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Is the set of polynomials satisfying $ \{a \in \mathbf{R}^k | p(0) = 1, |p(t)| \leq 1 \text{ for } \alpha \leq t \leq \beta \}$, where $\{ p(t) = a_1 + a_2t + \dots + a_k t^{k-1} \}$, convex?

How to show that? Any help or hints is appreciated.

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    It seems to me a question is asking whether the set is convex or not, not a polinomials themselves. Am I wrong? – 2017-02-11
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    I think you are on the wrong track. The problem is about showing a *set of polynomials* is convex, that a convex combination of two vectors in this set is again a polynomial belonging to that set. Typically this is an exercise in using the *triangle inequality* on those polynomials. – 2017-02-11

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Points in this set $S$ are just polnomials $p,q\in S$ that individually satisfy $|p(t)|\leq 1$, $|q(t)|\leq 1$ for $\alpha\leq t\leq \beta$, and $p(0) = q(0) = 1$.

Let $a+b = 1$, and $a,b\geq 0$. We thus have that $ap(t)+bp(t)$ is a convex combination of elements of $S$. If this is in $S$, then $S$ is convex.

There are two things to verify (which I'll hide by default to encourage you to try proving it): $$|ap(t)+bp(t)|\leq 1,\quad \alpha\leq t\leq \beta$$

This will just use the triangle inequality. We have $|ap(t)+bq(t)|\leq |a||p(t)|+|b||q(t)|\leq |a|+|b| = 1$, so this combination is less than $1$ for all $t\in[\alpha,\beta]$

$$ap(0)+bq(0) = 1$$

Here, we just note that $p(0) = q(0) = 1$, so $ap(0)+bq(0) = a+b = 1$ by our choice of $a,b$.

If you can show both of these are true for generic $p,q\in S$, then you'll be done.

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    there are some typos where $p(t)$ should be $q(t)$. – 2017-02-12