Is function $f(x)=\underset{t\in[a,b] }{\sup}p(t)-\underset{t\in[a,b] }{\inf}p(t)$ convex?

Question

Let

$$f(x) := \sup_{t \in[a,b]} p(t) - \inf_{t \in [a,b]} p(t)$$

where

$$p(t) := x_1 + x_2 t + x_3 t^2 + \cdots + x_n t^{n-1}$$

and $a, b \in \Bbb R$ and $a \lt b$. Is function $f$ convex?

Answer 1

Yes, $f$ is convex. Let us write $p(t)$ instead as $p(x;t)$ to make the dependence on $x$ and $t$ explicit. Then, for fixed $t$, $p(x;t)$ is linear in $x$. Further note that for every $x$, $\sup_t p(x;t) < \infty$ (by compactness of $[a,b]$) and finally conclude that $\sup_t p(x;t)$ is convex in $x$, since the supremum of convex (in particular linear) functions is also convex.

Handling the term $-\inf_t p(x;t)$ is analogous upon writing it as $\sup_t\{-p(x;t)\}$ and noting that $-p(x;t)$ is also linear in $x$ for any fixed $t$.