Any convex polynomial with degree $\geq 2$ is strictly convex.

Any convex polynomial $p(x)$ with degree $\geq 2$ is strictly convex.

clearly, $p(x)$ should be of even degree. Then $p''(x)\geq 0$. How to show $p$ is strictly convex then?

Solution 1:

$(I).$ Consider a polynomial $q(x)$ of degree $\ge 2.$ Suppose $a<b<c$ and $q(a)=q(b)=q(c)=0.$ Now $q$ is not constantly $0$ on $(a,b)$ because deg($q)\ge 2$. So take $x\in (a,b)$ with $q(x)\ne 0.$ By the MVT there exist $x',x''$ with $a<x'<x<x''<b$ such that $$0\ne \frac {q(x)}{x-a}=\frac {q(x)-q(a)}{x-a}=q'(x')$$ and $$0\ne\frac {-q(x)}{b-x}=\frac {q(b)-q(x)}{b-x}=q'(x'').$$ Now one of $q'(x'),q'(x'')$ is $>0.$ So let $x'''\in \{x',x''\}$ such that $q'(x''')>0.$

Similarly there exists $y'''\in (b,c)$ with $q'(y''')<0.$

By the MVT there exists $z\in (x''',y''')$ with $$q''(z)=\frac {q'(y''')-q'(x''')}{y'''-x'''}<0$$ so $q$ is not convex.

$(II).$ Suppose by contradiction that $p$ is a convex polynomial of degree $\ge 2$ but $p(x)$ is not strictly convex. Then there exist $a,b,c$ with $a<b<c$ and $p(b)=rp(a)+(1-r)p(b)$ where $r=\dfrac {c-b}{c-a}.$

Let $\ell (x)$ be linear with $\ell (a)=p(a)$ and $\ell(c)=p(c).$ Then $\ell (b)=p(b).$ Now $q(x)=p(x)-\ell(x)$ is convex with deg$(q)\ge 2$ and $q(a)=q(b)=q(c)=0,$ contrary to $(I).$

Remarks. (1). If deg($p)\le 1$ then the $q$ of $(II)$ is constantly $0$ so $(I)$ would not apply. (2). A direct proof by the same method but without using $\ell$ and the $q$ of $(II)$ is possible but with a lot more writing.