambda\in(0,1)$, $(1-\lambda)x + \lambda y \in \mathrm{int}(S)$. $\endgroup$
    – user332582
    Mar 6, 2020 at 15:58

1 Answer 1


This is not true. Take $S=\overline{B_1(0)}$ and $C=[-1,1]^2$. Then for $\lambda\in (0,1)$, the line $$ (-\lambda,\lambda) \times \{1\} $$ is a boundary segment of $(1-\lambda) S + \lambda C$: These points are convex combinations of $(0,1)\in S$ and $[-1,1]\times\{0\} \subset C$. In addition, $(1-\lambda) S + \lambda C$ clearly cannot contain points $(x_1,x_2)$ with $x_2>1$.


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