0

So, I'm trying to see if my approach can show that the set $S$ is not affine and is convex with a similar argument for both cases

$S = \left\{ \alpha \in \mathbb{R}^3 : \alpha_1 + \alpha_2 e^{-t} + \alpha_3 e^{-2t}\le 1.1 \right\} \\ \text{where } t \ge 1 $

My first observation is that, given $x \in S$ I can write it as:

$$ x \cdot \gamma(t) \le 1.1 $$

Where $ \gamma(t)^T = \left[ 1, e^{-t} , e^{-2t} \right]$.

With that, I can write an arbitrary pair $x$ and $y$ from S as the combination:

$$ \left[ \theta x + (1-\theta)y \right]\cdot \gamma(t) $$

If $\theta \in \mathbb{R} $ the combination is affine and if $0 \le \theta \le 1 $ the combination is convex. In my understanding, since $\theta x + (1-\theta)y$ is real and $x,y \in S$ I can bound them by 1.1.

How can I refine my argument to prove that:

  • S is convex when I construct the vector $\theta x + (1-\theta)y$ when $0 \le \theta \le 1 $
  • S is not affine when $\theta \in \mathbb{R} $

The reason for my thinking is that the difference of the conditions upon $\theta$ that helps me see each case, no?

Guilherme Marthe
  • 156
  • 1
    Your set is not well defined. You have put elements $\alpha$ of $\Bbb{R}^3$ in your set, but the expression after the colon does not restrict which $\alpha$ it could be. Should there be an equality or inequality involved? Or should it be $S = {\alpha_1 + \alpha_2 e^{-t} + \alpha_3e^{-2t} : \alpha \in \Bbb{R}^3}$? If so, then this would be equal to $\Bbb{R}$, which is definitely affine. – Theo Bendit Mar 12 '24 at 04:27
  • ooo you are absolutely right. The condition $x \cdot \gamma(t) \le 1.1$ defines the set. I've just edited it. – Guilherme Marthe Mar 20 '24 at 14:59

1 Answers1

1

Your thoughts so far are good. In general, the set $A = \{x \in \Bbb{R}^n : u \cdot x \le c\}$, where $u \in \Bbb{R}^n$ is non-zero and $c \in \Bbb{R}$, is a closed halfspace. Such sets are never affine; take any two vectors whose difference is not perpendicular to $u$, and the line connecting them cannot be contained in $A$.

More specifically, consider the points $x = \frac{cu}{\|u\|^2}$ and $y = \frac{(c-1)u}{\|u\|^2}$. Note that $$u \cdot x = \frac{c}{\|u\|^2}(u \cdot u) = c,$$ and similarly, $$u \cdot y = c - 1 \le c.$$ That is, $x, y \in A$. Now, let $$z = 2x - 1y = \frac{2cu}{\|u\|^2} - \frac{(c-1)u}{\|u\|^2} = \frac{(c+1)u}{\|u\|^2}.$$ Note that $z$ is an affine combination of $x$ and $y$, but $$z \cdot u = c + 1 > c,$$ hence $z \notin A$. Thus, $A$ is not affine.

On the other hand, halfspaces are convex. If we assume $x, y \in A$ are arbitrary, and $\theta \in [0, 1]$, then $$u \cdot (\theta x + (1 - \theta)y) = \theta(u \cdot x) + (1 - \theta)(u \cdot y) \le \theta c + (1 - \theta)c = c.$$ Hence, $\theta x + (1 - \theta)y \in A$ as required.

Theo Bendit
  • 50,900