Is $[0,1]\cdot A$ a convex set? Let $A\subset\mathbb{R}^d$ be a convex set. We define \begin{align}[0,1]\cdot A:=\{x\in\mathbb{R}^d:x=\lambda a \text{ for a }\lambda\in[0,1] \text{ and }a\in A\}.\end{align} Is this set convex?
I was thinking that in the case $d=1$ the statement is true, because $A$ is an interval and, hence, $[0,1]\cdot A$ is an interval too. But is it also true in higher dimension $d>1$?
I want to say, this is not a homework. I want to show that a specific problem in the theory of optimal control can be solved by considering a replacement problem.
 A: Let $x=\lambda a$ and $y=\mu b$ with $a,b\in A$ and $\lambda,\mu\in[0,1]$.
You want to prove that $tx+(1-t)y\in [0,1]A$ for all $t\in[0,1]$.
Now set $$\eta=\lambda t+\mu(1-t).$$
$\eta$ is  in $[0,1]$.
and set $$s=\frac{\lambda t}{\lambda t+\mu(1-t)}$$
also $s$ is in $[0,1]$.
With these constants you have
$tx+(1-t)y=s\eta a+(1-s)\eta b$
since $\eta A$ is convex then $s\eta a+(1-s)\eta b\in \eta A$ for all $s\in[0,1]$, so $tx+(1-t)y=s\eta a+(1-s)\eta b\in [0,1]A$, as wanted.
A: For a geometric approach, let $x = \lambda a \in [0,1] \cdot A$ and $ y = \mu b \in [0,1] \cdot A$.
Consider the triangle $T$ determined by $a$, $b$ and the origin. As $A$ is convex, the edge between $a$ and $b$ lies completely in $A$. Hence for any point $z$ on the edge between $a$ and $b$ the line between $z$ and the origin lies completely in $[0,1] \cdot A$. Therefore, the whole triangle T lies in $[0,1] \cdot A$, and in particular the line between $x$ and $y$ lies in $[0,1] \cdot A$, which means that $[0,1] \cdot A$ is convex.
