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Given $x,y \in \mathbb{R}^n$, let´s denote $$ [x,y] = \left\{ {u \in \mathbb{R}^n :u = tx + \left( {1 - t} \right)y\,,\,\,\,0 \leqslant t \leqslant 1} \right\} $$

Let $X , Y$ subsets of $\mathbb{R}^n $; let´s denote $$ X*Y = \bigcup\limits_{x\,\, \in X\,,\,\,y\, \in \,Y} {[x,y]} $$

Prove that if $X , Y$ are convex sets then also is $ X*Y $ . Obviously the only non-trivial case is where we take $u_0,u_1$ with $$ u_i = \alpha _i x_i + \beta _i y_i\,, \quad i = 1,2\,,\quad \alpha _1 + \beta _1 = \,\alpha _2 + \beta _2 = 1\,,\quad x_i \in X\,,\,y_i \in Y $$ and we want to see if for any $ {0 \leqslant t \leqslant 1} $ , $ tu_1 + \left( {1 - t} \right)u_2 \, \in X*Y $, i.e. $$ \exists \,x^* \in X\,,\,y^* \in Y\,,\,0 \leqslant t^* \leqslant 1:\,\,t^* x^* + \left( {1 - t^* } \right)y^* = tu_1 + \left( {1 - t} \right)u_2 $$ Note at least that $$ tu_1 + \left( {1 - t} \right)u_2 = t\left( {\alpha _1 x_1 + \beta _1 y_1 } \right) + \left( {1 - t} \right)\left( {\alpha _2 x_2 + \beta _2 y_2 } \right) \;:$$ it´s a convex combination of $ x_1,x_2,y_1,y_2$ , thus $X*Y$ at least is contained in the convex hull of $ X \cup Y $ But I don´t know how to prove that $X*Y$ is convex

PD: sorry for the blablabla

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  • $\begingroup$ Have you tried showing that the convex hull of $X\cup Y$ is a subset of $X*Y$? $\endgroup$ – Brian M. Scott Mar 4 '12 at 0:11
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Set $A=t\alpha_1+(1-t)\alpha_2$ and $B=t\beta_1+(1-t)\beta_2$. Then $A+B=t+(1-t)=1$ and $$\begin{align}& t\left( {\alpha _1 x_1 + \beta _1 y_1 } \right) + \left( {1 - t} \right)\left( {\alpha _2 x_2 + \beta _2 y_2 } \right) \\=& t\alpha_1x_1+(1-t)\alpha_2 x_2 + t\beta_1y_1+(1-t)\beta_2 y_2 \\=& A\left(\frac{t\alpha_1}{A} x_1+\frac{(1-t)\alpha_2}A x_2\right) + B\left(\frac{t\beta_1}{B}y_1 + \frac{(1-t)\beta_2}B y_2\right) \end{align} $$

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