Given following set $S = \{ \alpha \in \mathbf{R}^3 \mid \alpha _1 + \alpha _2e^{-t} + \alpha _3e^{-2t} \leq 1.1 \; \mbox{for}\; t \geq 1\}$. I more or less can prove and understand why it is not affine and why it is convex, but I can't prove why it is not polyhedron? This is because it is not linear inequality?
Thanks in advance.
Update: Don't quite understand why this question should be closed, so additional details: I take edx course provided by Stanford, authored by Stephen Boyd, convex optimization. I did this question wrong and because of lack of explanation I decided to ask here. I have hard times to prove that set $S$ is not polyhedron.
Best Answer
Here is a very tedious answer. I imagine there is a much slicker solution, but it escapes me.
Note that we can write $S = \{ x| (1,t,t^2)^T x\le {11 \over 10}, t \in (0,1] \}$. Since $S$ is the intersection of closed halfplanes it is convex and closed.
Let $S_0 = \{ x \in S | x_1 = 0 \}$ and note that if $S$ was polyhedral then $S_0$ would be too. Hence it suffices to show that $S_0$ is not polyhedral.
Just to reduce noise (I am switching use of $x$ here), let $S_0' = \{ (x,y)| tx+t^2 y \le 1.1, t \in (0,1]\} $.
Note that if $(x,y) \in S_0'$ then $(x-h,y) \in S_0'$ for all $h \ge 0$. Furthermore there is some $l>0$ such that $(x+l,y) \notin S_0'$. In addition, for any $y$ there is some $x$ such that $(x,y) \in S_0'$. Hence we can characterise $S_o'$ by computing $f(y) = \max_{(x,y) \in S_0'} x$ (the $\max$ exists because $s_0'$ is closed) and write $S_0' = \{(x,y) | x \le f(y) \}$.
We can write $tx+t^2y \le 1.1$ as $x \le {1.1 \over t} - ty$ and so we see that $f(y) = \inf_{t \in (0,1]} ({1.1 \over t}-t y)$.
If $y \ge 0$ then $t \mapsto {1.1 \over t}-t y$ is decreasing and so $f(y) = 1.1-y$.
If $y < 0$ then $t \mapsto {1.1 \over t}-t y$ is unimodal on $(0,\infty)$ and has a unique $\min$ in $t^* = \sqrt{1.1 \over -y}$.
In particular, for $y \ge - 1.1$, $f(y) = 1.1-y$ and for $y < -1.1$ we have $f(y) = 2 \sqrt{-1.1y}$.
It is straightforward to show from this that $S_0'$ is not polyhedral.