My professor went over the following theorem
Consider a univariate polynomial $p(x)$. Then,
- If $p(x)$ has degree $2d$, then $p(x)$ is nonnegative on $[-1,1]$ if and only if there exists sum-of-squares polynomials $s_1(x)$ of degree $2d$ and $s_2(x)$ of degree $2d-2$ such that $$p(x)=s_1(x)+(1-x^2)s_2(x)$$
- If $p(x)$ has degree $2d+1$, then $p(x)$ is nonnegative on $[-1,1]$ if and only if there exists sum-of-squares polynomials $s_1(x)$ of degree $2d$ and $s_2(x)$ of degree $2d$ such that $$p(x)=(1-x)s_1(x)+(1+x)s_2(x)$$
I have been unable to prove it despite my best efforts. I am aware of the result for a nonnegative polynomial on the whole real line, but I do not see a way to apply that to this case. My professor recommends induction but I have tried inducting on the degree of the polynomial to no success. Some help would be very much appreciated.
The hint suggests induction on the degree $n$ of $p(x)$. The following proves the induction step for the case $n=2d$ even, and the other case should work out similarly.
It is enough to consider the open interval $(-1,1)$, since both the premise and the conclusion extend to the closed interval $[-1,1]$ by continuity.
Any real root of $p(x)$ in $(-1,1)$ must have even multiplicity. If there are no real roots of odd multiplicity outside $(-1,1)$ then $p(x) \ge 0$ on $\mathbb R$, and is therefore a sum of squares. Otherwise, suppose there exists a real root $\alpha$ of odd multiplicity outside $(-1,1)$. It can be assumed WLOG that $\alpha \le -1$, otherwise the same argument would apply to $p(-x)$ with root $-\alpha$. For convenience of calculations below, let $\alpha = -1 - 2\beta$ with $\beta \ge 0$.
Then $x + 1 + 2 \beta\gt 0$ for $x \in (-1,1)$, so $q(x) = \dfrac{p(x)}{x + 1 + 2 \beta} \ge 0$ on $(-1,1)$, and by the induction hypothesis for $q(x)$ of odd degree $2d-1$ there exist sum-of-square polynomials $t_1,t_2$ such that: $$ \begin{align} q(x) &= (1-x)\,t_1(x) + (1+x)\,t_2(x) \\ \iff\;\; p(x) &= (x + 1 + 2\beta)(1-x)\,t_1(x) + (x + 1 + 2 \beta)(1+x)\,t_2(x) \\ &= \big(\underbrace{1 - x^2 + 2\beta\,(1-x)}_{= \beta(1-x)^2+(1-x^2)(1+\beta)}\big)\,t_1(x) + \big((1+x)^2 + \underbrace{2\beta\,(1+x)}_{= \beta\big((1+x)^2+(1-x^2)\big)}\big)\,t_2(x) \\ &= \underbrace{\beta(1-x)^2\,t_1(x) + (1+\beta)(1+x)^2\,t_2(x)}_{= s_1(x)} + (1-x^2)\underbrace{\big((1+\beta)t_1(x) + \beta\,t_2(x)\big)}_{=s_2(x)} \end{align} $$