Dear Convex Analysis Experts,
Is $\left\{a \in \mathbb R^K \: | \: f(0)=1, \: \lvert f(x)\rvert \leq 1 \text{ for } \alpha \leq x \leq \beta\right\}$, where $f(x) = \sum_{i=0}^{K-1} a_i x^i$, convex?
If yes or no, then why? In general, how to analyze whether a set is convex or not? (probably not a straight-forward answer).
Thank you.
Yes it is. It is the standard proof of checking definitions.
Take $a,b$ in your set, and let $\lambda \in [0,1]$ be a parameter. Define $f_c(x):=\sum_{i=0}^{K-1}c_ix^i$. Then note that $f_c(x)$ is linear in the subscript $c$, and in particular $$ f_{\lambda a + (1-\lambda)b}(x)= \sum_{i=0}^{K-1}\lambda a_ix^i + (1-\lambda)b_ix^i = \lambda f_a(x) + (1-\lambda)f_b(x),$$ so that for $\|F\|:=\sup_{x\in[\alpha,\beta]}|F(x)|$,
$$ \|f_{\lambda a + (1-\lambda)b} \| \leq \lambda \|f_a\| + (1-\lambda)\|f_b\| \leq 1.$$ Also, $f_{\lambda a + (1-\lambda)b}(0) = 1.$
Therefore the line segment $[a,b]$ is contained in your set for any two points $a,b$ already in the set; this is called convex.