I have a convex set $ A \subset \mathbb{R^n}$ and $K= \{x \in \mathbb {R^n}\mid \exists t>0( t^{-1} \text { for } x \in A)\}.$
I want to show, that $K$ is convex.
My idea:
$y \in K \Rightarrow \exists t>0: t^{-1} y \in A$ and $z \in K \Rightarrow \exists t>0: t'^{-1} z \in A$.
Thus $A$ convex, there is a $t''>0: t''^{-1}( \lambda t^{-1} y + (1- \lambda) t'^{-1} z) \in A $ with $ \lambda \in [0,1]$.
How can I argue now, that this convex combination is in $K$?
The best way to do things, is to note that if $a \in A$, then every element of the form $at, t > 0$ belongs in $K$, since $(at)t^{-1} = a \in A$. Conversely, if $l \in K$, then for some $t > 0$ we have $lt^{-1} = b \in A$, or that $l=tb$ for some $b \in A$ and $t > 0$.
All this tells us that $K = \{at : a \in A, t > 0\}$. If $A$ is convex, we claim this is convex.
With this description, indeed if $at \in K$ and $bt' \in K$, then if $\lambda \in [0,1]$ we want to show that $l = at\lambda + bt'(1-\lambda) \in K$.
For this, note that $t \lambda \geq 0$ and $t'(1-\lambda) \geq 0$, with $k=t\lambda + t'(1-\lambda) > 0$, since one of $\lambda,1-\lambda$ is strictly positive.
Then, $$l = \frac{l}{k}k = k\left(a\frac{t\lambda}{k} + b \frac{t'(1-\lambda)}{k}\right)$$
is the product of $k>0$ and a convex combination of $a$ and $b$ which lies inside $A$ since it is convex. By definition of $K$, we get that $l \in K$, or that $K$ is convex.