Is the set $\{a \in \mathbb{R}^k : p(0) = 1, |p(t)| \le 1$ for $0 \le t \le 1 \},$ where $p(t) = a_1 + a_2t + ... + a_kt^{k-1}$ convex?

Question

I'm having some trouble with this problem. I tried visualizing the problem first but wasn't sure if I was right. Since $|p(t)| \le 1$, I know that $1 \le p(t) \le 1$. So the function is always between 1 and 1. Would the set be convex because of this? Any help/hints would be appreciated.

Thanks

Answer 1

Here is one way; write the constraints as linear functions.

Let $p_t(a) = \sum_k a_k t^{k-1}$. Note that for a fixed $t$, the map $a \mapsto p_t(a)$ is a linear map.

The first constraint is $p_0(a) = 1$.

Fix some $t \in [0,1]$, then note that $|p_t(a)| \le 1$ is equivalent to $p_t(a) \le 1$ and $-p_t(a) \le 1$.

Finally, write the set as $\{ a | p_0(a) = 1 \} \cap_{t \in [0,1]} \{ a | p_t(a) \le 1 \} \cap_{t \in [0,1]} \{ a | -p_t(a) \le 1 \} $.

Since all sets involved are convex, the intersection is convex.

Answer 2

The convexity follows from basic definitions. Let $a, b \in \mathbb{R}^k$ be from the set and let $0\leq \lambda \leq 1$, we need to show that $(1-\lambda) a + \lambda b$ is also from the set in question. Define $p_a(t) = a_1 + a_2 t + ... + a_k t^k$, then $$ p_{(1-\lambda) a + \lambda b} (t) = (1-\lambda) p_a (t) + \lambda p_b(t). $$ Hence, in view of the choice of $a$ and $b$ we get $p_{(1-\lambda) a + \lambda b}(0) = (1-\lambda)p_a(0) + \lambda p_b(0) = 1$, and $|p_{(1-\lambda) a + \lambda b}(t)| \leq (1-\lambda) |p_a(t)| + \lambda |p_b(t)| \leq 1 - \lambda + \lambda = 1$ for all $0\leq t \leq 1$, proving the convexity.