Proving $K= \{x \in \mathbb R^n: \exists t>0 (t^{-1}, \text { for } x \in A\},$ given convex set $A\subset \mathbb R$

Question

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 $K$?

What does ($t^{-1}$ for $x\in A$)? – Mostafa Ayaz Jan 18 '19 at 16:10 — Mostafa Ayaz, Jan 18 '19 at 16:10
It should be $..\exists t>0 : t^{-1} \cdot x \in A$ – Sven Jan 18 '19 at 16:12 — Sven, Jan 18 '19 at 16:12
@SvenMath so you should have $ty$ and $tz$ then? – LinAlg Jan 18 '19 at 16:16 — LinAlg, Jan 18 '19 at 16:16

score 2 · Accepted Answer · answered Jan 18 '19 at 16:19

2

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.

answered Jan 18 '19 at 16:19

Sarvesh Ravichandran Iyer

76,105

Thank you for your proof. I need some time to unterstand what you have done. – Sven Jan 18 '19 at 16:37
You are welcome. I will edit my answer if required. – Sarvesh Ravichandran Iyer Jan 18 '19 at 16:38
No you musn't do that:) – Sven Jan 18 '19 at 16:39
Ok , I will not do so. – Sarvesh Ravichandran Iyer Jan 18 '19 at 16:40
Why is the last term a convex combination of a and b. Do you consider $ \frac{ t \lambda}{k} \in [0,1] \ and \ \frac{ t'(1- \lambda)}{k} \in [0,1] $ – Sven Jan 18 '19 at 16:48
Exactly, that is the case. The sum of those terms is $1$. – Sarvesh Ravichandran Iyer Jan 18 '19 at 16:49
I see. Then you can use the condition, that A is convex and because elements of K have the form $ at$, you get the result. – Sven Jan 18 '19 at 16:52
Thank you again:) I understand your proof:) That was very helpful:) – Sven Jan 18 '19 at 16:53
Brilliant. Imagine $A$ to be some convex set. Try to imagine what $K$ would be like. I think the set $K$ is called a convex cone : it is a sort of "convex conification" of $A$, if you like. – Sarvesh Ravichandran Iyer Jan 18 '19 at 16:55
Yeah, that's a good point. If $x \in K$, then $\lambda x \in K$ for $ \lambda >0$. That is a convex cone as you said. The tricky part was to consider the form of elements in K not in A:) – Sven Jan 18 '19 at 16:59

Proving $K= \{x \in \mathbb R^n: \exists t>0 (t^{-1}, \text { for } x \in A\},$ given convex set $A\subset \mathbb R$

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